JEE Chemistry 2026 Question with Solution

Given: Weak acid equilibrium $\mathrm{HA \rightleftharpoons H^+(aq) + A^-(aq)}$, $pK_a = 4$, and initial concentration $[\mathrm{HA}]_0 = 10\ \text{mM} = 0.01\ \text{M}$.

Find: The pH of the solution.

Use the acid dissociation relation:

$$K_a = \frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$$

From $pK_a = 4$,

$$K_a = 10^{-pK_a} = 10^{-4}$$

Since the degree of dissociation is negligible with respect to unity, take $[\mathrm{HA}] \approx [\mathrm{HA}]_0 = 0.01\ \text{M}$ and $[\mathrm{H^+}] = [\mathrm{A^-}] = x$. Then,

$$K_a = \frac{x^2}{0.01}$$

So,

$$x = \sqrt{K_a \times 0.01} = \sqrt{10^{-4} \times 10^{-2}} = \sqrt{10^{-6}} = 10^{-3}\ \text{M}$$

Thus,

$$\text{pH} = -\log_{10}[\mathrm{H^+}] = -\log_{10}(10^{-3}) = 3$$

Therefore, the pH of the solution is $3$.

Given: $pK_a = 4$ and concentration $C = 0.01\ \text{M}$.

Find: pH.

For a weak acid with small dissociation,

$$[\mathrm{H^+}] = \sqrt{K_a C}$$

Here $K_a = 10^{-4}$ and $C = 10^{-2}$, so

$$[\mathrm{H^+}] = \sqrt{10^{-4} \times 10^{-2}} = \sqrt{10^{-6}} = 10^{-3}$$

Hence,

$$\text{pH} = 3$$

This shortcut works because the weak acid approximation gives $[\mathrm{HA}] \approx C$. The correct answer is $3$.