The ionization constant of acetic acid is $1.74 \times 10^{-5}$. Calculate the degree of dissociation of acetic acid in its $0.05\, M$ solution. Calculate the concentration of acetate ion in the solution and its $pH$.
Method $1$
$(1)$ $CH _{3} COOH \longleftrightarrow CH _{3} COO ^{-}+ H ^{-+}$ $K_{a}=1.74 \times 10^{-5}$
$(2)$ $H _{2} O + H _{2} O \longleftrightarrow H _{3} O ^{+}+ OH ^{-}$ ${K_w} = 1.0 \times {10^{ - 14}}$
Since $K a\,>\,>\,K_{w,}$
$CH _{3} COOH + H _{2} O \longleftrightarrow CH _{3} COO ^{-}+ H _{3} O ^{+}$
$C_{i}=$ $0.05$ $0$ $0$
$0.05-.05 \alpha$ $0.05 \alpha$ $0.05 \alpha$
$K_{a}=\frac{(.05 \alpha)(.05 \alpha)}{(.05-0.05 \alpha)}$
$=\frac{(.05 \alpha)(0.05 \alpha)}{.05(1-\alpha)}$
$=\frac{.05 \alpha^{2}}{1-\alpha}$
$1.74 \times 10^{-5}=\frac{0.05 \alpha^{2}}{1-\alpha}$
$1.74 \times 10^{-5}-1.74 \times 10^{-5} \alpha=0.05 \alpha^{2}$
$0.05 \alpha^{2}+1.74 \times 10^{-5} \alpha-1.74 \times 10^{-5}$
$ D =b^{2}-4 a c $
$=\left(1.74 \times 10^{-5}\right)^{2}-4(.05)\left(1.74 \times 10^{-5}\right) $
$=3.02 \times 10^{-25}+.348 \times 10^{-5} $
$\alpha =\sqrt{\frac{K_{a}}{c}} $
$\alpha=\sqrt{\frac{1.74 \times 10^{-5}}{.05}}$
$=\sqrt{\frac{34.8 \times 10^{-5} \times 10}{10}}$
$=\sqrt{3.48 \times 10^{-6}}$
$= CH _{3} COOH \longleftrightarrow CH _{3} COO ^{-}+ H ^{+}$
$\alpha 1.86 \times 10^{-3}$
$\left[ CH _{3} COO ^{-}\right]=0.05 \times 1.86 \times 10^{-3}$
$=\frac{0.93 \times 10^{-3}}{1000} $
$=.000093$
Method $2$
Degree of dissociation,
$ \alpha =\sqrt{\frac{K_{a}}{c}} $
$c =0.05 M $
$ K_{a} =1.74 \times 10^{-5} $
Then, $\alpha=\sqrt{\frac{1.74 \times 10^{-5}}{.05}}$
$\alpha=\sqrt{34.8 \times 10^{-5}}$
$\alpha=\sqrt{3.48} \times 10^{-4}$
$\alpha=1.8610^{-2}$
$CH _{3} COOH \longleftrightarrow CH _{3} COO ^{-}+ H ^{+}$
Thus, concentration of $CH _{3} COO -= c.a$
$=.05 \times 1.86 \times 10^{-2}$
$=.093 \times 10^{-2}$
$=.00093 \,M$
Since $\left[ oAc ^{-}\right]=\left[ H ^{+}\right]$
$\left[ H ^{+}\right]=.00093=.093 \times 10^{-2}$
$ pH =-\log \left[ H ^{+}\right] $
$=-\log \left(.093 \times 10^{-2}\right) $
$\therefore pH =3.03 $
Hence, the concentration of acetate ion in the solution is $0.00093 \,M$ and its $Ph$ is $3.03$
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