Chemistry - Ionic Equilibrium Question with Solution | TestHub

${\text{K}}_{\text{eq.}}=\frac{\left[\text{HCN}\right]\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CN}}^{-}\right]\left[{\text{CH}}_3\text{COOH}\right]};$

${\text{K}}_{{\text{CH}}_3\text{COOH}}=\frac{\left[{\text{CH}}_3{\text{COO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_3\text{COOH}\right]};$

${\text{K}}_{\text{HCN}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{CN}}^{-}\right]}{\left[\text{HCN}\right]}$

${\text{pK}}_{\text{b}}\text{of}{\text{CN}}^{-}=4.69\therefore {\text{pK}}_{\text{HCN}}=9.31;{\text{K}}_{\text{HCN}}=4.9\times {10}^{-10}$

${\text{pK}}_{{\text{CH}}_3\text{COOH}}=\text{4.75} ; {\text{K}}_{{\text{CH}}_3\text{COOH}}=\text{1.77}\times 10^{-5}$

$\therefore {\text{K}}_{\text{eq.}}=\frac{{\text{K}}_{{\text{CH}}_3\text{COOH}}}{{\text{K}}_{\text{HCN}}}=\frac{\text{1.77}\times 10^{-5}}{\text{4.9}\times 10^{-10}}=\text{3.7}\times 10^4$