The pH of a 0.02 M NH_4Cl solution will be | vedclass

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(B) $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$.The formula for the $pH$ of such a salt solution is:$pH = \frac{1}{2} [pK_w

Vedclass · Accepted Answer

$5.35$

Vedclass · Answer

(B) $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$.The formula for the $pH$ of such a salt solution is:$pH = \frac{1}{2} [pK_w - pK_b - \log C]$Given:$C = 0.02 \ M = 2 	imes 10^{-2} \ M$$K_b = 10^{-5} \implies pK_b = 5$$K_w = 10^{-14} \implies pK_w = 14$Substituting the values:$pH = \frac{1}{2} [14 - 5 - \log(2 	imes 10^{-2})]$$pH = \frac{1}{2} [9 - (\log 2 + \log 10^{-2})]$$pH = \frac{1}{2} [9 - (0.301 - 2)]$$pH = \frac{1}{2} [9 - (-1.699)]$$pH = \frac{1}{2} [10.699]$$pH = 5.3495 \approx 5.35$

The $pH$ of a $0.02 \ M \ NH_4Cl$ solution will be [given $K_b \ (NH_4OH) = 10^{-5}$ and $\log \ 2 = 0.301$ ]

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