Question

Using the average initial pH of your acetic acid solutions (2.96), and the average molarity of those solutions(.0602), calculate a value of Ka for acetic acid. Report your result to 3 significant figures.

Answer 1

Answer : The value of Ka for acetic acid is,  $2.03\times 10^{-5}$

Explanation :

The chemical formula of acetic acid is, $CH_3COOH$.

The chemical equilibrium reaction will be:

$CH_3COOH\rightleftharpoons CH_3COO^-+H^+$

Given:

pH = 2.96

First we have to calculate the concentration of hydrogen ion.

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

$2.96=-\log [H^+]$

$[H^+]=1.096\times 10^{-3}M$

That means,

$[H^+]=[CH_3COO^-]=1.096\times 10^{-3}M$

$[CH_3COOH]=0.0602-(1.096\times 10^{-3})=0.0591M$

The expression for reaction is:

$K_a=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$

$K_a=\frac{(1.096\times 10^{-3})\times (1.096\times 10^{-3})}{0.0591}$

$K_a=2.03\times 10^{-5}$

Thus, the value of Ka for acetic acid is,  $2.03\times 10^{-5}$