Question

Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid at 25 oC. b) Determine the percent dissociation for the solution.

Answer 1

Answer: a) The $K_a$ of acetic acid at $25^0C$ is $1.82\times 10^{-5}$

b) The percent dissociation for the solution is $4.27\times 10^{-3}$

Explanation:

$CH_3COOH\rightarrow CH_3COO^-H^+$

 cM              0             0

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

So dissociation constant will be:

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

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

Also $pH=-log[H^+]$

$2.87=-log[H^+]$  

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

$[CH_3COO^-]=1.35\times 10^{-3}M$

$[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M$

Putting in the values we get:

$K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}$

$K_a=1.82\times 10^{-5}$

b)  $\alpha=\sqrt\frac{K_a}{c}$

$\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}$

$\alpha=4.27\times 10^{-5}$

$\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}$