Question

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

• A. $1.25\times {10}^{-3}$
• B. $1.32\times {10}^{-5}$
• C. $1.32\times {10}^{-3}$
• D. None of these

Answer 1

The correct option is C $1.32\times {10}^{-3}$

$C{H}_{3}C{H}_{2}COOH+H_{2}O\rightleftharpoons C{H}_{3}C{H}_{2}CO{O}^{-}+H_3{O}^{+}$

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

from Ostwald's Dilution law,

$\alpha =\sqrt{Ka/C}=\sqrt{\frac{1.35\times {10}^{-5}}{0.05}}$

$\alpha =0.016248$

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

$=0.0008124=8.124\times {10}^{-4}$

$pH=-\log \left[H_3{O}^{+}\right]=-\log (8.124\times {10}^{-4})$

$=4-0.9098=3.0902\simeq 3.09$

When the $so{l}^n$ contains $0.01M$ of $HCl$,

$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right[H^{+}]}{\left[C{H}_{3}C{H}_{2}COOH\right]}$

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

Now, $\left[H_3{O}^{+}\right]=0.01M$ from $HCl$. In the presence of $0.01M$ $HCl$ dissociation of propanoic acid decreases.

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

$C\alpha =6.6\times {10}^{-5}$

$\therefore$ degree of ionization, $\alpha =\frac{6.6\times {10}^{-5}}{0.05}$

$1.32\times {10}^{-3}$