CBSE Class 11 Chemistry Equilibrium Worksheet Set 07

Ionic Equilibrium Worksheet

Question 1. Define Brønsted–Lowry acids and bases. Write the conjugate base of HCO₃⁻ and the conjugate acid of NH₃, and name a species that is both an acid and a base.
Answer:
A Brønsted acid donates a proton (H⁺) and a Brønsted base accepts a proton.
Conjugate base of HCO₃⁻ = CO₃²⁻ (remove H⁺).
Conjugate acid of NH₃ = NH₄⁺ (add H⁺).
H₂O (also HCO₃⁻ and H₂PO₄⁻) can act as an acid and as a base, so it is amphoteric.
In simple words: An acid gives a proton; the leftover is its conjugate base.

Teacher's Note:
1. A conjugate pair differs by exactly one H⁺.
2. A strong acid has a very weak conjugate base.

Question 2. Classify the following species as Lewis acids or Lewis bases and explain: NH₃, BF₃, SnCl₄, Ni²⁺, CO.
Answer:
Lewis acids accept an electron pair: BF₃ (vacant p orbital on B), SnCl₄ (Sn can accept more electron pairs) and Ni²⁺ (vacant orbitals).
Lewis bases donate an electron pair: NH₃ (lone pair on N) and CO (lone pair on C).
Example of a Lewis acid–base reaction: BF₃ + :NH₃ → F₃B←NH₃.
In simple words: Acid: takes a pair of electrons. Base: gives a pair of electrons.

Teacher's Note:
1. Look for a lone pair (base) or a vacant orbital (acid).
2. All Brønsted bases are Lewis bases, but not every Lewis acid is a Brønsted acid.

Question 3. Explain why the pH of 0.1 M acetic acid (K_a = 1.8 × 10⁻⁵) is higher than the pH of 0.1 M HCl, and calculate both values.
Answer:
HCl is a strong acid and is ionised completely: [H⁺] = 0.1 M, so pH = 1.
Acetic acid is a weak acid and is ionised only slightly: [H⁺] = √(K_ac) = √(1.8 × 10⁻⁵ × 0.1) = 1.34 × 10⁻³ M, so pH = 2.87.
A smaller [H⁺] means a higher pH.
In simple words: The weak acid releases far fewer H⁺ ions, so its solution has a higher pH.

Teacher's Note:
1. Use [H⁺] = √(K_ac) for a weak acid.
2. pH = −log[H⁺].

Question 4. Calculate the pH of (a) 0.0025 M Ba(OH)₂ solution and (b) the solution obtained by diluting 100 mL of 0.1 M HCl to 1 litre.
Answer:
(a) Ba(OH)₂ gives 2OH⁻: [OH⁻] = 2 × 0.0025 = 5.0 × 10⁻³ M. pOH = 2.30, so pH = 14 − 2.30 = 11.70.
(b) Moles of HCl = 0.1 × 0.1 = 0.01 mol in 1 L, so [H⁺] = 0.01 M and pH = 2.
In simple words: For a base find pOH first and subtract from 14; for a strong acid [H⁺] equals its molarity.

Teacher's Note:
1. Check whether the base gives one or two OH⁻.
2. A tenfold dilution raises the pH by one.

Question 5. The ionic product of water at 373 K is 5.5 × 10⁻¹³. Calculate the pH of pure water at this temperature and state whether it is acidic, basic or neutral.
Answer:
[H⁺] = [OH⁻] = √K_w = √(5.5 × 10⁻¹³) = 7.42 × 10⁻⁷ M.
pH = −log(7.42 × 10⁻⁷) = 6.13.
Since [H⁺] = [OH⁻], the water is still neutral. Neutral does not always mean pH 7.
In simple words: At a higher temperature K_w is larger, so neutral water has a lower pH than 7.

Teacher's Note:
1. Neutral means [H⁺] = [OH⁻].
2. pK_w is about 12.26 at this temperature.

Question 6. Calculate the pH and the degree of ionisation of 0.2 M hydrofluoric acid. (K_a = 3.2 × 10⁻⁴)
Answer:
[H⁺] = √(K_ac) = √(3.2 × 10⁻⁴ × 0.2) = √(6.4 × 10⁻⁵) = 8.0 × 10⁻³ M.
pH = −log(8.0 × 10⁻³) = 2.10.
α = [H⁺] ÷ c = 8.0 × 10⁻³ ÷ 0.2 = 0.04 (4%). This is below 5%, so the approximation 1 − α ≈ 1 is acceptable.
In simple words: Only 4 out of every 100 HF molecules ionise, which is what 'weak' means.

Teacher's Note:
1. Check that α < 0.05 after using the approximation.
2. α = √(K_a ÷ c) gives the same value.

Question 7. Calculate the pH of 0.10 M ammonia solution. (K_b = 1.8 × 10⁻⁵)
Answer:
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
[OH⁻] = √(K_bc) = √(1.8 × 10⁻⁵ × 0.10) = 1.34 × 10⁻³ M.
pOH = 2.87, so pH = 14 − 2.87 = 11.13.
In simple words: A weak base gives only a little OH⁻, so the pH is well below that of a strong base of the same concentration (13).

Teacher's Note:
1. Use K_b for a base and find pOH first.
2. α = √(K_b ÷ c) = 1.3%.

Question 8. A 0.01 M solution of a weak monobasic acid has a pH of 4. Calculate its K_a and its degree of ionisation.
Answer:
[H⁺] = 10⁻⁴ M and [A⁻] = [H⁺] = 10⁻⁴ M.
K_a = [H⁺][A⁻] ÷ [HA] = (10⁻⁴)² ÷ 0.01 = 1.0 × 10⁻⁶.
α = 10⁻⁴ ÷ 0.01 = 0.01 (1%).
In simple words: Square the H⁺ concentration and divide by the concentration of the acid.

Teacher's Note:
1. The acid left unionised is almost the whole 0.01 M.
2. pK_a = 6.

Question 9. Calculate the approximate pH of 0.05 M H₂S. (K_a1 = 1.0 × 10⁻⁷, K_a2 = 1.3 × 10⁻¹³) Why can the second ionisation be ignored?
Answer:
K_a2 is about a million times smaller than K_a1, so almost all the H⁺ comes from H₂S ⇌ H⁺ + HS⁻.
[H⁺] = √(K_a1c) = √(1.0 × 10⁻⁷ × 0.05) = √(5.0 × 10⁻⁹) = 7.07 × 10⁻⁵ M.
pH = 4.15.
In simple words: The second proton is much harder to remove, so it adds almost no H⁺.

Teacher's Note:
1. K_a1 > K_a2 for every polyprotic acid.
2. [S²⁻] ≈ K_a2.

Question 10. Calculate the pH of 0.10 M acetic acid (K_a = 1.8 × 10⁻⁵) before and after dissolving 0.10 mol of sodium acetate in 1 L of the solution. Name the effect observed.
Answer:
Before: [H⁺] = √(1.8 × 10⁻⁵ × 0.10) = 1.34 × 10⁻³ M, so pH = 2.87.
After: [acid] = [salt] = 0.10 M, so pH = pK_a + log(0.10 ÷ 0.10) = pK_a = −log(1.8 × 10⁻⁵) = 4.74.
The pH rises because the acetate ion suppresses the ionisation of acetic acid. This is the common ion effect.
In simple words: Adding a salt with a common ion pushes the weak-acid equilibrium back and lowers [H⁺].

Teacher's Note:
1. Once the salt is added, use the Henderson equation.
2. The solution is now an acidic buffer.

Question 11. How would you prepare 1 L of an acetate buffer of pH 5.0 from 0.10 M acetic acid (K_a = 1.8 × 10⁻⁵)? Find the concentration of sodium acetate needed.
Answer:
pK_a = −log(1.8 × 10⁻⁵) = 4.74.
pH = pK_a + log([salt] ÷ [acid]): 5.0 = 4.74 + log([salt] ÷ 0.10), so log([salt] ÷ 0.10) = 0.26.
[salt] ÷ 0.10 = 1.8, so [CH₃COONa] = 0.18 mol L⁻¹.
Dissolve about 0.18 mol (14.8 g, molar mass 82 g mol⁻¹) of sodium acetate in 1 L of 0.10 M acetic acid.
In simple words: Add about 1.8 times as much acetate as acid to raise the pH 0.26 units above the pK_a.

Teacher's Note:
1. pH = pK_a + log(salt ÷ acid) is the Henderson equation.
2. Choose an acid whose pK_a is close to the target pH.

Question 12. Explain, with equations, how an acetic acid–sodium acetate buffer resists a change of pH when a small amount of (i) HCl and (ii) NaOH is added.
Answer:
The buffer contains CH₃COOH (weak acid) and CH₃COO⁻ (from the salt).
(i) Added H⁺: it combines with the acetate ion, CH₃COO⁻ + H⁺ → CH₃COOH. The strong acid is replaced by a weak acid, so the pH changes very little.
(ii) Added OH⁻: it reacts with the acid, CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. The strong base is replaced by a weak base, so the pH changes very little.
The ratio [salt] ÷ [acid] changes only slightly, so by the Henderson equation the pH hardly changes.
In simple words: One component of the buffer mops up added acid and the other mops up added base.

Teacher's Note:
1. A buffer works best when [salt] = [acid].
2. It loses its action if one component is used up.

Question 13. Classify the salts NH₄Cl, Na₂CO₃, KNO₃ and CH₃COONH₄ as giving acidic, basic or neutral solutions in water. Explain with equations.
Answer:
NH₄Cl (strong acid + weak base): NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. The solution is acidic.
Na₂CO₃ (weak acid + strong base): CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻. The solution is basic.
KNO₃ (strong acid + strong base): neither ion hydrolyses, so the solution is neutral.
CH₃COONH₄ (weak acid + weak base): both ions hydrolyse and K_a of CH₃COOH is about equal to K_b of NH₃, so the solution is nearly neutral.
In simple words: The stronger parent of the salt decides whether the solution is acidic or basic.

Teacher's Note:
1. Ions of strong acids and strong bases do not hydrolyse.
2. Hydrolysis changes the pH only when a weak parent is involved.

Question 14. The solubility of Ag₂CrO₄ in water is 6.5 × 10⁻⁵ mol L⁻¹ at 298 K. Calculate its solubility product.
Answer:
Ag₂CrO₄(s) ⇌ 2Ag⁺ + CrO₄²⁻. If the solubility is S, [Ag⁺] = 2S and [CrO₄²⁻] = S.
K_sp = (2S)² × S = 4S³ = 4 × (6.5 × 10⁻⁵)³ = 4 × 2.75 × 10⁻¹³ = 1.1 × 10⁻¹².
In simple words: Write the ion concentrations in terms of S, then multiply with the powers from the formula.

Teacher's Note:
1. For A₂B-type salts K_sp = 4S³.
2. K_sp is dimensionless in most school problems, but its unit is mol³ L⁻³ here.

Question 15. The K_sp of BaSO₄ is 1.1 × 10⁻¹⁰. Compare its solubility in pure water with its solubility in 0.1 M Na₂SO₄ and name the effect involved.
Answer:
In pure water S² = 1.1 × 10⁻¹⁰, so S = 1.05 × 10⁻⁵ mol L⁻¹.
In 0.1 M Na₂SO₄, [SO₄²⁻] ≈ 0.1 M, so [Ba²⁺] = K_sp ÷ 0.1 = 1.1 × 10⁻¹⁰ ÷ 0.1 = 1.1 × 10⁻⁹ mol L⁻¹.
The solubility drops about ten-thousand-fold. This is the common ion effect (SO₄²⁻ is common to both).
In simple words: Extra sulphate ions push BaSO₄ out of solution, so less of it dissolves.

Teacher's Note:
1. Neglect the sulphate from BaSO₄ itself, which is tiny.
2. K_sp stays the same; only the solubility changes.