Question

Calculate the hydrolysis constant, of $0.25\text{M}NaCN$ solution
$K_a\left(HCN\right)=4.8\times {10}^{-10}\text{M}$.

• A. $3.02\times {10}^{-5}\text{M}$
• B. $3.02\times {10}^{-4}\text{M}$
• C. $2.08\times {10}^{-5}\text{M}$
• D. $4.05\times {10}^{-5}\text{M}$

Answer 1

The correct option is C $2.08\times {10}^{-5}\text{M}$

The hydrolysis reaction of $C{N}^{-}$ is
$C{N}^{-}+H_{2}O\rightleftharpoons HCN+O{H}^{-}$
The value of hydrolysis constant is
$K_h=\frac{\left[HCN\right]\left[O{H}^{-}\right]}{\left[C{N}^{-}\right]}=\frac{K_w}{K_a}=\frac{(1.0\times {10}^{-14}{\text{M}}^2)}{(4.8\times {10}^{-10}\text{M})}=2.08\times {10}^{-5}\text{M}$