Question

The acidic ingredient in vinegar is acetic acid. The pH of vinegar is around 2.4, and the molar concentration of acetic acid in vinegar is around 0.85 M. Based on this information, determine the value of the acid ionization constant, Ka, for acetic acid.

Answer 1

Answer: The value of acid ionization constant $K_a$ for acetic acid is $1.87\times 10^{-5}$

Explanation:

$CH_3COOH\rightarrow H^+CH_3COO^-$

 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= 0.85 M and $pH$ = 2.4

$pH=-log[H^+]$

$[H^=}=c\times \alpha=10^{-2.4}=3.98\times 10^{-3}$

$K_a=?$

Putting in the values we get:

$K_a=\frac{(3.98\times 10^{-3})^2}{(0.85-3.98\times 10^{-3})}=1.87\times 10^{-5}$

Thus the value of acid ionization constant $K_a$ for acetic acid is $1.87\times 10^{-5}$