JEE Main 2026 — Apr 4, Shift 1 — Chemistry solution

JEE Main2026Apr 4, Shift 1Chemistry

The pH a solution obtained by mixing $5$ mL of $0.1$ M $N H_{4} OH$ solution with $250$ mL of $0.1$ M $N H_{4} Cl$ solution is _____ $\times 1 0^{- 2}$ . (Nearest integer)

Given : $p K_{b} (N H_{4} OH) = 4.74$

$lo g 2 = 0.30$

$lo g 3 = 0.48$

$lo g 5 = 0.70$

Your answer

Solution

This is a basic buffer ( $N H_{4} OH$ + $N H_{4} Cl$ ). Use the Henderson-Hasselbalch form for a base:

$pOH = p K_{b} + lo g \frac{[ salt ]}{[ base ]} .$

Since volumes are common, the ratio of concentrations equals the ratio of millimoles:

mmol NH $_{4}$ OH $= 5 \times 0.1 = 0.5$ ; mmol NH $_{4}$ Cl $= 250 \times 0.1 = 25$ .

$\frac{[ N H _{4} Cl ]}{[ N H _{4} OH ]} = \frac{25}{0.5} = 50.$

$lo g 50 = lo g 100 - lo g 2 = 2 - 0.30 = 1.70.$

$pOH = 4.74 + 1.70 = 6.44.$

$pH = 14 - 6.44 = 7.56 = 756 \times 1 0^{- 2}$ .

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