Answer
The percent ionization of lactic acid in that solution is equal to 3.3%
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$HC_3H_5O_3(aq) + H_2O(l) \lt -- \gt C_3H_5O_3^-(aq) + H_3O^+(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $C_3H_5O_3^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M
-$[H_3O^+] = [C_3H_5O_3^-] = 0 + x = x$
-$[HC_3H_5O_3] = [HC_3H_5O_3]_{initial} - x$
For approximation, we are going to consider $[HC_3H_5O_3]_{initial} = [HC_3H_5O_3]$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_3H_5O_3^-]}{ [HC_3H_5O_3]}$
$Ka = 1.4 \times 10^{- 4}= \frac{x * x}{ 0.125}$
$Ka = 1.4 \times 10^{- 4}= \frac{x^2}{ 0.125}$
$x^2 = 0.125 \times 1.4 \times 10^{-4} $
$x = \sqrt { 0.125 \times 1.4 \times 10^{-4}} = 4.18 \times 10^{-3} $
Percent ionization: $\frac{ 4.18 \times 10^{- 3}}{ 0.125} \times 100\% = 3.3\%$
%ionization $\lt$ 5% : Right approximation.
