Question

Determine the acid dissociation constant for a 0.10 m acetic acid solution that has a ph of 2.87. Acetic acid is a weak monoprotic acid and the equilibrium equation of interest is

Answer 1

The given question is incomplete. The complete question is:

Determine the acid dissociation constant for a 0.10 M acetic acid solution that has a pH of 2.87. Acetic acid is a weak monoprotic acid and the equilibrium equation of interest is $CH_3COOH+H_2O\rightleftharpoons H_3O^+ +CH_3COO^-$

Answer: 0.000017

Explanation:

$CH_3COOH+H_2O\rightleftharpoons H_3O^+ +CH_3COO^-$

            cM                   0 M             0 M

          $c-c\alpha$               $c\alpha$       $c\alpha$  

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.10 M and $\alpha$ = dissociation constant = ?

$K_a=?$

Putting in the values we get:

$K_a=\frac{(0.10\times \alpha)^2}{(0.10-0.10\times \alpha)}$

$pH=-log[H^+]$

$2.87=-log[H^+]$

$[H^+]=1.35\times 10^{-3}$

$[H^+]=c\times \alpha$

$1.35\times 10^{-3}=0.10\times \alpha$

$\alpha=0.013$

$K_a=\frac{(0.10\times 0.013)^{2}}{0.10-0.10\times 0.013}$

$K_a=0.000017$

Thus the acid dissociation constant is 0.000017