Calculate the $\mathrm{pH}$ of $0.5 \mathrm{M}$ aqueous solution of $\mathrm{NaCN}$, the $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{CN}^{-}$ is $4.70$

Question

Aicharya · Accepted Answer

$\mathrm{NaCN}$ is a salt of strong base and weak acid ; $\mathrm{pH}$
$=7+\frac{1}{2} \mathrm{pK}_{\mathrm{a}}+\frac{1}{2} \log \mathrm{C}$
$\mathrm{pK}_{\mathrm{a}}$ for $\mathrm{HCN}=14-4.70=9.30$
$\therefore \mathrm{pH}=7+\frac{1}{2} \times 9.30+\frac{1}{2} \log 0.5 ; \mathrm{pH}=11.5$ *

Calculate the $\mathrm{pH}$ of $0.5 \mathrm{M}$ aqueous solution of $\mathrm{NaCN}$, the…

Solution