Question

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

Answer 1

$C{h}_{3}CH-2COOH$ Propanoic acid

$C{H}_{3}C{H}_{2}COOH+H_{2}O\leftrightharpoons C{H}_{3}C{H}_{2}CO{O}^{-}+H_3{O}^{+}$ $K_a=1.32\times {10}^{-5}$

At equlibrium

$0.05-C\alpha$ $C\alpha$ $C\alpha$

Ostwald's Dilution Law$⟶\alpha =\sqrt{\frac{K_d}{C}}=\sqrt{\frac{1.35\times {10}^{-5}}{0.05}}$

$\alpha =0.016248$

Degree of ionization is $16.25\times {10}^{-3}$

$\left[H_3{O}^{+}\right]=C\alpha =0.05\times 0.016248=8.124\times {10}^{-4}$

$pH=-log\left[H_3{O}^{+}\right]=C_{1}0.9098\cong 3.09$

$pH=3.09$

$K_a=\frac{\left[{CH}_3{CH}_{2}CO{O}^{-}\right]\left[H^{+}\right]}{\left[{CH}_2{CH}_{2}COOH\right]}$

$1.32\times {10}^{-5}=\frac{C\alpha \left(0.01\right)}{0.05-\alpha }$

$\left[H_3{O}^{+}\right]=0.0M$ From $HCl$

$C\alpha =\frac{1.32\times {10}^{-5}(0.05-\alpha )}{0.01}$

$C\alpha =6.60\times {10}^{-5}$

Degree of ionisation$=\frac{6.66\times {10}^{-5}}{0.05}=1.32\times {10}^{-3}$