Question

Calculate $\left[H_3{O}^{+}\right],\left[{OH}^{-}\right]$ and pH of $0.2M$ solution of $HCN$. ($K_a=7.2\times {10}^{-10}$)

Answer 1

$HCN\rightleftharpoons H^{+}+C{N}^{-}$

$0.2$

$0.2(1-\alpha )$ $0.2\alpha$ $0.2\alpha$

$K_a=\frac{\left(0.2\alpha \right)\left(0.2\alpha \right)}{0.2(1-\alpha )}$

Let $\alpha <<1$ so $1-\alpha =1$

$K_a=0.2{\alpha }^2=7.2\times {10}^{-10}$

$\alpha =6\times {10}^{-5}$

$\left[H^{+}\right]=0.2\alpha =1.2\times {10}^{-5}mol{L}^{-1}$

It is known that $pH=-log\left(\right[H^{+}\left]\right)=4.92$

As we know $\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}$

$\left[{OH}^{-}\right]=8.33\times {10}^{-10}mol{L}^{-1}$