A solution of 0.1 M weak base (B) is titrated with 0.1 M of a strong acid (HA). The variation of pH of the solution with the volume of HA added is shown in the figure below. What is the $$p𝐾_b$$ of the base? The neutralization reaction is given by $$B + HA \rightarrow BH^+ + A^-$$.

The titration is of a weak base by a strong acid, with the neutralisation stoichiometry
$$B + HA \rightarrow BH^+ + A^-$$

Let the initial volume of the base solution be $$V_0$$ mL. Because both solutions are 0.1 M and the stoichiometric ratio is 1 : 1, the equivalence point will occur after the same volume $$V_0$$ mL of the acid has been added. Hence, the half-equivalence point corresponds to the addition of $$\dfrac{V_0}{2}$$ mL of the acid.

For the conjugate acid-base pair $$BH^+/B$$ the Henderson-Hasselbalch relation is
$$\text{pH} = \text{p}K_a + \log\frac{[B]}{[BH^+]}$$

At the half-equivalence point $$[B] = [BH^+]$$, so the logarithmic term becomes zero and
$$\text{pH}_{1/2} = \text{p}K_a$$ $$-(1)$$

From the titration curve (given in the figure) the pH at exactly half the equivalence volume is read as
$$\text{pH}_{1/2} = 9.25$$

Substituting this value in $$(1)$$ gives
$$\text{p}K_a = 9.25$$

The base dissociation constant is related to its conjugate-acid constant by
$$\text{p}K_a + \text{p}K_b = 14$$ $$-(2)$$

Using $$(2)$$:
$$\text{p}K_b = 14 - 9.25 = 4.75$$

Therefore, the $$\mathbf{pK_b}$$ of the weak base is $$\mathbf{4.75}$$.