Question

Calculate the degree of ionization and $\left[H_3{O}^{+}\right]$ of $0.01MC{H}_{3}COOH$ solution. The equilibrium constant of acetic acid is $1.8\times {10}^{-5}$.

Answer 1

Given, $K=1.8\times {10}^{-5},C=0.01M$

We know that $K=C{\alpha }^2$

So $18\times {10}^{-6}={10}^{-2}{\alpha }^2$

$18\times {10}^{-6}={\alpha }^2$

$\alpha =\sqrt{18}\times {10}^{-2}=3\sqrt{2}\times {10}^{-2}$

$\alpha =4.2\times {10}^{-2}$

$\alpha =0.042$

Hence , degree of ionization is $0.042$

$\left[H^{+}\right]=C\alpha =0.01\left(0.042\right)$

$\left[H^{+}\right]=42\times {10}^{-5}$ $M$