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4 mole of $N H_{4} O H$ and 1 mole of $H_{2} S O_{4}$ are mixed with enough water to get 1L solution. The $p^{H}$ of the solution is [p^kb of $N H_{4} O H = 4.8$ ]

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$4.8$

$11.2$

$9.8$

$9.2$

answer is B.

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Detailed Solution

$2 N H_{4} O H + H_{2} S O_{4} \to {(N H_{4})}_{2} S O_{4} + 2 H_{2} O$

$∴ η$ . GE of salt formed = 2

$η$ .GE of $N H_{4} O H$ Of left end = 2
$∴$ p^OH = 4.8 + log $\log \frac{2}{2}$
= 4.8
$∴$ p^H= 9.2

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