ઍસિટિક ઍસિડનો આયનીકરણ અચળાંક $1.74 \times 10^{-5}$ છે. તેના $0.05$ $M$ દ્રાવણમાં વિયોજન અંશ ગણો. દ્રાવણમાં ઍસિટેટ આયનની સાંદ્રતા અને તેની $pH$ ગણો.

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 $3.03$