Question

For 0.50 M aqueous solution of sodium cyanide, $\left({\text{pK}}_{\text{b}}{\text{of CN}}^{-}\text{is}4.70\right)$, calculate hydrolysis constant.

• A. $1.03\times {10}^{-15}$
• B. $1.03\times {10}^{-12}$
• C. $1.99\times {10}^{-5}$
• D. None of the above

Answer 1

The correct option is C $1.99\times {10}^{-5}$

$\left[\text{NaCN}\right]=0.5$
${\text{pK}}_{\text{b}}{\text{of CN}}^{-}\text{is}4.70$
For salt of weak acid and strong base, hydrolysis constant $\left(K_h\right)$ is given by
${\text{K}}_{\text{h}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}$
$p{K}_a+p{K}_b=14$
$p{K}_a=14-4.70$
$p{K}_a=9.3$
$K_a={10}^{-9.3}=5.01\times {10}^{-10}$
${\text{K}}_{\text{h}}=\frac{{10}^{-14}}{5.01\times {10}^{-10}}$
${\text{K}}_{\text{h}}=1.99\times {10}^{-5}$