The pH of a 0.01 , M ammonium cyanide (NH_4CN | vedclass

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Question

(D) Ammonium cyanide $(NH_4CN)$ is a salt of a weak acid $(HCN)$ and a weak base $(NH_4OH)$. For such salts,the $pH$ is given by the formula: $pH = 7

Vedclass · Accepted Answer

$7.02$

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(D) Ammonium cyanide $(NH_4CN)$ is a salt of a weak acid $(HCN)$ and a weak base $(NH_4OH)$. For such salts,the $pH$ is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$. This formula shows that the $pH$ of the solution is independent of the concentration of the salt. Therefore,adding water (dilution) does not change the $pH$ of the solution. Thus,the $pH$ remains $7.02$.

The $pH$ of a $0.01 \, M$ ammonium cyanide $(NH_4CN)$ solution is $7.02$. If $5 \, L$ of water is added to this solution,the $pH$ will be .......

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