Answer
Percent ionization of $0.0075M$ butanoic acid: $4.5$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$HC_4H_7O_2(aq) + H_2O(l) \lt -- \gt C_4H_7O_2^-(aq) + H_3O^+(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $C_4H_7O_2^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M
-$[H_3O^+] = [C_4H_7O_2^-] = 0 + x = x$
-$[HC_4H_7O_2] = [HC_4H_7O_2]_{initial} - x$
For approximation, we are going to consider $[HC_4H_7O_2]_{initial} = [HC_4H_7O_2]$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_4H_7O_2^-]}{ [HC_4H_7O_2]}$
$Ka = 1.5 \times 10^{- 5}= \frac{x * x}{ 7.5\times 10^{- 3}}$
$Ka = 1.5 \times 10^{- 5}= \frac{x^2}{ 7.5\times 10^{- 3}}$
$x^2 = 0.0075 \times 1.5 \times 10^{-5} $
$x = \sqrt { 0.0075 \times 1.5 \times 10^{-5}} = 3.4 \times 10^{-4} $
Percent ionization: $\frac{ 3.4 \times 10^{- 4}}{ 7.5\times 10^{- 3}} \times 100\% = 4.5\%$
%ionization < 5% : Right approximation.