Question

The ionization constant of propanoic acid is $1.32\times {10}^{-5}$. Calculate the degree of ionization of the acid in its 0.05 M solution and also its $pH$. What will be its degree of ionization if the solution is 0.01M in $HCl$ ?

Answer 1

Let x be the degree of dissociation.
$x=\sqrt{\frac{K_a}{c}}=\sqrt{\frac{1.32\times {10}^{-5}}{0.05}}=0.0162$
$\left[H^{+}\right]=cx=0.05\times 0.0162=8.12\times {10}^{-4}$
$pH=log\left[H^{+}\right]=-log(8.12\times {10}^{-4})=3.09$
If the solution is 0.01 M in HCl also then,
$x=\sqrt{\frac{K_a}{c}}=\sqrt{\frac{1.32\times {10}^{-5}}{0.01}}=0.00132$