Key idea for titration of a weak base with a strong acid

For the reaction $B + HA \rightarrow BH^+ + A^-$ when the weak base is half-neutralised, we have: $[B] = [BH^+]$

Using the Henderson-type relation for a weak base buffer, $\mathrm{pOH} = \mathrm{p}K_b + \log\frac{[BH^+]}{[B]}$

At half-neutralisation, $[BH^+] = [B]$ so $\mathrm{pOH} = \mathrm{p}K_b$

Therefore, $\mathrm{p}K_b = 14 - \mathrm{pH} \quad \text{at half-equivalence point}$

Read the graph conceptually

Since both base and acid are of the same concentration, the equivalence volume is where moles of acid added equal initial moles of base.

Hence, the half-equivalence point occurs at half of that volume.

From the given titration curve, the pH at the half-equivalence point is approximately: $\mathrm{pH} \approx 11$

Calculate $\mathrm{p}K_b$

$\mathrm{p}K_b = 14 - 11 = 3$

Final answer

$\boxed{3}$

Comparison with stored correct answer

Stored correct answer = 3

Our derived answer matches it.