Question

Acetic acid (HC2H3O2, Ka=1.8x10^-5) is a weak acid. Calculate the pH of an aqueous solution of .25M acetic acid.

Answer 1

Answer: The pH of an aqueous solution of .25M acetic acid is 2.7

Explanation:

$HC_2H_3O_2\rightarrow H^+C_2H_3O_2^-$

 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= 0.25 M and $\alpha$ = ?

$K_a=1.8\times 10^{-5}$

Putting in the values we get:

$1.8\times 10^{-5}=\frac{(0.25\times \alpha)^2}{(0.25-0.25\times \alpha)}$

$(\alpha)=0.0084$

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

$[H^+]=0.25\times 0.0084=0.0021$

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

$pH=-log[0.0021]=2.7$

Thus pH is 2.7