Chapter 8 — Ionic Equilibrium Exercise Solutions
This chapter is formula-heavy. Write the equilibrium expression carefully before substitution, and check whether the acid/base is strong or weak.
Key Concepts Quick Recap
| Term | Definition |
|---|
| Strong acid/base | Completely ionised in water (HCl, H₂SO₄, NaOH) |
| Weak acid/base | Partially ionised; Ka or Kb << 1 |
| pH | −log[H⁺]; pH + pOH = 14 at 25 °C |
| Ka | Acid dissociation constant; higher Ka → stronger acid |
| pKa | −log Ka; lower pKa → stronger acid |
| Buffer solution | Resists change in pH on adding small amounts of acid/base |
| Acid buffer | Weak acid + its conjugate base (salt) |
| Basic buffer | Weak base + its conjugate acid (salt) |
| Solubility product (Ksp) | Product of ionic concentrations at saturation; if Q > Ksp → precipitation |
| Common ion effect | Addition of an ion common to the equilibrium suppresses ionisation |
| Hydrolysis | Salt of weak acid/strong base gives basic solution; vice versa acidic |
| Isoelectric point | pH at which amino acid carries no net charge |
Formula Recap
| Concept | Formula |
|---|
| pH of strong acid | pH = −log[H⁺] = −log[acid] |
| pH of strong base | pOH = −log[OH⁻]; pH = 14 − pOH |
| pH of weak acid | [H⁺] = √(Ka × C); pH = ½(pKa − log C) |
| pH of weak base | [OH⁻] = √(Kb × C); pOH = ½(pKb − log C) |
| Acid buffer | pH = pKa + log([A⁻]/[HA]) |
| Basic buffer | pOH = pKb + log([BH⁺]/[B]) |
| Ksp (AB type) | Ksp = [A⁺][B⁻] = s² |
| Ksp (AB₂ type) | Ksp = [A²⁺][B⁻]² = 4s³ |
I. Choose the Best Answer (MCQs)
1. The pH of 0.01 M HCl solution is: (a) 1 (b) 2 (c) 7 (d) 12
Answer(b) 2 HCl is a strong acid: [H⁺] = 0.01 = 10⁻² M. pH = −log(10⁻²) = 2.
2. A buffer solution is best prepared from: (a) Strong acid + strong base (b) Weak acid + its salt with strong base (c) Strong acid + its salt (d) Strong base + water
Answer(b) Weak acid + its salt with strong base Example: CH₃COOH + CH₃COONa. The buffer resists pH change — weak acid neutralises added base, salt anion neutralises added acid.
3. The solubility of BaSO₄ (Ksp = 1×10⁻¹⁰) in pure water is: (a) 1×10⁻⁵ mol/L (b) 1×10⁻¹⁰ mol/L (c) 1×10⁻⁴ mol/L (d) 1×10⁻²⁰ mol/L
Answer(a) 1×10⁻⁵ mol/L BaSO₄ ⇌ Ba²⁺ + SO₄²⁻. Ksp = s². s = √(1×10⁻¹⁰) = 1×10⁻⁵ mol/L.
4. Adding NaCl to a saturated solution of AgCl will: (a) Increase solubility of AgCl (b) Decrease solubility of AgCl (c) Not affect solubility (d) Increase pH
Answer(b) Decrease solubility of AgCl Common ion effect: adding Cl⁻ (from NaCl) shifts AgCl ⇌ Ag⁺ + Cl⁻ equilibrium to the left → AgCl precipitates more → solubility decreases.
5. The pH of a solution of NaCH₃COO (sodium acetate) is: (a) 7 (b) < 7 (acidic) (c) > 7 (basic) (d) 0
Answer(c) > 7 (basic) NaCH₃COO is a salt of weak acid (CH₃COOH) and strong base (NaOH). The acetate ion hydrolyses: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ → solution is basic.
6. The degree of ionisation of a weak acid increases when: (a) Temperature decreases (b) Concentration increases (c) Concentration decreases (more dilute) (d) A common ion is added
Answer(c) Concentration decreases Ostwald's dilution law: α = √(Ka/C). As C decreases, α increases. A weak acid is more completely ionised in dilute solution.
7. Which of the following pairs can act as a buffer? (a) NaCl + HCl (b) NH₃ + NH₄Cl (c) NaOH + NaCl (d) HCl + NaOH
Answer(b) NH₃ + NH₄Cl NH₃ (weak base) + NH₄Cl (its salt) forms a basic buffer. pH controlled near pKa of NH₄⁺ (≈ 9.25).
II. Short Answer Questions
Q1. Explain the Henderson-Hasselbalch equation with an example.
Answer
The Henderson-Hasselbalch equation gives the pH of a buffer without solving a quadratic:
For acid buffer: pH = pKa + log([salt]/[acid]) = pKa + log([A⁻]/[HA]) For basic buffer: pOH = pKb + log([salt]/[base]) = pKb + log([BH⁺]/[B])
Example: An acetate buffer prepared with 0.1 M CH₃COOH and 0.1 M CH₃COONa. (pKa = 4.74)
pH = 4.74 + log(0.1/0.1) = 4.74 + 0 = 4.74
When [salt] = [acid], pH = pKa → this is the buffer point (maximum buffer capacity).
Q2. What is the common ion effect? Illustrate with an example.
Answer
The common ion effect is the suppression of ionisation of a weak electrolyte by adding a strong electrolyte that has an ion in common with the weak electrolyte.
Example: Adding HCl to a solution of CH₃COOH:
- CH₃COOH ⇌ CH₃COO⁻ + H⁺
- HCl → H⁺ + Cl⁻ (completely)
- The added H⁺ from HCl shifts the equilibrium to the left → less ionisation of CH₃COOH → α decreases → pH decreases.
Application: Used to control solubility in qualitative analysis.
Q3. Explain how Ksp is used to predict precipitation.
Answer
Ionic product (Q) = product of actual ionic concentrations (raised to stoichiometric powers).
- If Q < Ksp: solution is unsaturated → no precipitation.
- If Q = Ksp: solution is just saturated → equilibrium.
- If Q > Ksp: solution is supersaturated → precipitation occurs until Q = Ksp.
Example: AgCl has Ksp = 1.8×10⁻¹⁰. Mix 0.001 M AgNO₃ with 0.001 M NaCl. Q = [Ag⁺][Cl⁻] = 10⁻³ × 10⁻³ = 10⁻⁶. Since 10⁻⁶ > 1.8×10⁻¹⁰ → AgCl precipitates.
Q4. Distinguish between pH and pOH. Write their relationship.
Answer
- pH = −log[H⁺]: measures acidity. pH < 7 (acidic), pH = 7 (neutral), pH > 7 (basic) at 25 °C.
- pOH = −log[OH⁻]: measures basicity. pOH < 7 (basic), pOH > 7 (acidic).
- Relationship: pH + pOH = 14 (at 25 °C, since Kw = [H⁺][OH⁻] = 10⁻¹⁴).
Worked Examples
Example 1 — pH of a strong acid and base
Strong acid: 0.01 M HCl → [H⁺] = 0.01 M = 10⁻² M → pH = 2
Strong acid: 0.001 M H₂SO₄ → [H⁺] = 2 × 0.001 = 0.002 M → pH = −log(0.002) = 2.70
Strong base: 0.001 M NaOH → [OH⁻] = 10⁻³ → pOH = 3 → pH = 14 − 3 = 11
Example 2 — pH of weak acid
Find the pH of 0.1 M acetic acid (Ka = 1.8×10⁻⁵).
CH₃COOH ⇌ CH₃COO⁻ + H⁺
[H⁺] = √(Ka × C) = √(1.8×10⁻⁵ × 0.1) = √(1.8×10⁻⁶) = 1.342×10⁻³ M
pH = −log(1.342×10⁻³) = 3 − log(1.342) = 3 − 0.128 = 2.87
Example 3 — Buffer pH
Find pH of buffer: 0.2 M CH₃COOH + 0.3 M CH₃COONa. (pKa = 4.74)
pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
= 4.74 + log(0.3/0.2)
= 4.74 + log(1.5)
= 4.74 + 0.176
= 4.92
Example 4 — Solubility from Ksp
Find the solubility of Ag₂CrO₄ in pure water. (Ksp = 1.1×10⁻¹²)
Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻
If solubility = s: [Ag⁺] = 2s, [CrO₄²⁻] = s
Ksp = [Ag⁺]²[CrO₄²⁻] = (2s)²(s) = 4s³
4s³ = 1.1×10⁻¹²
s³ = 2.75×10⁻¹³
s = (2.75×10⁻¹³)^(1/3) = 6.5×10⁻⁵ mol/L
Example 5 — Solubility in presence of common ion
Find solubility of AgCl in 0.1 M NaCl solution. (Ksp = 1.8×10⁻¹⁰)
AgCl ⇌ Ag⁺ + Cl⁻
[Cl⁻] from NaCl ≈ 0.1 M (large excess; assume s << 0.1)
Ksp = [Ag⁺][Cl⁻] = s × (0.1 + s) ≈ s × 0.1
s = Ksp / 0.1 = 1.8×10⁻¹⁰ / 0.1 = 1.8×10⁻⁹ mol/L
Compare with solubility in pure water: √(1.8×10⁻¹⁰) = 1.34×10⁻⁵ mol/L
Solubility decreases ≈ 10,000 times due to common ion effect.
Common Mistakes
- Confusing pH and pOH and their relationship to acidic/basic nature.
- Using Henderson-Hasselbalch with [acid] and [salt] reversed.
- Forgetting stoichiometric powers in Ksp expression (e.g. Ag₂CrO₄ gives [Ag⁺]²[CrO₄²⁻]).
- Not checking whether acid/base is strong or weak before choosing the formula.
- Applying √(Ka × C) for strong acids — it applies only to weak acids.
Exam Checklist
- Strong vs weak identification is done first.
- Ionisation equation is correctly written.
- Ksp expression has correct stoichiometric powers.
- Henderson-Hasselbalch ratio is [salt]/[acid] in correct order.
- pH range of answer is chemically sensible (0–14).
- Final pH has correct units (dimensionless).
