Question

The ionization constant of acetic acid is $1.74\times {10}^{-5}$. Calculate the degree of dissociation of acetic acid in its $0.05M$ solution. Calculate the concentration of acetate ion in solution and its $pH$.

Answer 1

Degree of Dissociation

$\alpha =\sqrt{\frac{Ka}{c}}$

Give :- $Ka=1.74\times {10}^{-5}$ $c=0.05M$

$\alpha =\sqrt{\frac{1.74\times {10}^{-5}}{0.05}}=\sqrt{34.8\times {10}^{-5}}$

$\alpha =\sqrt{3.48\times {10}^{-4}}=1.861\times {10}^{-2}$

$\alpha =1.86\times {10}^{-2}$

$C{H}_{3}CHOOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$

Thus concentration of $C{H}_{3}CO{O}^{-}=C.\alpha$

$=0.05\times 1.86\times {10}^{-2}$

$=9.3\times {10}^{-4}$

$\therefore$ concentration of $\left[C{H}_{3}CO{O}^{-}\right]=0.00093M$

$\because$ Concentration of $C{H}_{3}CO{O}^{-}=\left[H^{+}\right]\Rightarrow 0.00093$

& $pH=-10g\left[H^{+}\right]$

$\therefore pH=-10g\left[0.00093\right]=-10g[9.3\times {10}^{-4}]$

$\therefore pH=3.03$

Hence, the concentration of acetate ion in the solution $=0.00093M$ & $pH=3.03$