Answer
Percent ionization $\approx$ 4.9%.
Work Step by Step
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [HCOO^-] = x$
-$[HCOOH] = [HCOOH]_{initial} - x = 0.0751 - x$
For approximation, we consider: $[HCOOH] = 0.0751M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][HCOO^-]}{ [HCOOH]}$
$Ka = 1.8 \times 10^{- 4}= \frac{x * x}{ 7.51 \times 10^{- 2}}$
$Ka = 1.8 \times 10^{- 4}= \frac{x^2}{ 7.51 \times 10^{- 2}}$
$ 1.352 \times 10^{- 5} = x^2$
$x = 3.677 \times 10^{- 3}$
Percent ionization: $\frac{ 3.677 \times 10^{- 3}}{ 7.51 \times 10^{- 2}} \times 100\% = 4.896\%$
%ionization < 5% : Right approximation.
