The ionization constant of acetic acid is $1.74 \times 10^{-5}$. Calculate the degree of dissociation of acetic acid in its 0. 05 $\mathrm{M}$ solution. Calculate the concentration of acetate ion in the solution and its $\mathrm{pH}$.

Question

The ionization constant of acetic acid is $1.74 \times 10^{-5}$. Calculate the degree of dissociation of acetic acid in its 0. 05 $\mathrm{M}$ solution. Calculate the concentration of acetate ion in the solution and its $\mathrm{pH}$.,

MARKS App · Accepted Answer

$\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$ $\begin{aligned} &\text { As }\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]=\left[\mathrm{H}^{+}\right] \ &\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}^2\right]}=\frac{\left[\mathrm{H}^{+}\right]^2}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \ &\text { or } \quad\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{a}}\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \ &\quad=\sqrt{\left(1.74 \times 10^{-5}\right)\left(5 \times 10^{-2}\right)} \ &\quad=9.33 \times 10^{-4} \mathrm{M} \ &{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]=\left[\mathrm{H}^{+}\right]=9.33 \times 10^{-4} \mathrm{M}} \ &\mathrm{pH}=-\log \left(9.33 \times 10^{-4}\right)=4-0.9699=4-0.97=3.03 \end{aligned}$

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