Answer
The percent ionization of lactic acid in that solution is equal to 1.9%
Work Step by Step
1. Drawing the ICE table we get these concentrations at the equilibrium:
$HC_3H_5O_3(aq) + H_2O(l) \lt -- \gt C_3H_5O_3^-(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[HC_3H_5O_3] = 0.125 M - x$
$[C_3H_5O_3^-] = [NaC_3H_5O_3]_{initial}* + x = 7.5 \times 10^{-3}M + x$
* $NaC_3H_5O_3$ is a strong electrolyte, so $[C_3H_5O_3^-]_{initial} = [NaC_3H_5O_3]_{initial}$
$[H_3O^+] = 0 + x$
2. Calculate 'x' using the $K_a$ expression.
$ 1.4\times 10^{- 4} = \frac{[C_3H_5O_3^-][H_3O^+]}{[HC_3H_5O_3]}$
$ 1.4\times 10^{- 4} = \frac{( 7.5\times 10^{- 3} + x )* x}{ 0.125 - x}$
Considering 'x' has a very small value.
$ 1.4\times 10^{- 4} = \frac{ 7.5\times 10^{- 3} * x}{ 0.125}$
$ 1.4\times 10^{- 4} = 0.0600x$
$\frac{ 1.4\times 10^{- 4}}{ 0.0600} = x$
$x = 2.33\times 10^{- 3}$
Percent ionization: $\frac{ 2.33\times 10^{- 3}}{ 0.125} \times 100\% = 1.9\%$
