JEE Main 2026 — Jan 28, Shift 1 — Chemistry solution

JEE Main2026Jan 28, Shift 1Chemistry

Consider the dissociation equilibrium of the following weak acid $HA ⇌ H^{+} (a q) + A^{-} (a q)$

If the pK $_{a}$ 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]

Your answer

Solution

For a weak acid with degree of dissociation $α ≪ 1$ :

$[H^{+}] = K_{a} \cdot c$

Taking the $- lo g$ on both sides:

$pH = \frac{1}{2} [p K_{a} - lo g c]$

With p $K_{a} = 4$ and $c = 10$ mM $= 1 0^{- 2}$ M:

$pH = \frac{1}{2} [4 - lo g (1 0^{- 2})] = \frac{1}{2} [4 - (- 2)] = \frac{1}{2} (6) = 3$

$∴$ Answer: $3$ .

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