Answer
Percent ionization = $1.342 \%$
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.
-$[H_3O^+] = [Conj. Base] = x$
-$[Formic acid] = [Formic acid]_{initial} - x = 1 - x$
For approximation, we consider: $[Formic acid] = 1M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Formic acid]}$
$Ka = 1.8 \times 10^{- 4}= \frac{x * x}{ 1}$
$Ka = 1.8 \times 10^{- 4}= \frac{x^2}{ 1}$
$ 1.8 \times 10^{- 4} = x^2$
$x = 1.342 \times 10^{- 2}$
Percent ionization: $\frac{ 1.342 \times 10^{- 2}}{ 1} \times 100\% = 1.342\%$
%ionization < 5% : Right approximation.
