JEE Main 2025 Apr 7 Shift 2, Chemistry Q21: Buffer Solutions

JEE Main2025Apr 7, Shift 2Chemistry

Only litre buffer solution was prepared by adding 0.10 mol each of NH $_{3}$ and NH $_{4}$ Cl in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is ____ $\times 1 0^{- 2}$ , (Nearest integer)

(Given : $p K_{b}$ of NH $_{3}$ = 4.745 and $lo g_{10} 3$ = 0.477)

Your answer

Solution

For the initial NH $_{3}$ /NH $_{4}$ Cl buffer with equal moles of base and salt, the Henderson-Hasselbalch relation gives $pO H = p K_{b} + lo g \frac{[ N H _{4}^{+} ]}{[ N H _{3} ]} = 4.745 + lo g (1) = 4.745$ .

Adding 0.05 mol HCl converts 0.05 mol of NH $_{3}$ into NH $_{4}^{+}$ : NH $_{3}$ drops from 0.10 to 0.05 mol, and NH $_{4}^{+}$ rises from 0.10 to 0.15 mol.

The new $pO H^{'} = 4.745 + lo g \frac{0.15}{0.05} = 4.745 + lo g 3 = 4.745 + 0.477 = 5.222$ .

Change in pOH: $Δ pO H = pO H^{'} - pO H = 0.477$ . Since $p H + pO H = 14$ is constant, the change in pH has the same magnitude: $Δ p H = 0.477$ .

Expressed as $(Δ p H) \times 1 0^{2} = 47.7 \approx 48$ .

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JEE Main 2025 Apr 7 Shift 2, Chemistry Q21: Buffer Solutions

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