Solving Acid-Base Dissociation Problems (HL) | HL IB Chemistry Revision Notes 2025

Solving Acid-Base Dissociation Problems

In reactions of weak acids and bases, we cannot make the same assumptions as for the ionisation of strong acids and bases
For a weak acid and its conjugate base, we can use the equation:

K_w= K_a K_b

pK_w= pK_a + pK_b

pK_a= -logK_aK_a= 10^–pK_a

pK_b= -logK_bK_b= 10^–pK_b

The assumptions we must make when calculating values for K_a, pK_a, K_b and pK_bare:
- The initial concentration of acid ≈ the equilibrium concentration of acid
- [A^-] = [H⁺]
- There is negligible ionisation of the water, so [H⁺] is not affected
- The temperature is 298 K

Calculate the acid dissociation constant, K_a, at 298 K for a 0.20 mol dm^-3 solution of propanoic acid with a pH of 4.88.

Answer:

Step 1: Calculate [H⁺] using
- [H⁺] = 10^-pH
  - [H⁺] = 10^-4.88
  - [H⁺] = 1.3183 x 10^-5
Step 2: Substitute values into K_a expression
- K_a = $\frac{{[H^{+}]}^{2}}{[{CH}_{3} {CH}_{2} COOH]}$
  - K_a = $\frac{{(1.3182 \times 10^{- 5})}^{2}}{0.2}$
  - K_a = 8.70 × 10^-10

A 0.035 mol dm^-3 sample of methylamine (CH₃NH₂) has pK_b value of 3.35 at 298 K. Calculate the pH of methylamine.

Answer:

Step 1: Calculate the value for K_b using
- K_b= 10^–pK_b
  - K_b= 10^-3.35
  - K_b= 4.4668 x 10^-4
Step 2: Substitute values into K_b expression to calculate [OH^-]
- K_b= $\frac{{[{OH}^{-}]}^{2}}{[{CH}_{2} {NH}_{2}]}$
  - 4.4668 x 10^-4 = $\frac{[{OH}^{-}]}{0.035}$
  - [OH^–] = $\sqrt{(4.4668 \times 10^{- 4} \times 0.035)}$
  - [OH^–] = 3.9540 x 10^-3
Step 3: Calculate the pH
- [H⁺] = $\frac{K_{w}}{[{OH}^{-}]}$
  - [H⁺] = (1 x 10^-14) ÷ 3.9539 x 10^-3
  - [H⁺] = 2.5291 x 10^-12

- pH = -log [H⁺]
  - pH = -log 2.5291 x 10^-12
- pH = 11.60 to 2 decimal places

Step 3: Calculate pOH and therefore pH
- pOH = -log [OH^–]
  - pOH = -log 3.9540 x 10^-3
  - pOH = 2.4029
- pH = 14 - pOH
  - pH = 14 - 2.4030
  - pH = 11.60 to 2 decimal places