Equilibria

In this lesson, we will start by watching a video to learn about two types of reactions: reversible (can go back and forth with products and reactants existing simultaneously) and irreversible (once products are formed, they cannot turn back into reactants). We will see examples of these reactions in everyday life and use simple graphs to understand dynamic equilibrium, where reactions reach a state of constant behaviour. Through experiments with anhydrous copper sulphate, we will explore how reactions work. We will also learn to write formulas, called equilibrium constant expressions, to characterise these reactions. Finally, we will explore how temperature and/or pressure changes affects equilibrium using Le Chatelier’s principle, helping chemists to find the best conditions for reactions.

KEY IDEAS

• Equilibrium constant (K): A value that expresses the ratio of the concentrations of products to reactants at equilibrium, specific to a particular reaction at a given temperature or pressure.

• Equilibrium position: The relative concentrations of reactants and products in a system at equilibrium, which can shift in response to changes in conditions.

• Forward reaction: The reaction that converts reactants into products.

• Heterogeneous equilibrium: An equilibrium in which reactants and products are in different phases.

• Homogeneous equilibrium: An equilibrium in which all reactants and products are in the same phase.

• Reverse reaction: The reaction that converts products back into reactants.

1. Reversible Reactions

Some reactions happen only once. The reactants (starting materials) combine to make a product, and when the reactants are all used up, the reaction stops. In most situations, once a reaction is complete, the new substances (the products) stay as they are. However, with some chemical reactions, the products can actually change back into the original starting materials (the reactants). This means the entire process is happening in two directions at the same time: one process works to form the products, and the other simultaneously works to remake the reactants. This is what we call a reversible reaction.

A reaction is called reversible when both forward and backward reactions can happen at the same time under certain conditions like temperature and pressure. We show this with two arrows (⇌) pointing in opposite directions between the reactants and products.

e.g. N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

2. Irreversible Reactions

Most chemical reactions only go in one direction and cannot go back to the materials. These are called irreversible reactions. For example, when carbon burns air, it forms carbon dioxide (CO₂) and cannot turn back into carbon and oxygen. We show irreversible reactions with a single arrow (→) pointing from the reactants to product, like this:

C(s) + O2(g) → CO2(g)

3. Dynamic Equilibrium

Some reactions do not completely stop but reach a balance called chemical equilibrium. In these reactions, both the concentrations of the reactants and products stay constant because they are continuously changing back and forth at the same rate. For each reactant that is changed into a product, a product is changed back into a reactant. Despite there being no change in the overall concentrations of reactants and products the reactions are both still occurring.

Example of Dynamic Equilibrium: When nitrogen dioxide (NO₂) gas reacts to form dinitrogen tetroxide (N₂O₄) in a container, NO₂changes colour from dark brown to colourless. Eventually, the stops changing, showing that the reaction has reached equilibrium where concentration of NO₂and N₂O₄stays constant.

The reaction could better be described with the following equation:

NO₂(g) + NO₂(g) ⇌ N₂O₄(g) Dynamic Equilibrium means the forward and backward reactions happen at the rate, and the amounts of reactants and products stay constant in a closed system.

Characteristics of Dynamic Equilibrium

1. Happens in a closed system (no reactants or products can be added or removed).

2. Both reactants and products are present.

3. Forward and reverse reactions happen at the same rate.

4. Equilibrium can be reached from either direction.

5. Concentrations of reactants and products do not change over time.

Applications of Dynamic Equilibrium

1. Chemical Systems: Rechargeable batteries, like lithium-ion batteries, through dynamic equilibrium, storing and releasing energy.

2. Biological Systems: Our bodies keep glucose levels steady with equilibrium, as the liver stores and releases glucose as needed.

3. Environmental Processes: Dynamic equilibrium helps in carbon cycling photosynthesis, respiration, and decomposition.

Activity 3.1 Reversible and Irreversible Reactions

Go online and watch a brief video that demonstrates both reversible and irreversible reactions. (https://www.youtube.com/ watch?v= ty9TczsW5ew here is a possible example)

1. Watch for differences between reactions that can “go back and forth” and those only “go one way.”

2. After the video, define reversible and irreversible reactions.

3. Give examples of reversible and irreversible reactions

4. Use simple equations to show these forward and reverse reactions.

5. Explain dynamic equilibrium as a balance point where the forward and reactions happen at the same speed.

6. Draw a simple graph showing how the amounts of reactants and products constant over time in dynamic equilibrium.

a. Label the graph with time on the x-axis and concentration on the y-axis.

b. Show how the lines for reactants and products start changing but eventually out to a constant value at equilibrium.

Activity 3.2 Reversible Reaction with Anhydrous Copper (II) Sulfate and Water Objective: To explore a reversible reaction and understand its relevance through observation, recording, and analysis.

Materials Needed: Anhydrous Copper (II) Sulphate (white powder form), distilled Water, heat source (optional, for drying the copper (II) sulphate back to anhydrous form), spoon or spatula, glass beakers or test tubes, protective gloves and goggles Steps Experiment – Conducting the Reaction

1. Observe and describe anhydrous copper (II) sulphate

2. Add a few drops of distilled water to the anhydrous copper (II) sulphate and observe.

3. Note the colour change, texture, and other observations about the substance water is added.

4. Gently heat the blue hydrated copper (II) sulphate. As it dries, it will turn white again, indicating the removal of water.

5. Observe and record any changes during heating. Discuss how the colour returns white as the substance reverts to its original anhydrous state.

6. Compare observations before and after each step.

7. Share what you learned about reversible reactions.

1. The Law of Mass Action

The Law of Mass Action explains how substances in a reaction affect each other.

It helps us understand how reactions work and what happens when they reac balanced state, called equilibrium.

The Law of Mass Action states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants, each raised t power equal to the coefficient of that reactant in the balanced chemical equation.

Imagine a reaction where: aA + bB ⇌ cC + dD Here:

A and B are reactants; C and D are products Forward and Reverse Reaction Rates The forward rate (reaction moving from reactants to products) is: r_(f)= k_(f)[A]ᵃ[B]ᵇ where k_(f)is a constant specific to the reaction being conducted.

The reverse rate (reaction going back from products to reactants) is:

r_(b)= k_(b)[C]ᶜ[D]ᵈwhere k_(b)is a constant specific to the reaction being conducted Equilibrium Constant (K) When the reaction reaches equilibrium, the forward and reverse rates are equal:

k_(f)[A]ᵃ[B]ᵇ= k_(b)[C]ᶜ[D]ᵈ Rearranging this gives us the equilibrium constant (K), which shows the between products and reactants:

K_(c) = [C]ᶜ[ D]ᵈ_ [A]ᵃ[B]ᵇThis K_(c)value helps us know the proportions of products and reactants at equilibrium.

2. Equilibrium Constants: K_(c)and Kₚ For a reversible reaction like:

aA + bB ⇌ cC + dD at equilibrium, the ratio of concentrations of products and reactants stays the at a certain temperature. This is called the equilibrium constant in terms concentration, written as Kc:

K_(c) = [C]ᶜ[ D]ᵈ_ [A]ᵃ[B]ᵇHere:

[A] and [B] are the concentrations of the reactants.

[C] and [D] are the concentrations of the products.

Kc represents the balance point of the reaction.

For gases, we can use their partial pressures instead of concentrations. This gives us a new equilibrium constant, called Kp:

K_(P) = P_(C) ᶜ× P_(D) ᵈ_ P_(A) ᵃ× P_(B) ᵇHere, P represents the partial pressures of each gas, and Kp tells us the balance of pressures at equilibrium for gases.

Example 1

To determine the equilibrium concentration of hydrogen gas (H₂) for the reaction:

N₂+ 3H₂⇌ 2NH₃ Given:

Equilibrium constant, K_(c)= 13.7 Equilibrium concentration of nitrogen, [N₂] = 1.88 mol/dm³ Equilibrium concentration of ammonia, [NH₃] = 6.62 mol/dm³

Solution

From the balance equilibrium reaction, K_(c) = [ NH₃]²_ [ N₂] [ H₂]³Substituting for the known equilibrium concentrations and the Kc, this becomes 13.7 = [6.62]²_ [1.88 ] [ H₂]³Solve for [ H₂]³ [ H₂]³= (6.62)²_ 1.88 × 13.7 [H₂] = ³√1.7015 = 1.19 mol /dm³Example 2 Consider the following equilibrium reaction at a certain temperature:

N₂+ 3H₂⇌ 2NH₃ At equilibrium, the partial pressures are measured as follows:

P_(N2) = 0.5 atm, P_(H2) = 1.5 atm, P_(NH3) = 2.0 atm Calculate the equilibrium constant Kₚfor this reaction.

Solution

K_(P) = P_(N)_(H)₃ ²_ P_(H)₂ ³× P_(N)₂ Substitute the given partial pressures into the expression:

K_(P) = ^((2.0)2)_ (1.5)³× 0.5 = 2.3 atm⁻²Notice the units for K_(P)here. These are determined by the powers in the numerator and denominator of the equilibrium constant equation and will be unique to each reaction. Some will have units of atm⁻², atm⁻¹, atm, atm²or even no units at all!

Atm raised to any power is theoretically possible here.

This principle also applies to the units for a K_(c)value. Where the units of K_(c)will be (moldm⁻³) raised to any theoretical power.

Significance of Equilibrium Constants: Kc and Kp

In chemical reactions, equilibrium constants help us understand how much of the reactants turn into products when the reaction reaches a balance, known as equilibrium.

a. If K_(c)or Kₚis greater than 1, The reaction favours products, meaning more products are formed than reactants left.

b. If K_(c)or Kₚis less than 1, The reaction favours reactants, meaning more reactants remain than formed.

c. If K_(c)or Kₚequals 1, The reaction is balanced, with similar concentrations of reactants and products.

Equilibrium Constants (Kc and Kp) in Real Life

Equilibrium constants, K_(c)and Kₚ, help scientists and engineers predict how chemical reactions behave. Here are some real-world examples:

a. Making Ammonia (Fertilizer Production) Role of Kc and Kp: These constants help determine the best temperature pressure to produce the most ammonia, which is essential for fertilizers.

b. Airbag Safety Role of Kc and Kp: Understanding these constants ensures the reaction fast enough to protect passengers.

c. Fuel production (petroleum refining):

Role of Kc and Kp: These constants guide how to adjust conditions to get the mix of fuel products.

By understanding K_(c)and Kₚ, we can control chemical reactions to make useful products safely and/or efficiently.

3. Relationship between K_(c)and Kₚusing ideal gas equation.

The relationship between K_(c)and Kₚfor a particular reaction follows the fact that for an ideal gas, PV = nRT, dividing both sides by volume, (V) PV = nRT Suppose the balanced equation for a hypothetical equilibrium reaction has the general form aA + bB ⇌ cC + dD Kₚ = P_(C) ᶜ. P_(D) ᵈ_ P_(A) ᵃ. P_(B) ᵇ……………………………………equation 1 As number of moles divided by volume can be thought of as concentration the partial pressures can be expressed as shown below P_(A) = C_(A) RT = [A]RT, P_(B) = C_(B) RT = [B]RT, P_(C) = C_(C) RT = [C]RT, P_(D) = C_(D) RT = [D]RT Substituting P= [ ]RT into equation 1 Kₚ = [C]ᶜ(RT)ᶜ[D]ᵈ(RT)ᵈ_______________ [A]ᵃ(RT)ᵃ[B]ᵇ(RT)ᵇ= [C]ᶜ[D]ᵈ(RT)⁽ᶜ⁺ᵈ⁾____________ [A]ᵃ[B]ᵇ(RT)⁽ᵃ⁺ᵇ⁾but K_(c)expression for the same equilibrium reaction,K_(c) = [C]ᶜ[D]ᵈ_______ [A]ᵃ[B]ᵇKₚ = K_(c) (RT)⁽ᶜ⁺ᵈ⁾−(a+b) ∆ n = (c + d) − (a + b) Kₚ = K_(c) (RT)∆n ∆ n = change in stoichiometric number of moles of products and reactants NB: a and b are the coefficients of reactants A and B and c and d are the coefficients of products C and D.

NB: The equilibrium constants K_(c)and Kₚhave units unless the units cancel, that is when the sum of the powers in the numerator and the denominator are the same.

Example 3

Write the relationship between K_(c)and Kₚfor each of the following equilibrium reactions:

a. A(g) ⇌ B(g) + C(g),

b. 2A (g) + B (g) ⇌ 4C (g)

c. 2 SO₂(g)+ O₂(g) ⇌ 2S O₃(g)

Solution

a. A(g) ⇌ B(g) + C(g), Kₚ = K_(c) (RT)∆n ∆ n = (1 + 1) − (1) = 1 Kₚ = K_(c) RT

b. A (g) + B (g) ⇌ 4C (g) Kₚ = K_(c) (RT)∆n ∆ n = 4 − (1 + 1) = 2 Kₚ = K_(c)RT²c. 2 SO₂(g)+ O₂(g) ⇌ 2S O₃(g) Kₚ = K_(c) (RT)∆n ∆ n = 2 − (2 + 1) = − 1 Kₚ = K_(c) (RT)⁻¹Kₚ = K_(c)_ RT

4. K_(c)and Kₚfor homogeneous and heterogeneous equilibrium reactions Homogeneous equilibrium reactions A homogeneous equilibrium is simply a reversible reaction where everything involved is in the same physical state. This means that all the initial materials (reactants) and all the final materials (products) are either all gases mixed together, or they're all dissolved in the same solution (liquid). They all share the same phase.

For example, 2 SO₂(g)+ O₂(g) ⇌ 2S O₃(g) K_(c)Uses concentrations (mol/dm³) of the reactants and products in the expression.

K_(c) = [S O₃]²_ [SO₂]²[ O₂] KₚUses the partial pressures of the reactants and products when all species are gases.

Kₚ = P_(S)_(O)₃ ²_ P_(S)_(O)₃ ²P_(O)₂ Heterogeneous equilibrium reactions A heterogeneous equilibrium is a reversible reaction where the materials are not all in the same state. This means the reactants and products are a mix of different phases, such as solids, liquids, and gases all reacting together. They don't share the same physical state.

Equilibrium Constants

In these reactions, pure solids and pure liquids are not included in the expression for K_(c)or Kₚbecause their concentrations remain constant.

Kc: Includes only the concentrations of gaseous or aqueous species.

Example: For the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g) K_(c) = [[CaO][ CO₂]_ [CaC O₃] K_(c) = [ CO₂] Kp: Includes only the partial pressures of gases.

Kₚ = P_(CO)₂

Example 4

Write the K_(c)and Kₚexpressions for the following processes:

a. C(s) + H₂O(g) ⇌ CO(g) + H₂(g)

b. NH₄Cl(s) ⇌ NH₃(g) + HCl(g)

Solution

a. Here, carbon (C) is a solid, so it is not included in the K_(c)expression. Only reactants and products are included:

K_(c) = [[CO][H₂]_ [H₂ O] Kₚ = P_(CO) × P_(H)₂_ P_(H)₂_(O)

b. Ammonium chloride (NH₄Cl) is a solid, so it is not included in the K_(c) expression. O the concentrations of the gaseous products, NH₃and HCl, are included:

K_(c) = [HCl][NH3] Kₚ = P_(HCl) × P_(NH)₃

5. Calculations involving Kₛₚ of sparingly soluble salts The solubility product constant (Kₛₚ) simply tells us the maximum amount of a substance that can dissolve in water before it starts forming a solid and dropping to the bottom. It measures how much that substance is willing to dissolve. It is used for salts that do not dissolve completely. A higher Kₛₚmeans the salt dissolves more. If more than 1 gram of a salt can dissolve in 100 cm³ of water, it is considered soluble.

When a salt dissolves in water, it breaks apart into ions and reaches a balance, or equilibrium, between the solid salt and the ions in the solution. For example, if a salt is written as AₓB_(y), it dissolves like this:

AₓB_(y)(s) ⇌ xAʸ⁺(aq) + yBˣ⁻(aq) The Kₛₚformula for this salt is:

Ksp = [ Aʸ+]ˣ[ Bˣ−]ʸTo find Kₛₚ, use the concentrations of the ions, raised to the numbers in the formula. The solid salt itself is not included in the Kₛₚcalculation because its concentration doesn’t change.

Example 5

The solubility of calcium fluoride (CaF₂) is 0.015 moldm⁻³. Calculate the Kₛₚof calcium fluoride.

Solution

The dissociation of calcium fluoride in water is represented by the equation:

CaF₂(s) ⇌ Ca²⁺(aq) + 2F−(aq) Let the solubility of CaF₂be S = 0.015 moldm⁻³.

This means:

[Ca²+] = S = 0.015 moldm⁻³ [F–] = 2S = 2 × 0.015 = 0.030 moldm⁻³ Now, the Kₛₚexpression is:

Kₛₚ= [Ca²⁺][F−]² Substitute the values:

Kₛₚ = (0.015)(0.030)²= 0.015 × 0.0009 = 1.35×10⁻⁵ Thus, the Kₛₚof CaF₂is 1.35 × 10⁻⁵mol³dm⁻⁹Example 6 The solubility of lead (II) iodide (PbI₂) is 1.1 × 10⁻⁴moldm⁻³. Calculate the Kₛₚof lead (II) iodide.

Solution

The dissociation of lead(II) iodide (PbI₂) is:

PbI₂(s) ⇌ Pb²⁺(aq) + 2I−(aq) Let the solubility of PbI₂be S = 1.1 × 10⁻⁴ This means:

[Pb²+] = S= 1.1 × 10⁻⁴moldm⁻³ [I–] = 2S = 2 × 1.1×10⁻⁴moldm⁻³ Now, the Kₛₚexpression is:

Kₛₚ= [Pb²⁺][I−]² Substitute the values:

Kₛₚ= (1.1 × 10⁻⁴)(2.2 × 10⁻⁴)²= (1.1 × 10⁻⁴)(4.84 × 10⁻⁸) = 5.32 × 10⁻¹²mol³dm⁻⁹

Example 7

Given that the Ksp of PbCl2 is 1.6×10⁻⁵, calculate the solubility of PbCl₂in moldm⁻³.

Solution

The dissociation of PbCl₂in water is given by:

PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl−(aq) Let the solubility of PbCl₂be S moldm⁻³. At equilibrium:

The concentration of Pb²+ will be S.

The concentration of Cl– will be 2S (because each PbCl₂produces 2 Cl– ions).

The expression for Kₛₚis:

Kₛₚ = [Pb²⁺][Cl−]² Substitute the concentrations in terms of S:

Kₛₚ = (S)(2S)²= 4S³ Now, substitute the given value of Kₛₚ 1.6 × 10⁻⁵= 4S³ Solve for S:

S³= 1.6 × 10⁻⁵________ 4 = 4.0 × 10⁻⁶S = ³√4.0 × 10⁻⁶= 1.58 × 10⁻²mol dm⁻³So, the solubility of PbCl₂is 1.58 × 10⁻²mol dm⁻³Example 8 The Ksp of BF3 is 3.9×10⁻³. Calculate the solubility of BF₃in moldm⁻³.

Solution

The dissociation of BF₃in water is:

BF₃(s) ⇌ B³⁺(aq) + 3F−(aq) Let the solubility of BF₃be S moldm⁻³. At equilibrium:

The concentration of B3+ will be S.

The concentration of F- will be 3S (since three moles of F– are produced for each mole of BF₃).

The expression for Ksp is:

Kₛₚ= [B³⁺][F−]³ Substitute the concentrations in terms of S:

Kₛₚ= (S)(3S)³= 27S⁴ Now, substitute the given value of Kₛₚ:

3.9 × 10⁻¹⁰= 27S⁴ Solve for S:

S⁴= 3.9 × 10⁻¹⁰_________ 27 = 1.44 × 10⁻¹¹S = ⁴√__________ 1.44 × 10⁻¹¹= 1.94 × 10⁻³mol dm⁻³Activity 3.3 Deriving the Expression for the Equilibrium Constant (K_(c)or Kₚ) Steps

1. Using the reversible reactions:

a. aA + bB ⇌ cC + dD ,

b. 2N H₃ ⇌ N₂ + 3 H₂

c. C O₂ + H₂ O ⇌ H₂ C O₃ deduce the expression for the equilibrium constant (K_(c)or Kₚ) in mixed ability groups.

2. Present your findings and discuss any differences or misunderstandings.

Activity 3.4 Discussing the meaning of the equilibrium constant Steps

1. Discuss what the value of the equilibrium constant K_(c)or Kₚtells them about a reaction.

2. What does a large value of K mean about the products vs. the reactants?

3. What about a small value of K?

4. How does the value of K influence how we predict the outcome of a reaction?

Activity 3.5 Relationship Between Kₚand K_(c)using the Ideal Gas Equation Steps

1. Using the ideal gas equation: PV = nRT.

Discuss how PV = nRT relates to Kₚby introducing the formula:

Kₚ= K_(c)(RT)^(Δn)where Δn is the change in the number of moles of gas between products and reactants

2. Write the Kₚand K_(c)expressions for the reaction N₂(g) + 2 O₂(g)⇄ 2N O₂(g)

3. Calculate the Kp using the given value of Kc= 5.21 × 10 −5 moldm⁻³, temperature (T) = 500 °C, gas constant ® = R = 8.31 kPa dm³mol⁻¹K⁻¹Activity 3.6 Sparingly soluble salts and Calculations Involving Kₛₚbased on solubility data Steps

1. Using PbS O₄(s) as an example of sparingly soluble salts:

PbS O₄(s) ⇄ Pb²⁺(aq) + S O₄ ²⁻(aq) Given the equilibrium concentrations: [S O₄ ²⁻] = 1.48 × 10⁻⁴moldm⁻³, [Pb²⁺] = 5.7 × 10⁻⁵moldm⁻³2. Calculate the Kₛₚof PbS O₄.

3. Compare your results with others.

Le Chatelier’s Principle

Imagine a see-saw or a balance scale. When it is perfectly balanced, we say it is in “equilibrium.” But if someone pushes down on one side, the see-saw tilts. To balance it again, you need to make changes to bring it back to even.

In a chemical reaction, “equilibrium” means the forward and backward reactions are happening at the same rate, so everything stays balanced. If something changes—like adding more weight (or a reactant/product) or changing how hard someone pushes (like temperature or pressure)—the reaction shifts to balance things out again. This is what Le Chatelier’s Principle is all about!

Factors That Affect Equilibrium

a. Concentration (Amount of Stuff) If you add more of one side (reactants or products), the reaction balances it by making more of the opposite side.

Example: Imagine a teeter-totter where you add more toys on one side. To balance it, you move more toys to the other side.

If a reaction is: A + B ⇌ C + D, and you add more A, the reaction will shift to the right (using up A and making more C and D). If you take away some C, the reaction will also shift to the right to make up for the lost C.

b. Pressure and Volume Imagine blowing air into a balloon. If you make the balloon smaller by squeezing (reducing the volume), the air inside feels squished, and the pressure goes up. If let the balloon expand (increasing the volume), the air spreads out, and the goes down.

In chemical reactions with gases, the same idea works! How much space the have (volume) and how squished they are (pressure) can change the balance of reaction.

What Happens When Pressure or Volume Changes?

More Pressure (Smaller Volume)

If you squeeze the gases into a smaller space (increase pressure), the reaction will reduce the pressure.

Example: In a reaction like: xA ⇌ yB If one side of the reaction has fewer gas molecules, the reaction will shift towards side to lower the pressure.

Less Pressure (Bigger Volume)

If you give the gases more room (increase volume), the reaction increases the.

pressure.

Example: The reaction will shift to the side with more gas molecules to balance out.

No Effect if the Gases are Equal If both sides of the reaction have the same number of gas molecules, changing the pressure or volume will not change the balance.

c. Temperature and Equilibrium Two Types of Reactions

i. Endothermic Reactions (Heat is Absorbed):

Example: A + B + Heat ⇌ C If you turn up the temperature (add heat), the reaction shifts make more C (products) thereby lowering the temperature.

If you cool it down (remove heat), the reaction shifts backward to make more A and B (reactants) to produce more heat.

ii. Exothermic Reactions (Heat is Released)

Example: D + E ⇌ F + Heat If you turn up the temperature (add heat), the reaction gets rid of the extra by shifting backward to make more D and E (reactants).

If you cool it down (remove heat), the reaction shifts forward to make more F (products) because that releases heat.

d. Catalyst Simplified A catalyst speeds up a chemical reaction without getting used up. It makes the reaction happen faster by creating a lower energy pathway for the molecules to react.

Even though the catalyst helps the reaction go faster, it doesn’t change the final balance of the reaction (the equilibrium). It doesn’t affect the concentrations of reactants and products at the end; it just helps the reaction reach that point more quickly.

Activity 3.7 Exploring factors that affect Chemical Equilibrium Materials Needed: Flashcards with equilibrium scenarios (e.g., adding reactants, changing pressure, etc.), a simple balance scale or see-saw diagram to represent equilibrium Step

1. State the factors that can disturb equilibrium.

2. From the given flashcards with specific scenarios, predicting changes when:

a. Add more reactant.

b. Increase temperature in an exothermic reaction.

c. Decrease volume of a gas mixture.”

3. For each scenario predict which side reaction will shift to.

Activity 3.8 Applying Le Chatelier’s Principle to Optimise Reaction Conditions Steps

1. Use the simple reaction: A(g) + B(g) ⇌ C(g) + 2D(g) Discuss the following conditions:

a. What happens if more of A is added to the reaction?

b. What happens if C is removed from the reaction?

c. If pressure is increased, what will happen to the reaction?

2. Use the simple reaction: N₂(g)+ 3 H₂(g) ⇌ 2N H₃(g) ∆ H = − x kJmol − 1 Discuss the following conditions:

a. What happens when temperature increased?

b. What happens to equilibrium position?

c. What effect will increase in temperature have on Kₚ?

The principle of chemical equilibrium and rates of reaction are used by chemists to help make industrial processes more efficient. By understanding how reactions behave and how to control certain factors, industries can increase the volume of desired products. Le Chatelier’s Principle helps predict how changes in temperature, pressure, or concentration affect a reaction at equilibrium, allowing chemists to make adjustments that increase production.

1. The Haber Process (Making Ammonia)

The Haber process is used to make ammonia (NH₃), which is important for fertilizers. The reaction is:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJmol⁻¹This reaction is exothermic, meaning it releases heat. To make more ammonia, chemists:

Lower the temperature (even though it slows the reaction, it favours the production ammonia).

Increase the pressure (this helps shift the reaction toward ammonia, since there fewer molecules on the product side).

Use an iron catalyst to speed up the reaction.

2. The Contact Process (Making Sulfuric Acid)

In the contact process, sulphur is used to make sulfuric acid, an important chemical industries. The key reaction in this process is:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −196 kJmol⁻¹To maximize the production of SO₃(which is used to make sulfuric acid), chemists:

Increase the pressure (since there are fewer gas molecules on the product side).

Lower the temperature (this favours the production of SO₃).

Use a vanadium(V) oxide catalyst to speed up the reaction.

3. Catalytic Converters in Cars

In cars, harmful gases like carbon monoxide (CO), hydrocarbons (CₓH_(y)), and oxides (NO and NO₂) are produced when fuel is burned. Catalytic converters help these harmful gases into less harmful ones like carbon dioxide (CO₂), nitrogen (N₂), and water (H₂O).

The reduction process converts NO to N₂and O₂:

2NO(g) ⇌ N₂(g) + O₂(g) The oxidation process converts CO to CO₂:

2CO(g) + O₂(g) ⇌ 2CO₂(g) To make these reactions work better, high temperatures are used (because reactions are endothermic, meaning they need heat). Catalysts like platinum (Pt) speed up these reactions. Le Chatelier’s principle explains that increasing concentration of harmful gases makes the reaction move forward to create harmful gases.

4. Petroleum Refining

In petroleum refining, catalytic cracking and reforming are used to turn long- hydrocarbons into useful fuels like gasoline and diesel.

Catalytic cracking breaks down large molecules into smaller ones (this process heat).

C₁₂H₂₆(g) ⇌ C₆H₁₄(g) + C₇H₁₆(g) Catalytic reforming rearranges molecules to make high-octane gasoline.

This also needs heat.

Both of these reactions are controlled by Le Chatelier’s principle, and conditions temperature are adjusted to get the right products.

5. Water Treatment (Ion Exchange)

In water treatment, hard water (which contains calcium and magnesium ions) can made soft by removing these ions. One way to do this is with ion exchange, calcium ions react with chloride ions:

Ca²⁺(aq) + 2Cl−(aq) ⇌ CaCl₂(s) By adding brine (NaCl), which increases the chloride ions, the reaction shifts produce more calcium chloride (CaCl₂). This helps remove calcium and ions from the water, making it softer.

Activity 3.9 Mini Research Work

Go online and find other sources about how Le Chatelier’s principle is applied in industrial processes.

Activity 3.10 Group Presentations on Economic Reaction Optimization

Objective: In small groups, students work together to learn how chemists use special processes to produce important chemicals like ammonia (Haber Process) and sulphuric acid (Contact Process) efficiently.

Steps

1. a. Research the Haber Process for ammonia production using books, handouts, online resources to gather information.

b. Research the Contact Process for sulphuric acid production using books, handouts, or online resources to gather information.

Research tasks

i. Explain the main reaction in each process.

ii. Describe how Le Chatelier’s Principle helps maximise the yield.

iii. Identify the factors that are controlled in each process (temperature, pressure, catalyst).

iv. Create a simple diagram or flowchart to show the steps in the process.

2. Create a short presentation for the class.

The presentation should include:

a. What the process is used for.

b. How Le Chatelier’s Principle is applied to optimise the reaction.

c. What conditions (temperature, pressure, catalysts) are used to get the highest yield.

d. Use charts, drawings, or a simple poster or a PowerPoint slide to help explain.

3. Class reflections:

a. Why do you think chemists want to control the temperature and pressure in these reactions?

b. How does Le Chatelier’s Principle help in getting the best results in these processes?

c. How does understanding these processes help us in real life?

Review Questions 3.1

1. What is a reversible reaction?

2. Give an example of an irreversible reaction.

3. What is meant by “dynamic equilibrium” in a reversible reaction?

4. Describe the visual change observed when water is added to anhydrous copper (II) sulphate.

5. Why doesn’t dynamic equilibrium occur in an open system?

6. How is a reversible reaction represented in a chemical equation?

7. Compare reversible and irreversible reactions in terms of product formation energy change.

8. Explain how dynamic equilibrium is crucial in biological systems, providing example.

Review Questions 3.2

1. Given the reaction: 2A(g) + B(g) ⇌ 3C(g) If the equilibrium concentrations of the species are: [A] = 0.2 moldm⁻³, [B] = 0.5 moldm⁻³, and [C] = 1.0 moldm⁻³, calculate the constant K_(c).

2. What does a large value of the equilibrium constant (K_(c)) indicate about the position equilibrium in a chemical reaction?

3. For the following reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) At equilibrium, the concentrations of the species are: [N₂] = 0.1 moldm⁻³, [H₂] = 0.3 moldm⁻³, [NH₃] = 0.2 moldm⁻³Calculate the equilibrium constant K_(c).

4. Given the equilibrium reaction:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) If the equilibrium constant K_(c)is very small, what can you conclude about concentrations of reactants and products at equilibrium?

5. The following reaction is at equilibrium: 2NO₂(g) ⇌ N₂O₄(g) At equilibrium, the concentration of NO₂is 0.3 moldm⁻³and N₂O₄is 0.1 moldm⁻³. If initial concentration of NO₂was 0.5 moldm⁻³, calculate the equilibrium constant K_(c). Assume the reaction reached equilibrium and no other substances are involved.

6. For the reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) If the equilibrium constant K_(c)is much greater than 1, what would happen to system if the volume of the container is decreased? Justify your answer using Le Chatelier’s principle.

7. For the reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) At a certain temperature, Kₚis found to be 2.0×10³If the change in the number moles of gas, Δn is 0, what is the relationship between Kₚ?

8. The solubility of lead(II) iodide (PbI₂) in water is 1.2 × 10⁻³mol/dm-

3. Calculate the KₛₚPbI₂. Then, if the solubility is increased to 2.4 × 10⁻³moldm⁻³, calculate the new Kₛₚexplain why Kₛₚdoes not change with a change in solubility.

Review Questions 3.3

1. State Le Chatelier’s Principle.

2. What is the main product of the Haber Process and what are its uses?

3. If the concentration of a reactant in a system at equilibrium is increased, which direction will the equilibrium shift according to Le Chatelier’s Principle?

4. In the Haber Process, what are the key factors that affect the production of ammonia, and how do they influence the reaction?

5. Consider the following equilibrium reaction:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) What will happen to the position of equilibrium if the temperature is increased? Assume the reaction is exothermic.

6. In the Contact Process for making sulphuric acid, what are the main (temperature, pressure, and catalyst) that affect the reaction, and how are optimised for maximum yield?

7. In the Contact Process, how does the presence of a catalyst influence the rate reaction, and how does it help optimise the production of sulfuric acid?

8. The following reaction is at equilibrium:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) If the pressure is increased by reducing the volume of the container, which direction ill the equilibrium shift? Explain your reasoning based on Le Chatelier’s Principle.

9. Explain how Le Chatelier’s Principle applies to the change in the value of equilibrium constant (K) when the temperature is changed.

10. Explain how Le Chatelier’s Principle is applied in the Haber Process to ammonia production. What changes in temperature and pressure are made, and why?

In section, we will explore the basics of acids and bases in chemistry. You will learn about different acid-base theories like Arrhenius, Brønsted -Lowry, and Lewis. We will understand the physical and chemical properties of acids and bases, including their everyday uses. This section also covers salts—what they are, how they’re made, and their practical uses. Students will practice acid-base titration methods, like simple titration and back titration.

KEY IDEAS

• Analyte is a solution of unknown concentration (in the flask).

• Arrhenius acids are substances that produce hydrogen ions (H+) when dissolved in water.

• Arrhenius bases are substances that produce hydroxide ions (OH–) when dissolved in water.

• Brønsted -Lowry base a substance that accepts a proton (H+) from another substance.

• Brønsted-Lowry acid a substance that donates a proton (H+) to another substance.

• Conjugate acid-base pair is two species that differ by a single proton (H+).

• Endpoint is a point where the indicator changes colour, showing neutralization.

• Indicator is a substance that changes colour to show when a reaction is complete.

• Lewis acid a substance that accepts a pair of electrons to form a covalent bond.

• Lewis base a substance that donates a pair of electrons to form a covalent bond.

• Titrant is a solution of known concentration (in the burette).

• Titre value is the volume of titrant used to reach the endpoint.

Acids and bases are two important types of chemicals used in everyday items like cleaning products, medicines, and food. Long ago, people could tell them apart by taste and feel: acids have a sour taste, like in vinegar or lemon juice, and bases feel slippery or soapy, like some cleaners.

1. Arrhenius concept of acids and bases The first scientist to provide a more thorough chemical explanation (as opposed to taste/feel) of the characteristics of bases and acids was Arrhenius. He defines an acid as a substance that produces hydrogen ions or hydronium ions when dissolved in water (aqueous solution). Using H₂SO₄as an example, in water, the molecules of H₂SO₄break up (dissociate) to give its ions.

H₂SO₄(aq) → 2H + + SO₄ ²⁻ H₂SO₄+ H₂O→ H₃O + + SO₄ ²⁻ Therefore, sulphuric acid contains hydrogen ions or hydronium ions. All other aqueous solutions of acids also contain hydrogen ions and this is what gives them their acidity.

A base, according to Arrhenius, is a compound or substance that produces hydroxide ions when dissolved in water (aqueous) solution.

Using NaOH as an example, in aqueous solution NaOH molecules dissociate to give Na+ and OH− ions.

NaOH (aq) → Na+ + OH−

Activity 4.1 Exploring Arrhenius Theory of Acids and Bases Material needed: Computer, and videos to guide their research.

Steps

1. Introduce Arrhenius Theory

a. Give a simple explanation of Arrhenius acids.

b. Give 2 examples with diagrams.

2. In small groups, work together to explore and discuss Arrhenius acids and bases.

3. Discuss and share ideas on what makes Arrhenius’ theory useful. How does it help explain basic reactions?

4. In your groups, research the strengths (e.g., clear definitions for acids and bases in water) and limitations (e.g., theory only applies to water-based solutions, does not explain all acid-base reactions).

Each group should prepare a short presentation with their findings, using visual aids like charts, chemical reactions, or videos to make it engaging and easy to understand.

Limitation of Arrhenius’ concepts Arrhenius definition of acids and bases are limited because they apply only to aqueous solutions.

2. Brønsted -Lowry concept of acids and bases Two scientists, Brønsted and Lowry, gave a broader definition of acids and bases.

They said an acid-base reaction happens when a hydrogen ion is transferred from one compound to another. This can happen even when the reaction is not in water.

A hydrogen ion (H+) is formed when a hydrogen atom loses its electron, so H+ is like a proton. Brønsted and Lowry defined an acid as something that gives away a proton (H+) and a base as something that accepts a proton. We call these Brønsted -Lowry acids and Brønsted -Lowry bases.

HN O₃ +H₂ O → H₃ O++ N O₃ − Here HNO₃is acting as a Brønsted -Lowry acid, and H₂O is acting as a Brønsted

-Lowry base.

H₂ O + N H₃ → N H₄ + + OH− Here NH₃is acting as a Brønsted -Lowry base, and H₂O is acting as a Brønsted

-Lowry acid.

This means that in a Brønsted – Lowry acid – base reaction, the acid donates protons to the base.

Not all hydrogen atoms in a compound can be given away as ions, so not every compound with hydrogen is a Brønsted -Lowry acid. For example, methane (CH₄) is not acidic. Whether hydrogen can act as an acid depends on certain properties like polarity (how the atoms are attracted to each other) and the number of lone electron pairs on the atom it’s bonded to. Elements from groups 16 and 17, like oxygen and chlorine, when bonded with hydrogen, can form Brønsted -Lowry acids. Common examples of these acids in chemistry labs are HCl, HNO₃, H₂SO₄, CH₃COOH, H₃PO₄, and H₂CO₃.

The basicity of a compound depends on having lone pairs of electrons to accept protons. Elements from group 15 and group 16 often form bases. Common Brønsted -Lowry bases are NaOH, KOH, NH₃, C₅H₅N, and Na₂CO₃.

Brønsted -Lowry acids are grouped by how many protons they can donate.

Monoprotic acids donate one proton (like HCl and HNO₃).

Polyprotic acids donate more than one proton.

Diprotic acids (like H₂SO₄and H₂CO₃) donate two protons, while Triprotic acids (like H₃PO₄) donate three.

Brønsted -Lowry bases can also be monoprotic (accept one proton, like NaOH and NH₃) or diprotic (accept two protons, like CO₃²–). Some compounds can act differently in water, changing based on their surroundings.

When hydrogen chloride gas (HCl) dissolves in water, it gives a hydrogen ion to the water:

HCl(g) + H₂O(l) → H₃O+(aq) + Cl–(aq) In this reaction, HCl is a Brønsted -Lowry acid because it donates a hydrogen ion to water, while water acts as a Brønsted -Lowry base by accepting the hydrogen ion to form H₃O+ (hydronium ion).

On the other hand, when ammonia (NH₃) dissolves in water, it accepts hydrogen from water:

NH₃(aq) + H₂O(l) → NH₄ +(aq) + OH–(aq) Here, water donates a hydrogen ion to ammonia, making water a Brønsted

-Lowry acid and ammonia a Brønsted -Lowry base.

Since water can act as an acid with NH₃and as a base with HCl, it is called an amphiprotic compound (meaning it can act as both an acid and a base).

Conjugate Acid-Base Pairs

In the Brønsted -Lowry theory, acids and bases are connected through “conjugate pairs.” When an acid loses a hydrogen ion (proton), it forms its conjugate base.

The acid and its conjugate base differ by one proton. Similarly, when a base gains a proton, it forms its conjugate acid. Each Brønsted -Lowry acid has a conjugate base, and each base has a conjugate acid.

For example

1. H₂O + HCl → H₃O+ + Cl– Here, H₂O (base) and H₃O+ form one conjugate pair, while HCl and Cl– form another.

2. NH₃+ H₂O → NH₄ + + OH– NH₃and NH₄ + are one pair, and H₂O and OH– are another.

Activity 4.2 Understanding Brønsted-Lowry Theory of Acids and Bases

Activity Steps

1. Give a simple explanation of Brønsted -Lowry concept of acids and bases.

a. Use visuals like diagrams, charts, illustrate this idea, showing how protons from acids to bases.

2. In small groups,identify compounds that have hydrogen atoms.

a. Classify the compounds into simple acids (e.g., HCl), oxoacids (e.g., HNO₃), hydrated cations (e.g., Al(H₂O)₆³+).

b. Discuss which compounds could act as Brønsted -Lowry acids and bases.

3. Provide examples of acid-base reactions (e.g., HCl + NH₃→ NH₄ + + Cl–).

4. Demonstrate and discuss with the class how proton transfer occurs, which substance is the acid and which is the base in each example.

5. Discuss the strengths of Brønsted -Lowry Theory:

a. Discuss why this theory is helpful: for example, it explains more types of acid-base reactions than Arrhenius’ theory and works in non-aqueous (non-water) environments.

6. Discuss situations where the theory doesn’t fully explain certain reactions (e.g., reactions that don’t involve proton transfer).

a. Identify examples where the theory might not be complete.

7. Present your findings to the class, including examples of Brønsted -Lowry acids and bases, reactions with proton transfer, and key strengths and limitations.

3. Lewis Acids and Bases

The Lewis theory expands the idea by focusing on electron pairs rather than protons. According to Lewis, an acid is a substance that accepts an electron pair, while a base donates an electron pair to form a bond. For example:

H+ + NH3 → NH4+ Here, H+ (Lewis acid) accepts electrons from NH₃(Lewis base).

Another example is NH3 + BF3 → F3B–NH3, where NH₃donates an electron pair to BF₃, creating a bond.

A Lewis acid does not necessarily need a hydrogen, and this is what separates it from a Brønsted-Lowry acid.

Activity 4.3 Collaboratively Exploring Lewis Theory of Acids and Bases Materials needed: Chart papers or whiteboard, markers, handouts with examples Lewis Acids and Bases

Activity Outline

1. In small groups,brainstorm different examples of Lewis acids.

Starting points, such as:

a. Metal cations (e.g., Al³⁺)

b. Electron-deficient molecules like BF₃ Discuss why each example can act as a Lewis acid, focusing on its ability to accept electron pairs.

2. In small groups,brainstorm different examples of Lewis acids and L bases.

Starting points, such as:

a. Anions e.g.O²⁻

b. Electron-rich molecules like H₂O Discuss why each example can act as a Lewis bases focusing on its ability to donate electron pairs.

3. Using the chemical reaction provided identify Lewis acids and bases in each reaction:

a. Cu²⁺(aq) + 4NH₃(aq) → [Cu(NH₃)₄] 2+ (aq)

b. N H₃ + H+ → N H₄ +

c. B F₃ + N H₃ → B F₃ N H₃

4. Draw a diagram to show how ammonia donates an electron pair to the proton.

5. Explain that this is a Lewis acid-base reaction involving a coordinate covalent bond.

6. Brainstorm and discuss the strengths and limitations of the Lewis theory.

Activity 4.5 Identifying Conjugate Acid-Base Pairs

Materials Needed: Acid-base reaction cards (each showing a different acid- base reaction), coloured markers, and a chart with spaces for acids, bases, conjugate acids, and conjugate bases.

1. Organise yourselves into groups of no more than five. Each group will receive a reaction card.

2. Analyse the reaction card, identify the acid and base.

3. Determine the conjugate acid-base pairs.

4. Show the species donating a proton and which is accepting it.

1. Physical properties of acids and bases

a. Physical properties of acids

i. Acids have a sour, sharp taste.

ii. They change blue colour litmus paper to red.

iii. They change the colour of Methyl Orange/Yellow to Pink.

iv. Acidic substances change the colour of Phenolphthalein from deep pink colourless.

v. Have a wet feel.

vi. They are good conductors of electricity.

vii. Acids have a pH value of less than seven.

b. Physical Properties of bases

i. Bases have a bitter taste.

ii. They change red litmus paper to blue.

iii. Basic substances change phenolphthalein from colourless to pink.

iv. Feel slippery between the fingers (they have a soapy feel).

v. Bases conduct electricity in aqueous solution.

vi. Bases have a pH value greater than seven

2. Chemical properties of acids and bases

a. Chemical properties of acids

i. Acids react with some metals, (that are above hydrogen in the series, common examples are shown below) to produce hydrogen gas.

A typical reaction is that between hydrochloric acid and calcium:

H₂SO₄(aq) + Ca (s) → CaSO₄(aq) + H₂(g)

ii. Acids react with carbonates and bicarbonates, such as Na₂C O₃, C aC O₃, NaHC O₃, to produce carbon dioxide gas.

CaC O₃(s) + H₂SO₄(aq) → CaSO₄(aq) + H₂O (l) + C O₂(g)

iii. Acids react with bases to produce salt and water. This reaction is neutralisation.

Acid + base → salt + water HC l (aq) + NaOH (aq) → NaC l (aq) + H₂O (l)

b. Chemical properties of bases

i. The aqueous salt solution of heavy metals reacts with soluble bases (alkalis) form a precipitate (ppt) of the metallic hydroxide.

3NaOH (aq) + AlC l₃(aq) → A(OH)₃(s ) + 3NaC l (aq) White ppt

ii. Bases reacts with warmed ammonium salt to give off ammonia gas NaOH (aq) + NH₄C l (s ) → NaC l (aq) + H₂O (l) + NH₃(g)

iii. Bases reacts with acids to form salt and water KOH (aq) + HC l (aq) → KC l (s ) + H₂O (l)

3. Classification of household substances as acids and bases Household products are categorized according to their chemical and physical characteristics. The majority of these tests are qualitative and may be conducted outside of a laboratory. The common testing consists of

a. Taste (Do not taste any chemical in the laboratory) Acids are sour-tasting chemicals. Because they contain more acid than oranges, unripe oranges have a highly sour taste. Conversely, bases have a taste.

b. Litmus paper test Acidic substances or their solutions turn blue litmus paper red. Bases or their solutions turns red litmus paper blue.

c. The pH scale Aqueous solutions of acids have pH values less than 7. Whiles aqueous of bases have pH values greater than 7.

d. Universal indicator test A combination of many indicators that change colour in response to a substance’s pH level is known as a universal indicator. By comparing the colour change wit colour code that matches the pH value, one may determine if a chemical is basic, neutral, or acidic.

An acidic solution can become yellow, beige, orange, pink, or red when with the universal indicator. The resultant solution can be dark green, turquoise, pale blue, blue, dark blue, violet, or purple when combined with a basic

solution. Universal indicator turns green when using neutral solutions.

4. Application of neutralisation reactions in real life

a. Application of Neutralisation Reactions: Antacids When we have too much stomach acid, it can cause discomfort, like and acid reflux. Antacids are medicines that help by neutralising the stomach acid.

Our stomach naturally has hydrochloric acid (HCl) to help digest food, but much of it leads to problems. Antacids contain basic compounds magnesium hydroxide (Mg(OH)₂), aluminium hydroxide (Al(OH)₃), carbonate (CaCO₃), or sodium bicarbonate (NaHCO₃).

These compounds react with the excess hydrochloric acid in a reaction.

For example: Mg(OH)₂+ 2HCl → MgCl₂+ 2H₂O Here, magnesium hydroxide (from the antacid) reacts with hydrochloric acid form a salt (MgCl₂) and water (H₂O). The water reduces the acidity in stomach, providing relief from symptoms like heartburn.

People take antacids as needed, but it’s important to follow the correct to avoid side effects. Neutralisation reactions in antacids help to make stomach environment less acidic, bringing temporary relief from acidity.

b. Neutralisation in Wastewater Treatment Wastewater treatment uses neutralisation reactions to adjust the pH of water a neutral level before it’s released into the environment. Industrial, municipal, and agricultural activities can create wastewater that’s too acidic or basic, which can harm the environment if left untreated.

In wastewater treatment plants, acids or bases are added to neutralise water:

i. For acidic wastewater (low pH), a base like calcium hydroxide (Ca(OH)₂) added.

ii. For basic wastewater (high pH), an acid like sulphuric acid (H2SO4) hydrochloric acid (HCl) is used.

For example: 2HCl + Ca(OH)₂→ CaCl₂+ 2H₂O Here, hydrochloric acid (HCl) reacts with calcium hydroxide to form calcium chloride (salt) and water. This raises the pH of acidic water to make it neutral.

Neutralising wastewater protects aquatic life and keeps water ecosystems healthy by preventing harmful effects from untreated acidic or basic water.

c. Neutralisation in Agriculture In agriculture, neutralisation helps improve soil acidity for better plant growth. Acidic soil can limit plant growth and nutrient availability, so farmers agricultural lime (calcium carbonate, CaCO₃) to neutralise the acid in the soil.

The reaction is:

CaCO₃+ H₂SO₄→ CaSO₄+ H₂O + CO₂ Here, calcium carbonate reacts with sulphuric acid in the soil to form calcium sulphate, water, and carbon dioxide. This reaction reduces soil acidity and increases the pH, helping plants grow better and boosting crop yield.

d. Neutralisation in Cleaning In cleaning, neutralisation reactions help remove acidic or basic residues from surfaces, making cleaning more effective.

i. When an acidic cleaner (like citric acid) meets alkaline residues ( sodium hydroxide), they neutralise each other:

C₆H₈O₇(aq) + 3NaOH(aq) ⟶3 H₂O(l) + Na₃ C₆H₅O₇(aq) Citric acid sodium hydroxide water Sodium citrate

ii. When an alkaline cleaner (like sodium hydroxide) meets acidic (like hydrochloric acid), they also neutralise each other:

NaOH(aq) + HCl(aq) ⟶ H₂O(l) + ⟶NaCl(aq) sodium hydroxide hydrochloric acid water sodium chloride These reactions form water and harmless salts like sodium citrate or sodium chloride, which helps remove unwanted residues and ensures surfaces are clean.

e. Neutralisation in Food and Beverages In the food industry, neutralisation reactions help control pH levels to food safety and quality, especially in fermented foods like yogurt.

During yogurt fermentation, bacteria turn lactose (milk sugar) into lactic acid, making the mixture more acidic. To stabilise the pH and prevent too acidity, a base like calcium hydroxide (lime water) is used to neutralise lactic acid.

The reaction is:

2C₃H₆O₃(aq) + Ca (OH)₂(aq) ⟶ 2 H₂ O(l) + Ca₃C₃H₆O₃(aq) lactic acid calcium hydroxide water calcium lactate Here, lactic acid reacts with calcium hydroxide to form calcium lactate (a neutral salt) and water. This helps control the acidity in yogurt, ensuring it has the right taste and quality.

Activity 4.6 Investigating physical properties of acids and bases Conduct the following activity in small groups. Discuss your findings and create a poster or a presentation to share your findings with the entire class.

In your groups, investigate the physical properties of:

a. Sour taste of acids and slippery feel of bases

b. pH Testing

c. Feel Behaviour of Acids and Bases Materials needed: vinegar (acid) and baking soda, computer, sodium chloride

solution, lemon juice, test tube or beaker, test tube or beaker, virtual lab online simulation sites Step for a:Sour taste of acids and bases

1. Use a virtual lab or online simulation to explore taste of acids and bases.

Steps for b: pH Testing

1. Take about 5 cm³the test solution into a test tube or beaker.

2. Using the litmus paper test each substance and record the colour change.

3. Identify which substances are acidic, basic, or neutral.

Steps for c: Exploring the Feel Behaviour of Acids and Base

1. Pour dilute solutions of vinegar (acid) and baking soda (base) into separate beakers.

2. Dip your fingers into each solution rub the fingers and feel the texture of the solution.

3. Based on the feel effect state if the solution is acid or base.

Activity 4.7 Exploring chemical reactions of Acids and Bases

1. Reaction of Acids

a. Reaction of Acids with Metals Materials needed: Small pieces of zinc (Zn), iron (Fe), magnesium (Mg), and hydrochloric acid (HCl) or vinegar.

Steps

i. Place each metal in a separate container

ii. Add a small amount of acid and observe.

iii. Observe and record what happens in the reaction.

iv. What is the colour of the gas evolved?

v. Test the gas evolved with blue and red litmus paper.

vi. Test the gas evolved with lighted splint.

b. Reaction of Acids with Bases (Neutralisation) Materials needed: Vinegar (acid) and baking soda solution (base).

Steps

i. Measure equal volume of the acid and base into separate beakers

ii. Mix the two solutions and observe.

iii. Observe and record what happen in the reaction.

c. Reaction of Acids with Carbonates Materials needed: Vinegar or lemon juice and baking soda (sodium bicarbonate) chalk (calcium carbonate).

Steps

i. Add vinegar (25 cm³) to about 5.0 g baking soda or chalk.

ii. Observe and record what happen in the reaction.

iii. What is the colour of the gas evolved?

iv. Test the gas evolved with blue and red litmus paper.

v. Test the gas evolved with limewater.

d. Reaction of Acids with Basic Oxides Materials needed: Vinegar and a magnesium oxide powder (or similar basic oxide).

Steps

i. Measure about 25 cm³of vinegar into a beaker.

ii. Add about 5.0 g of magnesium oxide to the vinegar solution.

iii. Observe and record what happen in the reaction.

2. Reaction of Bases

a. Reaction of Bases with Acidic Oxides Materials needed: Limewater (aqueous calcium hydroxide) and carbon dioxide balloon or from exhaled breath.

Steps

i. Pour small quantity of limewater into a beaker.

ii. Fill a balloon with carbon dioxide.

iii. Release the carbon dioxide into the limewater.

iv. Observe and record what happen in the reaction.

b. Reaction of Bases with Ammonium Salts Materials needed: Ammonium chloride and a strong base, such as sodium hydroxide.

Steps

i. Mix ammonium chloride with sodium hydroxide carefully, in a well- ventilated area.

ii. What is the colour of the gas evolved?

iii. Test the gas evolved with blue and red litmus paper.

Activity 4.8 Classifying Household Items as Acids or Bases Materials needed: Litmus paper or pH strips, and a selection of household items (e.g., lemon juice, baking soda solution, vinegar, soap solution, milk, apple juice).

Steps Perform the test on each solution with litmus paper or pH strips and record whether it turns the paper red (acidic) or blue (basic). Record your test and observation in the table below:

Type Colour observed using Classify the solution as acid or base Blue litmus Red litmus Lemon juice, Vinegar apple juice Baking soda

solution, Soap solution

Activity 4.9 Application of neutralization in everyday life Materials needed: Visual aids (diagrams, images) of an insect bite, an acidic lake, a garden, and a stomach Steps

1. In small groups, assign each group one scenario:

• Treating insect bites

• Helping with stomach indigestion

• Reducing soil acidity for plants

• Balancing lake acidity to protect fish

2. What kind of substance (acid or base) is needed to fix this problem?

3. What everyday things can help us neutralize this issue (like baking soda for bites, antacids for indigestion)?

4. How do these substances make things safe or comfortable again?

5. Share your scenario and what you learned with the class.

Differences between strong acids and weak acids A strong acid can be distinguished from a weak acid by examining its pH value, electrical conductivity, and reactivity.

1. pH pH is the logarithmic value of the inverse of the hydrogen or hydronium ion concentration.

Consider the reaction: HA(aq) + H₂ O(l) ⇌ H₃ O+(aq) + A−(aq);

a. Strong acids fully dissociate in water, meaning they release a lot of H+ ions per mole into solution, results in a low pH (e.g., hydrochloric acid, HCl).

Equation: HCl(aq) → H+(aq) + Cl−(aq)

b. Strong bases fully dissociate in water, creating a high concentration of OH− ions per mole in solution, which gives a high pH (e.g., sodium hydroxide, NaOH).

Equation: NaOH(s) + H₂ O(l) → Na+(aq) + O H−(aq)

c. Weak acids only partially dissociate, releasing fewer H+ ions per mole into solution and resulting in a higher pH compared to strong acids (e.g., acetic acid, CH₃COOH).

Equation: C H₃ COOH(aq) ⇌ H+(aq) + C H₃ CO O−(aq)

d. Weak bases partially dissociate, resulting in fewer OH− ions per mole in

solution and a lower pH than strong bases (e.g., ammonia, NH₃).

Equation: N H₃(aq) + H₂ O(l) ⇌ N H₄ +(aq) + O H−(aq)

2. Electrical Conductivity (per mole of acid/base)

a. Strong acids have high conductivity because they produce more ions in

solution.

b. Strong bases have high conductivity due to the high concentration of OH− ions.

c. Weak acids have low conductivity due to fewer ions in solution.

d. Weak bases have lower conductivity as they produce fewer OH− ions.

3. Reactivity (per mole of acid/base)

a. Strong acids are more reactive with metals, releasing more hydrogen gas quickly.

b. Strong bases react more readily with acids to form water and salt, and they can very corrosive.

c. Weak acids react more slowly, producing less hydrogen gas over time.

d. Weak bases are less reactive than strong bases.

4. Enthalpy of Neutralization (ΔH)

a. Strong acids release more energy per mole when neutralized by a strong base because they fully dissociate.

Example Equation

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂ O(l)ΔH = − 57.1 kJmol − 1

b. Weak acids release less energy per mole, as some energy is needed to break bonds before neutralization can occur.

Example Equation

C H₃ COOH(aq) + OH−(aq) → H₂ O(l) + C H₃ CO O− ΔH = − 56.1 kJmol − 1

Activity 4.10 Introduce acids and bases using visual aids, interactive simulations Materials needed: Computer, vinegar, vinegar, hydrochloric acid-based cleaning product or drain cleaner, baking soda, stopwatch

Step 1

1. Watch a video online showing basic home substances that are acidic and basic.

2. Use online simulations or videos showing how acids and bases behave in water.

Step 2: Strong vs. Weak Acids and Bases

1. Mix 10 cm³vinegar (a weak acid) with 10 g of baking soda and observe the reactions.

2. Mix 10 cm³drain cleaner (a strong acid) with baking soda and observe the reactions.

Compare the two reactions. In which reaction are more bubbles produced in the same time interval.

Step 3: Exploration Stations (explore the strength of acids and bases) Organise yourselves into groups of no more than five. Conduct the following exploration stations in your respective groups.

Station 1: Conductivity Measurement

Materials: Provide a small circuit with an LED or bulb and probes to test solutions, beaker, measuring cylinder.

1. Measure 25 cm³drain cleaner into a beaker.

2. Insert two probes or electrodes connected with wire into the drain cleaner.

3. Connect bulb between the two probes or electrodes.

4. Observe the level of brightness of the bulb.

5. Measure 25 cm³vinegar into a beaker.

6. Repeat steps 2, 3, and 4.

7. Which solutions make the bulb light up more and why?

Station 2: pH Measurement Materials: pH strips, drain cleaner, vinegar, baking soda, beaker, distilled water

1. Measure 25 cm³drain cleaner into a beaker.

2. Dip a pH strip into the drain cleaner for a few seconds.

3. Remove the pH strip and allow the colour to develop.

4. Compare the colour with the pH scale provided to determine the approximate pH the solution.

5. Repeat steps 1, 2, 3, and 4 with vinegar, baking soda, distilled water.

Organise the solutions from most acidic to most basic based on their pH values.

SALTS A salt is a type of chemical compound formed when the hydrogen ions in an acid are replaced by metal or ammonium ions. Salts are important in everyday life and in many industries.

Basic Equation for Salt Formation: Acid + Base → Salt + Water Types of Salts There are many types of salts, each with unique properties and uses. Some common types of salts include:

1. Normal Salts

Formed when an acid and base completely neutralize each other.

Example: NaOH + HCl → NaCl + H₂O

2. Acidic Salts

Formed when only some hydrogen ions in an acid are replaced.

Example: NaOH + H₂SO₄→ NaHSO₄+ H₂O Acidic salts often come from reactions between a strong acid and a weak base, ammonium chloride (NH₄Cl).

3. Basic Salts

Formed when some hydroxide ions of a base are replaced.

Example: CH₃COOH + NaOH → CH₃COONa + H₂O

4. Double Salts

Made from two salts mixed in equal amounts. They have two types of cations.

Example: Potash alum, used in dyeing and water purification.

5. Complex Salts

Have a central metal atom bonded to other molecules or ions. They do not fully dissociate into ions in water.

Example: [Ni(CO)₄] or [Cr(NH₃)₆]Cl₃.

6. Hydrated Salts

Crystalline salts that hold water molecules within them.

Example: Magnesium tetraoxosulphate (VI) heptahydrate (Epsom salt) has trapped inside it and calcium tetraoxosulphate (VI) dihydrate

7. Deliquescent Salts

Salts that absorb water from the air and can even turn into a liquid.

Example: Calcium chloride and sodium hydroxide.

8. Hygroscopic Salts

Salts that absorb water from the air but don’t dissolve completely.

Example: Anhydrous copper sulphate and anhydrous cobalt chloride.

9. Efflorescent Salts

Salts that release water into the air, losing their water content.

Example: Sodium carbonate decahydrate and sodium sulphate decahydrate.

Preparing soluble salts Preparing Soluble Salts: Two Methods Method 1: Reacting an Acid with a Solid Metal, Insoluble Base, or Carbonate Materials needed Dilute sulfuric acid (H₂SO₄), an insoluble base (like copper (II) oxide) or an insoluble carbonate, beaker, Bunsen burner (or a warm water bath), stirring rod, evaporating dish, glass rod, filter paper Steps to Prepare the Salt

1. Pour some dilute sulfuric acid into a beaker.

2. Gently heat the beaker using a Bunsen burner or place it in a warm water bath (around 60°C). Be careful not to boil it!

3. Slowly add the copper (II) oxide (or other base) to the warm acid. Stir well until the base stops dissolving and a bit remains at the bottom (this means all the acid has reacted).

4. Pour the mixture through filter paper into an evaporating dish. This removes any leftover solid.

5. Heat the solution in the evaporating dish until some water evaporates and the solution becomes more concentrated. Be careful not to let it boil too much.

6. Dip a cold glass rod into the solution. If crystals form on the rod, the solution is ready.

7. Leave the solution in a warm place and let it sit undisturbed to form crystals.

8. Carefully pour off any leftover solution and let the crystals dry. You can pat them dry with filter paper if needed.

Preparing Soluble Salts: Reacting a Dilute Acid with an Alkali Materials needed Alkali (like sodium hydroxide solution), dilute acid (like hydrochloric acid), methyl orange indicator, conical flask, burette, evaporating dish Steps to Prepare the Salt

1. Measure sodium hydroxide solution into a conical flask using a pipette.

2. Add a few drops of methyl orange indicator to the alkali in the flask. This will help us see when the reaction is complete by changing colour.

3. Pour some hydrochloric acid into the burette and note the starting level.

4. Carefully add the acid from the burette into the flask with the alkali, swirling as you go. Stop when the colour changes from yellow to orange/red. This means the acid has neutralized the alkali.

5. Note how much acid you added by checking the level on the burette. This is the amount you need for the reaction.

6. Now, mix the same amount of acid and alkali in another container without adding any indicator.

7. Heat this mixture in an evaporating dish until some water evaporates, making a concentrated solution. Be careful not to let it boil too much.

8. As the solution cools, crystals will start to form. Leave it in a warm place for more crystals to grow.

9. Pour off any leftover liquid and let the crystals dry completely.

Uses of Salts

1. Medicine and Health

a. Saline Solution: A mixture of salt (sodium chloride) and water, used hospitals for:

i. Rehydration: Helps people who are dehydrated get fluids quickly.

ii. Wound Cleaning: Used to clean wounds and prevent infections.

2. Industry

a. Desalination: Salts are removed from seawater to make fresh water processes like reverse osmosis.

b. Chemical Production: Table salt (sodium chloride) is used to make chlorine, which is important for treating water and making plastics.

Soda ash ( carbonate) is used in making glass and detergents.

3. Agriculture

a. Fertilizers: Some salts provide nutrients for plants. For example, nitrate provides potassium and nitrogen, and ammonium sulphate nitrogen and sulphur.

b. Animal Nutrition: Livestock are given salt blocks to get important minerals sodium and iodine.

c. Pest Control: Some salts help control pests and diseases in plants.

4. De-Icing:

a. In winter, rock salt is spread on roads to melt ice and make surfaces safer walk or drive on. It prevents ice from forming by lowering the freezing point water.

Activity 4.11 Exploring Salt

Step

1. In small groups, brainstorm and think what comes to mind when you hear the word ‘Salt’. Use the following questions to guide your discussion:

a. What is salt made of?

b. Where do we use salt in everyday life?

c. How do salts form?

2. Present your findings to the entire class.

3. Using textbooks or the internet, explore the following types of salts, their characteristics and their uses. Present your findings to the class.

a. normal,

b. acidic,

c. complex, or

d. double salts

Activity 4.12 Preparing Salts and Exploring Their Uses

Part 1: Preparing Soluble Salts

1. Select an acid (like hydrochloric acid) and an insoluble reactant (such as a oxide or carbonate).

2. Pour the dilute acid into a beaker.

3. Gradually add the powdered reactant to the acid, stirring continuously.

Watch bubbles (effervescence) as the reaction occurs.

4. Keep adding until no more dissolves and some unreacted powder remains at bottom (this means there’s an excess).

5. Pour the mixture through filter paper to separate the solid from the liquid, leavin clear solution of the soluble salt.

Part 2: Preparing Insoluble Salts

1. Mix two solutions of soluble salts (like lead nitrate and potassium iodide) in a beaker.

2. A solid will form (the insoluble salt) and settle at the bottom.

3. Use filter paper to separate the solid precipitate from the liquid.

4. Gently rinse the solid with distilled water to remove any leftover solution.

5. The resulting solution should be handled with care as it contains lead.

Ensure a proper disposal route is available.

Part 3: Discuss Everyday Uses of Salts

1. Discuss common salts you know and how they are used, like:

a. Table Salt (for food)

b. Baking Soda (in baking and cleaning)

c. Epsom Salt (for baths and gardening)

2. Share your thoughts on how salts are used at home and in the community.

Part 4: Field Trip or Excursion to a Salt Mine Objective: To provide a hands-on experience and real-world understanding of salt mining processes.

1. Your teacher will arrange trip to a salt mine or mining facility. Please take safety precautions and wear appropriate attire. During your visit, make sure to observe the following and make notes:

a. Salt extraction processes

b. Mining methods (e.g., room and pillar, solution mining)

c. Safety procedures

d. Environmental considerations

2. Jot down your observations and share your findings with the class. Engage in a whole class discussion on:

a. Salt mining processes

b. Real-world applications of salt

c. Environmental and economic impacts

3. Connect field trip to classroom learning:

a. Relate salt mining to geological processes

b. Discuss chemical properties of salt

c. Explore industrial uses of salt

Acid-Base Titration

Definition Acid-base titration is a method used to find out the concentration of an unknown acid or base by reacting it with a solution of known concentration. Anindicator helps show when the reaction is complete by changing colour.

Apparatus Used in Titration

1. Conical flask: Holds the solution with the unknown concentration (called the analyte) and the indicator. It has a narrow neck for easy swirling.

2. Indicator: A dye that changes colour to show when the acid has neutralized the (or vice versa). Common indicators are phenolphthalein (turns pink in basic solutions) and methyl orange (turns red in acidic solutions).

3. White tile: Placed under the flask to help see the colour change clearly.

4. Beaker: It is usually used to hold excess solution or waste during the titration process.

5. Analyte: The analyte is the solution we want to learn more about in a experiment. It is usually placed in a conical flask, and an acid-base indicator is to it, which changes colour during the experiment.

6. Titrant: It is the solution whose concentration is accurately known and usually in the burette.

7. Burette: A long tube that accurately measures and dispenses the solution wit known concentration (called the titrant). It has a tap to control the flow of the titrant.

8. Pipette: Measures a specific volume of the analyte and transfers it to the conical flask. See Figure 4.1 for a diagram of pipette filler.

Figure 4.1: Pipette filler Steps for Using the Pipette Bulb

a. Insert the top (mouth) of the pipette into the bottom of the pipette bulb.

b. Squeeze the button labelled “A” (Air valve) while pressing the bulb to release any air.

c. Place the tip of the pipette into the solution.

d. Squeeze the button labelled “S” (Syphon valve) to draw liquid up into the until it reaches the desired level.

e. To release the liquid, press the button labelled “E” (Empty valve).

9. Retort stand: A retort stand is a lab tool with three parts: a base, a rod, and a clamp. It securely holds the burette in place, allowing it to stay directly above the conical for accurate pouring of liquids during experiments. See

Figure 4.2 for titration setup.

Figure 4.2: Titration set up.

10. Acid-Base Indicator

An acid-base indicator is a substance that changes colour to show when a reaction is complete (the endpoint) in a titration. The indicator is added to the solution in the conical flask, and it stays the same colour at first. When the acid and base in the solution balance out, the indicator changes colour to signal that the reaction is done.

a. Phenolphthalein: Colourless in acids, turns pink in a basic solution.

b. Methyl Orange: Red in acidic solutions and yellow in basic solutions.

These indicators help scientists know when an acid and a base have completely reacted with each other.

Most indicators do not change colour at a particular pH. They do so over a range of pH from two to three units. This is called the pH range which is different for various indicators.

Therefore, a pH range is the working range within which an acid-base indicator changes from one colour to another.

The table below gives the pH ranges of some acid-base indicators Indicator Colour change (acid to base) pH range Methyl orange Red to orange 3.1 – 4.4 Methyl red Red to yellow 4.4 – 6.0 Litmus Red to blue 5.0 – 8.0 Bromothymol blue Yellow to blue 6.0 – 7.6 Phenolphthalein Colourless to pink 8.3 – 10.0 Principles of Acid-Base Indicators

1. Phenolphthalein as an Acid-Base Indicator

Phenolphthalein is a weak acid that can change colour depending on the

solution’s pH.

It exists in two forms in solution: HIn (the acidic form with one colour) and In– ( basic form with a different colour).

In an acidic solution, there are more H+ ions, which keeps phenolphthalein in its HIn form, showing the acid colour.

In a basic solution, there are more OH– ions, which cause phenolphthalein to to its In– form, showing the base colour (pink in phenolphthalein’s case).

2. Other Indicators (Weak Bases)

Some indicators work as weak bases and also change colour based on pH.

In an acidic solution, H+ ions from the acid react with OH– ions from the indicator, causing the indicator to show its acid colour.

In a basic solution, there are more OH– ions, so the indicator shows its base colour.

In simple terms, acid-base indicators change colour because they have one colour acidic conditions and a different colour in basic conditions, helping us tell whethe solution is acidic or basic.

Choosing the Right Indicator for Titration

The choice of indicator in a titration depends on the type of acid and base used.

Here is a simple guide:

a. Strong Acid and Strong Base When both the acid and base are strong, the pH changes a lot near the endpoint, so many indicators will work well.

Suitable indicators: Litmus, phenolphthalein, methyl orange, or bromothymol blue.

b. Weak Acid and Strong Base (or Strong Acid and Weak Base) If one is weak and the other is strong, the pH changes less near the endpoint, so you need an indicator that matches the pH at the endpoint.

Examples: Use methyl orange with strong acids and phenolphthalein with strong bases.

c. Weak Acid and Weak Base When both are weak, the pH change is very small, so bromothymol blue is a good choice.

Key Terms in Titration

1. Titrant: The titrant is a solution with a known concentration that is added to the analyte to out its properties. It is placed in the burette during titration.

Characteristics of a Titrant

a. Its concentration is known and accurate.

b. It is as pure as possible.

c. It reacts completely with the analyte.

d. The reaction has a clear endpoint, shown by a colour change.

2. Analyte: The analyte is the solution with an unknown concentration. We want to learn about by reacting it with the titrant. The analyte is placed in the conical flask.

3. Equivalence Point: This is the point where the exact amount of titrant has been added to react with the analyte. It is a theoretical point in the reaction.

4. Endpoint: The endpoint is when the colour change happens, showing that the reaction complete. This is observed with the help of an indicator.

5. Titre Value: The titre value is the volume of the titrant used to reach the endpoint. It’s calculated subtracting the starting reading from the final reading on the burette.

6. Consistent Titre Values: These are titre values that are very close to each other (within ± 0.20 cm³), accuracy in measurements.

7. Average Titre: This is the average of the consistent titre values, providing the most measurement of titrant used.

8. Indicator: Indicators are substances that change colour based on the pH level, helping us when the titration reaction is complete.

Types of Acid-Base Titration

1. Simple (Direct) Acid-Base Titrations

In a simple titration, we directly add a known solution (titrant) to an unknown (analyte) to find its concentration.

This type of titration uses only one acid-base indicator to show the endpoint ( change) when the reaction is complete.

Simple titration helps us figure out the exact amount of acid or base in a solution.

Steps for Simple Acid-Base Titration

1. Clamp the burette to a stand, rinse it with distilled water and then with the titrant. Fill with the titrant and record the starting volume.

2. Use a pipette to measure a specific amount of the analyte (usually 20 or 25 cm³) place it in a conical flask.

3. Add 2-3 drops of an indicator to the analyte in the flask and swirl to mix it.

4. Start Titration: Slowly add the titrant from the burette to the analyte in the flask, swirling until a permanent colour change (endpoint) appears.

5. Record the titre values: Note the initial and final burette readings and calculate volume of titrant used.

6. Repeat the process to get consistent readings and calculate the average titre.

7. Use the balanced chemical equation to find the ratio between the titrant and analyte, and then calculate the analyte’s concentration.

Example 1

A solution of 0.2 moldm⁻³of HCl was titrated against 25 cm³of KOH using methyl orange as indicator. The results obtained in the titration experiment are as follows Titrant: 0.2 moldm⁻³HCl Analyte: KOH of unknown concentration Capacity of Burette: 50 cm³Capacity of pipette: 25 cm³Indicator used: 2 drops of methyl orange Colour change: Yellow to red Titration Table Burette Reading (cm³) 1st 2nd Final 24.60 24.70 Initial 0.00 0.00 Volume of HCl Used 24.60 24.70 Calculate, i The average titre value ii The molar concentration of NaOH iii The mass concentration of NaOH

Solution

i. Average titre = 24.70 + 24.60/2 = 24.65 cm³ii. From the titration reaction HCl + KOH → NaCl + H₂O The mole ratio is given by n(KOH)_ n(HCl) = 1/1 C_(KOH) V_(KOH)_ C_(HCl) V_(HCl) = 1/1 C_(KOH) = C_(HCl) V_(HCl)_ V_(KOH) = 0.2 mol dm⁻³× 24.65 cm³___________________ 25 cm³= 0.197 mol dm⁻³iii. M(KOH) = 39 + 16 + 1 = 56 gmol⁻¹Mass concentration, ρ = C × M = 0.197 mol dm⁻³× 56 gmol⁻¹= 11.0 g dm⁻³In addition to solution standardization, direct acid-base titrations have various applications. They can be used to determine:

i. The percentage purity of a substance

ii. The number of moles or ions present in a solution

iii. The molar mass of a substance

iv. The solubility of a substance

2. Back (Indirect) Titration

Back titration is a method used when we cannot directly measure an substance (let’s call it “A”) by titration. Here’s how it works:

a. React Substance A with Excess B We add more than enough of a known solution (B) to react with all of A, making A is fully neutralized.

b. Measure the Leftover B After A has reacted with some of B, we titrate the leftover B with another

solution find out how much of it is left.

c. Calculate Amount of A By subtracting the leftover amount of B from the original amount, we can how much B was used to react with A. This tells us how much A was present.

Back titration is useful for many things, including finding out:

a. How much calcium carbonate (CaCO₃) is in limestone or eggshells

b. The amount of ammonia in an ammonium salt

c. The amount of calcium hydroxide (Ca(OH)₂) in a sample of soil

Example 2

To check if a new brand of aspirin (Ecotrin) contains lower doses of acetylsalicylic acid, a chemist conducted an experiment. They crushed two aspirin tablets and mixed them with 25 cm³ of 1.00 moldm⁻³NaOH, warming the mixture before transferring it to a 250 cm³ volumetric flask.

0.050 moldm⁻³H₂SO₄was titrated against 25 cm³ portions of the sodium hydroxide

solution, using phenolphthalein as an indicator.

The following results were recorded in the experiment:

Burette reading/cm³1 2 3 Final reading 10.70 20.40 40.75 Initial reading 0.10 10.30 30.45 Volume of acid used 10.60 10.10 10.30 Calculate

a. The average titre value

b. Moles of H₂SO₄that reacted

c. Moles of excess NaOH present in the solution.

d. Moles of NaOH that reacted with acetylsalicyclic acid

e. Mass of acetylsalicyclic acid in one tablet.

f. If the mass of each tablet is 2 g, determine the percentage mass of pure acetylsalicyclic acid in each tablet.

The equations of the reaction involved are:

H₂ SO₄(aq)+ 2NaOH(aq) → Na₂ SO₄(aq)+ 2H₂ O(l) C₉ H₈ O₄ + NaOH → C₉ H₇ O₄ Na+ H₂ O [C = 12, H = 1, O = 16, Na = 23, S = 32]

Solution

a. Average titre = 10.10 + 10.30/2 = 10.20 cm³b. C_(H)₂_(SO)₄ = 0.050 mol dm⁻³; V_(H)₂_(SO)₄ = 10.20 cm³= 0.0102 dm³n_(H)₂_(SO)₄ = C_(H)₂_(SO)₄ V_(H)₂_(SO)₄ = 0.050 mol dm⁻³× 0.0102 dm³= 0.00051 mol

c. From the balanced equation H₂ SO₄(aq)+ 2NaOH(aq) → Na₂ SO₄(aq)+ 2H₂ O(l) The mole ratio for the titration is n_(NaOH)_ n_(H)₂_(SO)₄ = 2/1 n_(NaOH) = 2 × n_(H)₂_(SO)₄ = 2 × 0.00051 mol = 0.00102 mol Therefore, moles of NaOH that was present in the 25 cm³and reacted with the acid is 0.00102 mol .

If 25 cm³of NaOH contains 0.00102 mol Then 250 cm³of NaOH will contain = 250 cm³× 0.00102 mol/25 cm³= 0.0102 mol

d. From the relation n = CV Initial concentration of NaOH before it was mixed with the crushed tablets = 1.0 mol dm⁻³.

Volume of the solution that was mixed with the crushed tablets = 25 cm³= 0.025 dm³n_(NaOH) = 1.0 mol dm⁻³× 0.025 = 0.025 mol If 0.0025 mol of NaOH was mixed with the crushed tablets and 0.0102 mol remained, then, Mole of NaOH that reacted with the acetylsalicyclic acid = 0.025 – 0.0102 = 0.0148 mol

e. Acetylsalicyclic acid reacts with sodium hydroxide according to the equation C₉ H₈ O₄ + NaOH → C₉ H₇ O₄ Na+ H₂ O The mole ratio is given by nₐₛₚᵢᵣᵢₙ_ n_(NaOH) = 1/1 nₐₛₚᵢᵣᵢₙ = n_(NaOH) = 0.0148 mol but m = n × M M( C₉ H₈ O₄) = (12 × 9) + (1 × 8) + (16 × 4) = 180 g dm⁻³m( C₉ H₈ O₄) = 0.0148 × 180 = 2.66 g 2 tablets contain 2.66 g ∴ 1 tablet contains 2.66/2 = 1.33 g

f. If the mass of each tablet is 2 g % (mass) = mass of pure acetylsalicyclic acid________________________ mass of tablet × 100 = 1.33/2 × 100 = 66.5 %

3. Double-Indicator Titration

Double-indicator titration is a method used to find the concentration of substances in a mixture, using two indicators to show separate stages of the reaction.

a. In this method, each part of the mixture reacts differently with the titrant. One forms an acidic product, while the other forms a basic (alkaline) product.

b. Indicators like methyl orange and phenolphthalein are chosen to match the pH of each stage, changing colour to show when each part of the reaction complete.

Two Methods of Double-Indicator Titration

1. Continuous Method: Both stages of the reaction happen one after the other.

2. Discontinuous Method: Each stage is carried out separately.

Continuous Method for Double-Indicator Titration

1. Place the acid (titrant) in the burette.

2. Measure 20 or 25 cm³ of the basic mixture and pour it into a conical flask.

Add drops of phenolphthalein and swirl to mix.

3. First Titration (Phenolphthalein Endpoint)

Slowly add the acid from the burette until the colour changes. Record the volume acid used as x cm³. This step reacts with sodium carbonate (Na₂CO₃) to form bicarbonate (NaHCO₃).

4. Add two drops of methyl orange to the solution after the first titration.

5. Continue adding the acid until a second colour change occurs. Record this volume as y cm³. Here, the remaining sodium bicarbonate (NaHCO₃) reacts with acid to form sodium chloride, water, and carbon dioxide.

6. To find the volume of acid that reacted with the original sodium bicarbonate, the first reading x from the second reading y (y - x). These values can then be used calculate the concentration of each component in the mixture.

Example 3

A mixture of Na₂CO₃and NaHCO₃reacted with 0.1 moldm⁻³HCl.

The results obtained when phenolphthalein indicator was used is Burette Reading/cm³1 2 3 Final 13.00 47.10 13.00 Initial 0.00 34.00 0.00 Volume of acid used 13.00 13.10 13.00 The titration continued with methyl orange being added to the resulting solution.

The results obtained at the methyl orange endpoint is Burette Reading/cm³1 2 3 Final 34.00 21.00 34.00 Initial 13.00 0.00 13.00 Volume of acid used 21.00 21.00 21.00

a. Write a balanced chemical equation to represent the reaction that took place at the;

i. Phenolphthalein endpoint

ii. Methyl orange endpoint

b. Calculate the concentration in moldm⁻³of,

i. Na₂ CO₃

ii. NaHCO₃

c. NaHCO₃ In the mixture that was used for the titration Determine the mass concentration of;

i. Na₂ CO₃

ii. NaHCO₃ Determine the percentage composition of Na₂ CO₃ in the mixture.

Solution

a. The balanced chemical equation for

i. Phenolphthalein endpoint is represented by:

Na₂ CO₃(aq) + HCl(aq) → NaHCO₃(aq)+ NaCl(aq)

ii. Methyl orange end point:

NaH CO₃(aq) + HCl(aq) → NaCl(aq)+ H₂ O(l) + CO₂(g)

b. From the titration reaction

i. The average titre value at the phenolphthalein endpoint is given by x = 13.00 + 13.00/2 = 13.00 cm³V₁ = 13.00 cm³= volume of HCl required to neutralise half the Na₂ CO₃ in the mixture at the end-point.

V₁ = 1/2 = Na₂ CO₃ V₁ = 13.00 2V₁ = 2 × 13.00 = 26.00 cm³The volume of HCl required to neutralise completely Na₂ CO₃ is 2 V₁ The complete reaction of HCl with Na₂ CO₃ that requires 26.00 cm³of the titrant (HCl) is represented by the chemical equation:

Na₂ CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂ O(l)+ CO₂(g) Mole ratio for the complete reaction is C_(Na)₂_(C)_(O)₃ V_(Na)₂_(C)_(O)₃_ C_(HCl) V_(HCl) = 1/2 C_(Na)₂_(C)_(O)₃ = C_(HCl) × V_(HCl)_ 2V_(Na)₂_(C)_(O)₃ C_(HCl) = concentration of HCl = 0.100 mol dm⁻³, V_(HCl) = volume of HCl for complete reaction = 26.00 cm³V_(Na)₂_(C)_(O)₃ = volume of mixture pipetted for the reaction = 25.00 cm³C_(Na)₂_(C)_(O)₃ = concentration of Na₂ C O₃ = C_(Na)₂_(C)_(O)₃ = C_(HCl) × V_(HCl)_ 2V_(Na)₂_(C)_(O)₃ = 0.100 mol dm⁻³× 26 cm³___________________ 2 × 25 cm³= 0.0520 mol dm⁻³ii. The average titre value at the methyl orange endpoint is given by y = 21.00 + 21.00/2 = 21.00 cm³ 21.00 cm³of HCl reacted with the NaHC O₃that was formed from the Na₂C O₃and the original NaHC O₃that was present in the mixture.

If half of Na₂C O₃was neutralised by 13.00 cm³, then the remaining half that was present in solution will also react with the same volume of HCl.

V₂ = volume of HCl required to neutralise the remaining half of Na₂ CO₃ and NaHCO₃ in the mixture at the methyl orange end point.

V₂ = 1/2 Na₂CO₃ + NaHCO₃ V₂ = V₁ + NaHCO₃ V₂ − V₁ = NaHCO₃ V_(HCl) = V₂ − V₁ = 21.00 − 13.00 = 8.00 cm³V₂ − V₁ = Volume acid required to neutralise NaHCO₃ The reaction equation is given by:

NaH CO₃(aq) + HCl(aq) → NaCl(aq)+ H₂ O(l) + CO₂(g) C_(NaHCO)₃ V_(NaHCO)₃___________ C_(HCl) V_(HCl) = 1/1 C_(NaHCO)₃ = C_(HCl) V_(HCl)_ V_(NaHCO)₃ C_(NaHCO)₃ = C_(HCl)(V₂ − V₁)___________ V_(NaHCO)₃ = 0.100 mol dm⁻³× 8.00 cm³____________________ 25 cm³= 0.032 mol dm⁻³c. i. ρ(Na₂ C O₃) = c(Na₂ C O₃) × M(Na₂ C O₃) M(Na₂ C O₃) = (2 × 23) + 12 + (3 × 16) = 106 g mol⁻¹ρ(Na₂ C O₃) = c(Na₂ C O₃) × M(Na₂ C O₃) = 0.0520 mol dm⁻³× 106 g mol⁻¹= 5.51 = 5.51 g dm⁻³

ii. M(NaHC O₃) = 23 + 1 + 12 + (3x16) = 84 g mol⁻¹ρ(NaHC O₃) = c(NaHC O₃) × M(NaHC O₃) = 0.032 mol dm⁻³× 84 g mol⁻¹= 2.688 = 2.69 g dm⁻³d. %(Na₂CO₃) = m(Na₂CO₃)________________ Total mass of mixture × 100 Total mass of mixture = m(Na₂CO₃) + m(NaHCO₃) = 5.51 + 2.69 = 8.2 g % (Na₂CO₃) = m(Na₂CO₃)________________ Total mass of mixture × 100 = 5.51/8.2 × 100 = 67.2% Discontinuous Method for Double-Indicator Titration

a. Place the acid (HCl) in the burette.

b. Measure 20 or 25 cm³ of the basic mixture into a conical flask and add 2-3 drops of phenolphthalein.

c. Slowly add HCl from the burette until the colour changes. Record this volume as V₁cm³.

d. This volume of HCl (V₁) converts all sodium carbonate (Na₂CO₃) in the mixture into sodium bicarbonate (NaHCO₃).

e. Discard the solution from the first titration.

f. Take another 20 or 25 cm³ of the mixture in a new conical flask, add 2 drops of methyl orange, and swirl to mix.

g. Slowly add HCl from the burette until the colour changes. Record this new volume as V₂cm³.

h. This time, all sodium carbonate and sodium bicarbonate in the mixture react with HCl, creating sodium chloride, water, and carbon dioxide.

i. To find the amount of original sodium bicarbonate (NaHCO₃), use the formula V3 = V2 - 2V₁.

j. This final volume, V₃, helps calculate the amount of sodium bicarbonate initially present in the mixture.

Example 4

A sample of Na₂C O₃and NaHC O₃were mixed together to form a solution.

25 cm³of this solution was titrated against a 0.1 moldm⁻³HCl solution using phenolphthalein as an indicator.

The results obtained when phenolphthalein indicator was used is Burette Reading/cm³1 2 3 Final 10.60 21.30 32.00 Initial 0.00 10.60 21.30 Volume of acid used 10.60 10.70 10.70 Another 25 cm³portions of the same solution was also titrated against 0.1 moldm⁻³ HCl solution using methyl orange as an indicator.

Burette Reading/cm³1 2 3

Final 23.70 47.30 27.60 Initial 0.00 23.70 0.00

Volume of acid used 23.70 23.60 27.60

a. Determine the average titre value at the;

i. Phenolphthalein endpoint

ii. Methyl orange endpoint

b. Determine the volume of HCl that reacted with;

i. All of the Na₂C O₃present

ii. The original quantity of NaHC O₃present

c. Calculate the concentration in moldm⁻³of;

i. Na₂C O₃

ii. NaHC O₃

d. Determine the mass concentration of;

i. Na₂C O₃

ii. NaHC O₃

e. Determine the percentage composition of NaHC O₃in the mixture

Solution

a. Calculation of the average titre values at the various stages of the titration

i. At the phenolphthalein endpoint, the average titre value is given by;

= 10.70 + 10.70/2 = 10.70 cm³ii. At the methyl orange endpoint, the average titre value is given by;

= 23.60 + 23.60/2 = 23.60 cm³b. To determine the volume of HCl that reacted with the various components of mixture;

i. The equation for the complete reaction of Na₂C O₃with HCl goes by the equation:

Na₂ CO₃(aq) + 2HCl(aq) → 2NaCl(aq) + H₂ O(l)+ CO₂(g) V₁ = volume of HCl required to neutralise half the Na₂ CO₃ in the mixture at phenolphthalein end-point.

V₁ = 1/2 Na₂ CO₃ V₁ = 10.70 2V₁ = 2 × 10.70 = 21.40 cm³The volume of HCl required to neutralise completely Na₂ CO₃ is 2 V₁ = 21.40 cm³ii. At the methyl orange endpoint, two sets of reaction occurred where all components of the mixture were neutralised by the HCl. These reactions can represented by the equations;

V₂ = Volume of HCl required to neutralise Na₂ CO₃ and NaHCO₃ in the mixture at methyl orange end point.

Na₂ CO₃ + 2HCl → 2NaCl + H₂ O + CO₂ (1) NaHCO₃ + HCl → NaCl + H₂ O + CO₂ (2) V₂ = Na₂ CO₃ + NaHCO₃ V₂ = 2V₁ + NaHCO₃ V₂ − 2V₁ = NaHCO₃ V₂ − 2V₁ = 23.60 − 21.40 = 2.20 = Volume acid required to neutralize NaHCO₃

c. i. The equation for the reaction of Na₂ CO₃ at the phenolphthalein endpoint is given by:

Na₂ CO₃ + HCl → NaHCO₃ + NaCl C_(Na)₂_(CO)₃ V_(Na)₂_(CO)₃_ C_(HCl) V_(HCl) = 1/1 C_(Na)₂_(CO)₃ = C_(HCl) V_(HCl)_ V_(Na)₂_(CO)₃ C_(Na)₂_(CO)₃ = 0.100 mol dm⁻³× 10.70 cm³_____________________ 25 cm³= 0.0428 mol dm⁻³ii. The reaction that took place between the original NaHC O₃in the mixture and HCl is represented by:

NaHCO₃ + HCl → NaCl + H₂ O + CO₂ This required 2.20 cm³of the HCl.

The mole ratio of this reaction is C_(NaHCO)₃ V_(NaHCO)₃___________ C_(HCl) V_(HCl) = 1/1 C_(NaHCO)₃ = C_(HCl) V_(HCl)_ V_(NaHCO)₃ = 0.100 mol dm⁻³× 2.20 cm³____________________ 25 cm³= 0.0088 mol dm⁻³d. i. ρ(Na₂ C O₃) = c(Na₂ C O₃) × M(Na₂ C O₃) M(Na₂ C O₃) = (2 × 23) + 12 + (3 × 16) = 106 g mol⁻¹ρ(Na₂ C O₃) = c(Na₂ C O₃) × M(Na₂ C O₃) = 0.0428 mol dm⁻³× 106 g mol⁻¹= 4.54 g dm⁻³ii. M(NaHC O₃)= 23 + 1 + 12 + (3 × 16) = 84 g mol⁻¹ρ(NaHC O₃) = c(NaHCO₃) × M(NaHCO₃) = 0.0088 mol dm⁻³× 84 g mol⁻¹= 0.739 g dm⁻³e. Total mass of mixture = m(Na₂ C O₃) + m(NaHCO₃) = 4.54 + 0.739 = 5.28 g %NaHC O₃ = m(NaHC O₃)_________________ Total mass of mixture × 100 = 0.739/5.28 × 100 = 14.0%

Activity 4.13 Using titration to determine the concentration of an analyte i solution.

Steps

a. Simple Acid-Base Titration

1. Set up the titration apparatus with the acid in the burette and the base in the flask.

2. Add an indicator, like phenolphthalein, to the conical flask.

3. Slowly add the acid from the burette to the base, swirling until the colour shows the endpoint.

4. Record your data and use it to calculate the concentration of the base.

5. Applying the results to determine:

i. Percentage purity

ii. Percentage water of crystallization

iii. Relative atomic mass A. Determining Percentage Purity A sample of impure Na₂CO₃weighing 2.50 g is dissolved in water and made up to 250 cm³. A 25 cm³ portion of this solution is titrated with 0.100 moldm⁻³ HCl. The average volume of HCl required for neutralization is 23.50 cm³.

Calculate the percentage purity of the Na₂CO₃in the sample.

Data given:

Mass of impure Na₂CO₃ = 2.50 g Total volume of solution = 250 cm³ Volume of solution used in titration = 25 cm³ Concentration of HCl = 0.100 moldm⁻³ Average volume of HCl used = 23.50 cm³ Steps

1. Write the balanced equation The reaction between sodium carbonate and hydrochloric acid is:

Na₂CO₃+ 2HCl → 2NaCl + H₂O + CO₂ From the equation, 1 mole of Na₂CO₃reacts with 2 moles of HCl.

2. Calculate the Moles of HCl Used

Given:

Concentration of HCl= 0.100 moldm⁻³Volume of HCl used = 23.50 cm³ = 0.0235 dm³ Using the formula:

n(HCl) = C_(HCl) × V_(HCl) = 0.10 × 0.0235 = 0.00235 mol

3. Calculate the Moles of Na₂CO₃ in 25 cm³ of Solution n_(Na)₂_(C)_(O)₃_ n_(HCl) = 1/2 n_(Na)₂_(C)_(O)₃ = n_(HCl)_ 2 = 0.00235 mol/2 = 0.00118 mol

4. Calculate the Moles of Na₂CO₃ in the 250 cm³ Solution Since the titration used only a 25 cm³ portion of the 250 cm³ solution, we scale up:

25 cm³of Na₂ CO₃ contain 0.00118 mol ∴ 250 cm³of Na₂ CO₃ = 250 cm³_______ 25 cm³× 0.00118 mol = 0.0118 mol

5. Calculate the Mass of Pure Na₂CO₃

Molar mass of Na₂CO₃ = 106 g/mol m(Na₂ C O₃) = n(Na₂ C O₃) × M(Na₂ C O₃) = 0.0118 × 106 = 1.25 g

6. Calculate the percentage purity The sample weighed 2.5 g, so:

percentage purity = Mass of pure Na₂ C O₃________________ mass of sample × 100 = 1.25/2.5 × 100 = 50.0% B. Determining Percentage of Water of Crystallization A hydrated sample of magnesium sulphate (MgSO₄⋅xH₂O) weighing 4.00 g is dissolved in water, and the solution is titrated with 0.100 moldm⁻³³ EDTA

solution (a reagent that binds with Mg²⁺). It requires 35.0 cm³ of EDTA to react with all the Mg²⁺ions in the sample. Calculate the percentage of water of crystallisation in the hydrated salt.

Note

EDTA binds with Mg²⁺ions in a 1:1 ratio.

Data Mass of hydrated MgSO₄⋅xH₂O = 4.00 g Concentration of EDTA = 0.100 moldm⁻³ Volume of EDTA used = 35.0 cm³ Molar mass of MgSO₄ = 120.4 gmol⁻¹ Molar mass of H₂O = 18 gmol⁻¹Steps

1. Calculate the Moles of Mg²⁺Ions

Given:

Concentration of EDTA (C_(EDTA)) = 0.100 moldm⁻³Volume of EDTA used = 35.0 cm³ = 0.035 dm³ Since EDTA binds with Mg²⁺ions in a 1:1 ratio n(Mg²⁺) = C_(EDTA) × V_(EDTA) = 0.100 × 0.035 = 0.0035 mol

2. Calculate the Mass of anhydrous MgSO₄ The molar mass of anhydrous MgSO₄is 120.4 gmol⁻¹m(MgS O₄) = n(MgS O₄) × M(MgS O₄) = 0.0035 × 120.4 = 0.421 g

3. Calculate the Mass of Water of Crystallisation

The total mass of the hydrated sample is 4.0 g, so:

Mass of water of crystallisation = Total mass of sample − mass of MgSO₄ = 4.0 − 0.421 = 3.58 g

4. Calculate the percentage of water of crystallisation Percentage of water of crystallisation = Mass of water of crystallisation_______________________ Total mass of sample × 100 = 3.58/4.0 × 100 = 89.5% C. Determining Relative Atomic Mass A 0.50 g sample of a metal carbonate, MCO₃, is dissolved in water, and the

solution is made up to 100 cm³. A 25 cm³ portion of this solution is titrated with 0.100 moldm⁻³hydrochloric acid. The average volume of HCl required for neutralisation is 24.0 cm³. Calculate the relative atomic mass of the metal M.

Data Mass of MCO₃ = 0.50 g Total volume of solution = 100 cm³ Volume of solution used in titration = 25 cm³ Concentration of HCl = 0.100 moldm⁻³ Average volume of HCl used = 24.0 cm³ Steps

1. Write the Balanced Equation

The reaction between metal carbonate (MCO₃) and hydrochloric acid HCl) is:

MC O₃ + 2HCl → M Cl₂ + H₂ O + C O₂ From the equation, 1 mole of MCO₃reacts with 2 moles of HCl.

2. Calculate the Moles of HCl Used in the Titration Given:

Concentration of HCl = 0.100 moldm⁻³Volume of HCl used = 24.0 cm³ = 0.024 dm³ Using the formula:

n(HCl) = C(HCl) × V(HCl) = 0.100 × 0.024 = 0.0024 mol

3. Calculate the Moles of MCO₃ in 25 cm³ of Solution Since 1 mole of MCO₃reacts with 2 moles of HCl, we find:

n(MC O₃)_ n(HCl) = 1/2 n(MC O₃) = n(HCl)______ 2 = 0.0024/2 = 0.0012 mol

4. Find the Total Moles of MCO₃ in the 100 cm³ solution 25 cm³of MC O₃ contains 0.0012 mol ∴ 100 cm³of MC O₃ = 100 cm³_______ 25 cm³× 0.0012 mol = 0.0048 mol

5. Calculate the Molar Mass of MCO₃

The sample of MCO₃weighs 0.50 g. Using the total moles calculated:

Molar Mass of MC O₃ = m(MC O₃)_________ n(MC O₃) = 0.50/0.0048 = 104.17

6. Calculate the Relative Atomic Mass of Metal M Molar mass of MC O₃ = Relative atomic mass of M + 12 + 48 104.17 = M + 60 M = 104.17 − 60 = 44.17 The relative atomic mass of metal M is approximately 44.17.

Activity 4.14 Indirect titration: Determining the Calcium Carbonate Content in Limestone To determine the amount of calcium carbonate (CaCO₃) in a 2.5 g sample of limestone, a student adds 50 cm³ of 0.5 moldm⁻³HCl to the powdered limestone sample. This mixture is left to react until no more gas evolved. The remaining unreacted HCl is then titrated with 0.25 moldm⁻³NaOH, requiring 20.0 cm³ of NaOH to reach the endpoint. Calculate the percentage of calcium carbonate in the limestone sample.

Steps

1. Calculate the Initial Moles of HCl Added

Given:

Volume of HCl = 50 cm³ = 0.05 dm³ Concentration of HCl = 0.5 moldm⁻³Using the formula:

Moles of HCl = Concentration × Volume Moles of HCl = 0.5 × 0.05 = 0.025 mol This is the initial amount of HCl added to the limestone sample.

2. Calculate the Moles of NaOH Used in the Back Titration Given:

Volume of NaOH used = 20.0 cm³ = 0.02 dm³ Concentration of NaOH = 0.25 moldm⁻³Moles of NaOH = concentration × Volume Moles of NaOH = 0.25 × 0.020 = 0.0050 mol Since HCl and NaOH react in a 1:1 molar ratio, the moles of unreacted HCl in the solution is also 0.005 mol

3. Calculate the moles of HCl that reacted with CaCO₃ Subtract the unreacted HCl from the initial HCl to find the moles of HCl that with CaCO₃:

Moles of HCl that reacted = 0.025−0.005 = 0.020 mol

4. Calculate the Moles of CaCO₃in the Sample

The reaction between CaCO₃and HCl is:

CaCO₃+ 2HCl → CaCl₂+ H₂O + CO₂ From the balanced equation, 1 mole of CaCO₃reacts with 2 moles of HCl.

Therefore, the moles of CaCO₃that reacted is:

Moles of CaCO₃ = Moles of HCl that reacted/2 = 0.020/2 = 0.010 mol

5. Calculate the Mass of CaCO₃in the Sample

Using the molar mass of CaCO₃ = 100 gmol⁻¹Mass of CaCO₃ = Moles of CaCO₃× Molar Mass of CaCO₃ Mass of CaCO₃ = 0.01×100 = 1.0 g

6. Calculate the Percentage of CaCO₃in the limestone sample Percentage of CaC O₃ = Mass of CaC O₃____________ Mass of sample × 100 = 1.0/2.5 × 100 = 40%

Activity 4.15 Titration with pH Monitoring Materials Needed pH meter or pH sensor with data logger, burette, pipette, conical flask, beakers, Strong acid (e.g., HCl), strong base (e.g., NaOH), weak acid (e.g., CH₃COOH

- acetic acid), weak base (e.g., NH₃- ammonia), distilled water, white tile, graph paper or computer software for plotting volume vs. pH graphs Steps

1. Preparation and Setup

a. Set up the burette on a retort stand.

b. Calibrate the pH meter.

c. Rinse and fill the burette with the titrant (acid or base) specific to each titration type.

d. Pipette a measured volume (e.g., 25 cm³) of the analyte into the conical flask.

e. Place the conical flask on a white tile under the burette.

2. For each titration, follow these general steps but with specific combinations for type of acid and base.

a. Strong Acid vs. Strong Base (e.g., HCl and NaOH)

i. Place the pH sensor in the conical flask with the analyte (NaOH).

ii. Record the initial pH of the solution before adding any HCl.

iii. Add the HCl titrant from the burette in small increments (e.g., 1.0 cm³).

iv. After each addition, swirl the flask gently and record the pH.

v. As the pH approaches neutral (around pH 7), reduce the increments to 0.20 cm³ for more precise measurements.

vi. Continue until the pH stabilizes after each addition.

vii. Plot the volume of HCl added on the x-axis and pH on the y-axis to determine sharp jump in pH around the equivalence point (expected to be near pH 7 strong acid vs. strong base).

b. Strong Acid vs. Weak Base (e.g., HCl and NH₃)

i. Use NH₃as the analyte and HCl as the titrant.

ii. Follow the same process, adding HCl in small increments and recording the pH.

iii. This titration will show a more gradual increase in pH until equivalence, which expected to be below pH 7 due to the acidic nature of the solution neutralization.

iv. The volume vs. pH graph should show a less sharp change at the point compared to strong acid vs. strong base.

c. Strong Base vs. Weak Acid (e.g., NaOH and CH₃COOH)

i. Use CH₃COOH as the analyte and NaOH as the titrant.

ii. Add NaOH in small increments, recording the pH after each addition.

iii. The equivalence point will be slightly above pH 7, as the acetate ion (CH₃COO−) produces a basic solution.

iv. The volume vs. pH graph will have a less steep curve around the point.

d. Weak Base vs. Weak Acid (e.g., NH₃and CH₃COOH)

i. Use NH₃as the analyte and CH₃COOH as the titrant.

ii. Follow the same process, adding CH₃COOH in small increments and recording the pH.

iii. The equivalence point will be near neutral, but the pH change will be very gradual, with no sharp increase.

iv. The volume vs. pH graph will have a gentle slope, making it harder to identify a distinct equivalence point.

3. For each titration, plot the volume of titrant added (x-axis) vs. pH (y-axis).

a. Identify the Equivalence Point: Look for the point on each graph where the pH changes most sharply.

b. Compare Graphs: Analyse how the steepness of the curve at the equivalence point differs for each type of titration:

i. Strong acid vs. strong base: Very sharp pH change around equivalence.

ii. Strong acid vs. weak base: Moderate pH change with an equivalence below pH 7.

iii. Strong base vs. weak acid: Moderate pH change with an equivalence above pH 7.

iv. Weak acid vs. weak base: Very gradual pH change with no sharp point.

4. Discussion and Conclusion

a. Discuss the pH at the equivalence points for each titration type and why they differ.

b. Discuss how the shape of each graph relates to the strength of the acid and base in the reaction.

Review Question 4.1

1. What is the Arrhenius definition of an acid and a base?

2. What is a conjugate acid-base pair in the Brønsted-Lowry theory?

3. What makes a substance a Lewis acid or a Lewis base?

4. How do the Arrhenius, Brønsted-Lowry, and Lewis definitions of acids and differ?

5. Why can the Brønsted-Lowry concept be applied in both water and non- environments, unlike the Arrhenius concept?

6. Give an example of a substance that fits each acid definition (Arrhenius, Brønsted- Lowry, and Lewis).

7. Classify each of the following species in aqueous solution as a Brønsted- Lowry acid base:

a. HCl

b. NO₂

c. CH₃COOH

8. Identify the Brønsted-Lowry acid and its conjugate base in the following reactions

a. H₂SO₄+ HNO₃→ H₂NO₃+ HSO₄ −

b. HCl + H₂O → H₃O + + Cl –

c. NH₃+ H₂S → HS− + NH₄ +

9. How would you determine if a given substance is an Arrhenius acid, a Brønsted-L acid, or a Lewis acid? What steps would you take?

10. Explain why the Lewis theory is considered the broadest definition of acids and bases. Can it describe all the situations that the Arrhenius and Brønsted-Lowry definitions cover?

11. Design an experiment that would allow you to identify whether a substance behaves an Arrhenius acid, a Brønsted-Lowry acid, or a Lewis acid.

12. Analyse the limitations of each acid-base theory and discuss scenarios where theory might be more useful than the others.

Review Question 4.2

1. What is a common physical property of acids?

2. What colour does litmus paper turn in the presence of a base?

3. Name a physical property of bases.

4. What is the pH range for acids and bases?

5. Explain why acids are able to conduct electricity.

6. Describe the chemical reaction that occurs when an acid is neutralized by a base.

7. Compare the physical properties of acids and bases based on taste, feel, and on indicators.

8. How can you use pH indicators to differentiate between an acid and a base?

9. A pH meter was used to check the pH level of four solutions labelled as W, X, Y and Z described in the table below W X Y Z pH 7 3 14 1

a. Which of the solution(s) will taste sour?

b. Identify the solution that will have the same pH as a dilute Na₂CO₃

c. Which solution is the most acidic?

d. Which solution will have no effect on litmus paper?

10. How does the chemical property of neutralisation help in treating an acidic stomach?

11. Analyse how the property of acids and bases as electrolytes is applied in technology.

12. Discuss why alkaline (basic) solutions are often used for cleaning purposes.

13. Evaluate the environmental impact of acid rain on soil and water bodies propose a method to counteract these effects using bases.

14. Propose a solution for farmers facing soil acidity issues by applying the neutralisation and evaluating the pros and cons of different bases that could used.

Review Question 4.3

1. What is a strong acid?

2. What is the pH range for a strong acid compared to a weak acid?

3. How do strong and weak acids differ in terms of electrical conductivity?

4. What does the term “extent of dissociation” mean in acids and bases?

5. Why does a strong acid conduct electricity better per mole than a weak acid?

6. Compare the rate of reaction of strong acids and weak acids with metals.

7. Describe the difference in enthalpy change during neutralisation between a strong and a weak acid.

8. How would you explain the differences in electrical conductivity between solutions hydrochloric acid (strong acid) and acetic acid (weak acid) at the same concentration?

9. Why does the enthalpy change of neutralisation differ between a strong acid like HCl and a weak acid like CH₃COOH?

10. Explain why strong bases like sodium hydroxide have a higher pH than weak bases ammonia, even at the same concentration.

11. Design an experiment to compare the electrical conductivity of strong and weak and bases. What materials would you need, and what results would you expect?

12. Evaluate the enthalpy changes of neutralisation for reactions involving strong weak acids and propose an explanation for any differences in the values.

13. Predict the pH, conductivity, and rate of reaction with magnesium of a 1M

solution HCl compared to a 1M solution of CH₃COOH, and explain the reasoning.

Review Question 4.4

1. What is a salt in chemistry?

2. What are the two main components that form a salt?

3. Give an example of a common salt.

4. How is a salt formed during a reaction between an acid and a base?

5. What is the difference between a normal salt and an acidic salt?

6. How can we classify salts based on their properties?

7. What makes a double salt different from a complex salt?

8. Explain why a basic salt might form when a weak acid reacts with a strong base.

9. How would you identify an efflorescent salt, and what conditions affect it?

10. Compare and contrast deliquescent and hygroscopic salts in terms of their behaviour with moisture in the air.

11. Why are hydrated salts important in certain industries?

12. Design an experiment to distinguish between an acidic salt and a basic salt. Explain the steps and expected results.

13. Analyse the environmental impact of efflorescent and deliquescent salts in industrial applications.

14. Predict the effects of using a double salt instead of a normal salt in a specific application, like water treatment or dyeing.

Review Questions 4.5

1. Describe the process of an acid-base titration.

2. Define “water of crystallisation” and give an example.

3. How would you calculate the percentage purity of a sample of impure NaOH if 0.50 g of the sample requires 25 cm³of 0.10 M HCl for complete neutralisation?

4. A hydrated salt has a mass of 4.0 g before heating and 2.8 g after heating.

C the percentage of water of crystallisation.

5. Explain why phenolphthalein is often used as an indicator in acid-base involving strong bases.

6. Assess potential sources of error in determining percentage purity through titration and suggest methods to minimise them.