CBSE Class 11 Chemistry Equilibrium Notes Set 08

These Class 11 CBSE Chemistry notes for Chapter 6 Equilibrium (ionic equilibrium) cover strong and weak electrolytes, the Arrhenius, Brønsted–Lowry and Lewis ideas of acids and bases, the ionisation of water, the pH scale, ionisation constants and Ostwald's dilution law, the common ion effect, buffer solutions with the Henderson equation, an elementary treatment of salt hydrolysis, and solubility product. Twelve solved examples show the numerical working for pH, K_a, buffers and K_sp.

Important Notes and Formulas

Electrolytes and Ionic Equilibrium

Substances whose aqueous solutions or melts conduct electricity because of ions are electrolytes. A strong electrolyte ionises almost completely in water (HCl, HNO₃, NaOH, NaCl). A weak electrolyte ionises only partly, so the solution contains ions and unionised molecules in equilibrium (CH₃COOH, HCN, NH₃, H₂CO₃). The equilibrium between unionised molecules and ions in solution is called ionic equilibrium.

Acids and Bases

Conjugate acid–base pairs. When an acid loses a proton, the species left is its conjugate base. When a base gains a proton, the species formed is its conjugate acid. The two differ by one H⁺. The stronger the acid, the weaker its conjugate base.

Species that can both donate and accept a proton are amphoteric (more precisely amphiprotic): H₂O, HCO₃⁻, HSO₄⁻, H₂PO₄⁻ and NH₃. PO₄³⁻ is not amphiprotic because it can only accept a proton.

Strength of an acid depends on the ease of losing H⁺. Across a period it increases with the electronegativity of the atom bonded to H (CH₄ < NH₃ < H₂O < HF). Down a group it increases as the H–X bond becomes weaker (HF < HCl < HBr < HI). Among oxoacids of one element the strength rises with the oxidation state of the central atom (HClO < HClO₂ < HClO₃ < HClO₄).

Ionisation of Water and the pH Scale

Water ionises very slightly: H₂O + H₂O ⇌ H₃O⁺ + OH⁻. The ionic product of water is

\( K_w = [H_3O^+][OH^-] = 1.0\times10^{-14}\ \text{at 298 K} \)

In pure water [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol L⁻¹. K_w increases with temperature, because ionisation is endothermic. At a higher temperature pure water still has [H⁺] = [OH⁻] and is neutral, but its pH is below 7.

\( pH = -\log_{10}[H^+]\qquad pOH = -\log_{10}[OH^-]\qquad pH + pOH = pK_w = 14\ \text{(at 298 K)} \)

1. For a strong monoprotic acid of concentration c, [H⁺] = c, and for a strong base of the type MOH, [OH⁻] = c. For Ca(OH)₂ or Ba(OH)₂, [OH⁻] = 2c.
2. A change of one pH unit means a tenfold change in [H⁺].
3. Diluting a strong acid 10 times raises pH by one unit, but a solution of an acid can never be made basic by dilution. For a very dilute acid (about 10⁻⁸ M) the H⁺ from water must also be counted, so the pH is slightly below 7 and not 8.
4. For a diprotic strong acid such as H₂SO₄, [H⁺] is taken as twice the concentration in simple problems.

Ionisation Constants of Weak Acids and Bases

For a weak acid HA ⇌ H⁺ + A⁻ of initial concentration c and degree of ionisation α, the equilibrium concentrations are c(1 − α), cα and cα.

\( K_a = \frac{[H^+][A^-]}{[HA]} = \frac{c\alpha^2}{1-\alpha} \)

Ostwald's dilution law. If α is very small (less than about 5%), 1 − α ≈ 1, and

\( K_a = c\alpha^2,\qquad \alpha = \sqrt{\frac{K_a}{c}},\qquad [H^+] = c\alpha = \sqrt{K_a\,c} \)

On dilution α increases (K_a is constant), but [H⁺] decreases. For a weak base BOH ⇌ B⁺ + OH⁻ the same relations hold with K_b: [OH⁻] = cα = √(K_bc). A larger K_a (or a smaller pK_a = −log K_a) means a stronger acid.

Conjugate pair relation: for an acid HA and its conjugate base A⁻, K_a × K_b = K_w, or pK_a + pK_b = pK_w = 14. A strong acid has a very weak conjugate base.

Polyprotic Acids

Acids that can give more than one proton, such as H₂CO₃, H₂S, H₂C₂O₄ and H₃PO₄, ionise in steps. Each step has its own constant and always K_a1 > K_a2 > K_a3, because it is harder to remove H⁺ from a negatively charged ion. For H₃PO₄, K_a1 = 7.5 × 10⁻³, K_a2 = 6.2 × 10⁻⁸ and K_a3 = 4.2 × 10⁻¹³. For a weak diprotic acid with K_a1 much larger than K_a2, almost all H⁺ comes from the first step, so [H⁺] = √(K_a1c) and [A²⁻] ≈ K_a2.

Common Ion Effect

The ionisation of a weak electrolyte is suppressed when a strong electrolyte with an ion in common is added. Adding CH₃COONa to CH₃COOH shifts CH₃COOH ⇌ CH₃COO⁻ + H⁺ to the left, so [H⁺] falls and pH rises. Adding NH₄Cl to NH₃ solution lowers [OH⁻] in the same way. The same effect lowers the solubility of a sparingly soluble salt (AgCl is less soluble in NaCl solution than in water).

Hydrolysis of Salts (Elementary Idea)

Hydrolysis is the reaction of the ions of a salt with water that changes the pH of the solution. The nature of the parent acid and base decides the result.

The hydrolysis constant of the anion of a weak acid is K_h = K_w ÷ K_a, and that of the cation of a weak base is K_h = K_w ÷ K_b.

Buffer Solutions

A buffer solution resists a change of pH when small amounts of acid or base are added or when it is diluted. An acidic buffer is a weak acid with its salt of a strong base (CH₃COOH + CH₃COONa). A basic buffer is a weak base with its salt of a strong acid (NH₃ + NH₄Cl). The acid or base reacts with one component, so the ratio [salt] ÷ [acid] changes only slightly.

\( pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}\qquad pOH = pK_b + \log\frac{[\text{salt}]}{[\text{base}]} \)

This is the Henderson–Hasselbalch equation. The buffer is most effective when [salt] = [acid], when pH = pK_a, and it works over the range pH = pK_a ± 1. The pH of a buffer does not change on moderate dilution, because the ratio stays the same. The pH of blood (7.4) is kept by the H₂CO₃ / HCO₃⁻ buffer.

Solubility Product

In a saturated solution of a sparingly soluble salt, the solid is in equilibrium with its ions: A_xB_y(s) ⇌ xA^y+(aq) + yB^x−(aq). The product of the ion concentrations, each raised to the power of its coefficient, is the solubility product K_sp (the solid is left out).

\( K_{sp} = [A^{y+}]^x\,[B^{x-}]^y = (xS)^x\,(yS)^y = x^x\,y^y\,S^{x+y} \)

Compare the ionic product Q (the same expression with the concentrations present) with K_sp: if Q < K_sp the solution is unsaturated and more solid can dissolve; if Q = K_sp the solution is saturated; if Q > K_sp the salt precipitates until Q = K_sp. Adding a common ion lowers the solubility of the salt (common ion effect), because S must adjust so that the product stays equal to K_sp. K_sp changes only with temperature.

Solved Examples

Example 1. Calculate the pH of (a) a solution of 0.3 g of Ca(OH)₂ in 500 mL of water and (b) the solution made by diluting 1 mL of 13.6 M HCl to 1 litre.
Answer:
(a) Molar mass of Ca(OH)₂ = 74 g mol⁻¹, so c = (0.3 ÷ 74) ÷ 0.5 = 8.11 × 10⁻³ M.
Ca(OH)₂ gives 2 OH⁻, so [OH⁻] = 1.62 × 10⁻² M, pOH = 1.79 and pH = 14 − 1.79 = 12.21.
(b) M₁V₁ = M₂V₂: 13.6 × 1 = M₂ × 1000, so [H⁺] = 1.36 × 10⁻² M and pH = 1.87.

Example 2. The degree of ionisation of 0.1 M bromoacetic acid is 0.132. Calculate its pH and pK_a.
Answer:
[H⁺] = cα = 0.1 × 0.132 = 1.32 × 10⁻² M, so pH = 1.88.
As α is large, use K_a = cα² ÷ (1 − α) = 0.1 × (0.132)² ÷ 0.868 = 2.0 × 10⁻³.
pK_a = −log(2.0 × 10⁻³) = 2.70.

Example 3. The pH of 0.005 M codeine solution (a weak base, B) is 9.95. Calculate its K_b and pK_b.
Answer:
pOH = 14 − 9.95 = 4.05, so [OH⁻] = 8.91 × 10⁻⁵ M, and [BH⁺] = [OH⁻].
K_b = [BH⁺][OH⁻] ÷ [B] = (8.91 × 10⁻⁵)² ÷ 0.005 = 1.6 × 10⁻⁶ (the ionised part is very small compared with 0.005 M).
pK_b = 5.8.

Example 4. The pH of a solution of a strong acid is 5.0. What will be the pH after diluting it 100 times?
Answer:
Initially [H⁺] = 10⁻⁵ M. After 100-fold dilution the acid gives 10⁻⁷ M.
This is the same as the H⁺ from water, so water cannot be ignored: total [H⁺] ≈ 10⁻⁷ + 10⁻⁷ = 2 × 10⁻⁷ M.
pH = −log(2 × 10⁻⁷) = 6.70 (the solution stays acidic; it does not become pH 7).

Example 5. The ionic product of water at 310 K is 2.7 × 10⁻¹⁴. What is the pH of neutral water at this temperature?
Answer:
In neutral water [H⁺] = [OH⁻] = √K_w = √(2.7 × 10⁻¹⁴) = 1.64 × 10⁻⁷ M.
pH = −log(1.64 × 10⁻⁷) = 6.78. The water is still neutral, although its pH is not 7.

Example 6. The pH of 0.08 M hypochlorous acid (HOCl) is 2.85. Calculate its ionisation constant.
Answer:
[H⁺] = antilog(−2.85) = 1.41 × 10⁻³ M, and [OCl⁻] = [H⁺].
K_a = [H⁺][OCl⁻] ÷ [HOCl] = (1.41 × 10⁻³)² ÷ 0.08 = 2.5 × 10⁻⁵.

Example 7. Calculate the approximate pH of 0.1 M H₂S. K_a1 = 1.0 × 10⁻⁷ and K_a2 = 1.3 × 10⁻¹³.
Answer:
K_a2 is far smaller than K_a1, so only the first ionisation counts: H₂S ⇌ H⁺ + HS⁻.
[H⁺] = √(K_a1 × c) = √(1.0 × 10⁻⁷ × 0.1) = 1.0 × 10⁻⁴ M, so pH = 4.0.
The sulphide ion concentration is [S²⁻] ≈ K_a2 = 1.3 × 10⁻¹³ M.

Example 8. What is the pH of the solution obtained by mixing 200 mL of HCl of pH 2.0 with 300 mL of NaOH of pH 12.0?
Answer:
HCl: [H⁺] = 10⁻² M, so 200 mL contains 2 millimoles. NaOH: [OH⁻] = 10⁻² M, so 300 mL contains 3 millimoles.
After neutralisation, excess OH⁻ = 3 − 2 = 1 millimole in 500 mL, so [OH⁻] = 2 × 10⁻³ M.
pOH = 2.70, so pH = 11.30.

Example 9. 0.1 M acetic acid (K_a = 1.0 × 10⁻⁵) is titrated with 0.05 M NaOH. Find the pH when (i) one-fourth and (ii) three-fourths of the acid has been neutralised.
Answer:
Partly neutralised acid and the sodium acetate formed make a buffer: pH = pK_a + log([salt] ÷ [acid]), with pK_a = 5.
(i) [salt] : [acid] = 1 : 3, so pH = 5 + log(1/3) = 5 − 0.477 = 4.52.
(ii) [salt] : [acid] = 3 : 1, so pH = 5 + log 3 = 5 + 0.477 = 5.48.

Example 10. Calculate the solubility of A₂X₃ in pure water if K_sp = 1.1 × 10⁻²³. Assume no hydrolysis.
Answer:
A₂X₃(s) ⇌ 2A³⁺ + 3X²⁻. If the solubility is S, [A³⁺] = 2S and [X²⁻] = 3S.
K_sp = (2S)²(3S)³ = 108S⁵ = 1.1 × 10⁻²³, so S⁵ = 1.02 × 10⁻²⁵.
S = 1.0 × 10⁻⁵ mol L⁻¹.

Example 11. Will a precipitate of AgCl form when 20 mL of 1.0 × 10⁻⁴ M AgNO₃ is mixed with 30 mL of 1.0 × 10⁻⁴ M NaCl? K_sp(AgCl) = 1.8 × 10⁻¹⁰.
Answer:
Total volume = 50 mL. [Ag⁺] = 1.0 × 10⁻⁴ × 20 ÷ 50 = 4.0 × 10⁻⁵ M and [Cl⁻] = 1.0 × 10⁻⁴ × 30 ÷ 50 = 6.0 × 10⁻⁵ M.
Q = [Ag⁺][Cl⁻] = 2.4 × 10⁻⁹, which is greater than K_sp = 1.8 × 10⁻¹⁰.
Since Q > K_sp, a precipitate of AgCl forms.

Example 12. Compare the solubility of AgCl (K_sp = 1.8 × 10⁻¹⁰) in pure water and in 0.1 M NaCl.
Answer:
In pure water, S² = 1.8 × 10⁻¹⁰, so S = 1.34 × 10⁻⁵ M.
In 0.1 M NaCl, [Cl⁻] ≈ 0.1 M (the Cl⁻ from AgCl is negligible), so S = K_sp ÷ 0.1 = 1.8 × 10⁻⁹ M.
The solubility falls by a factor of about 7000 because of the common ion Cl⁻.