Consider the dissociation equilibrium of the following weak acid $$HA\rightleftharpoons H^{+}(aq)+A^{-}(aq)$$If the pKa of the acid is 4, then the pH of 10 mM HA solution is __ .(Nearest integer)
[Given: The degree of dissociation can be neglected with respect to unity]

We need to find the pH of a 10 mM solution of weak acid HA with $$pK_a = 4$$.

Since $$pK_a = 4 \implies K_a = 10^{-4}$$ and the concentration is $$C = 10 \text{ mM} = 0.01 \text{ M} = 10^{-2} \text{ M}$$.

The dissociation equilibrium is given by $$HA \rightleftharpoons H^+ + A^-$$, and the acid dissociation constant is $$K_a = \frac{[H^+][A^-]}{[HA]}$$.

For a weak acid where the degree of dissociation can be neglected (as stated in the problem), the hydrogen ion concentration is $$[H^+] = \sqrt{K_a \times C}$$. Taking the negative logarithm gives $$pH = -\log[H^+] = -\frac{1}{2}\log(K_a \times C) = \frac{1}{2}(pK_a - \log C)$$.

Substituting the values yields $$pH = \frac{1}{2}(pK_a - \log C) = \frac{1}{2}(4 - \log(10^{-2}))$$, which becomes $$= \frac{1}{2}(4 - (-2)) = \frac{1}{2}(4 + 2) = \frac{1}{2} \times 6 = 3$$.

As a verification, one finds $$[H^+] = \sqrt{10^{-4} \times 10^{-2}} = \sqrt{10^{-6}} = 10^{-3}$$ M and the degree of dissociation $$\alpha = 10^{-3}/10^{-2} = 0.1$$. Since $$\alpha = 0.1$$ is small enough relative to unity (and the problem states it can be neglected), this is consistent.

The answer is 3.