Answer
$$Percent \space ionization = 5.8\%$$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
$HCHO_2(aq) + H_2O(l) \lt -- \gt CH{O_2}^-(aq) + H_3O^+(aq) $
\begin{matrix} & [HCHO_2] & [H_3O^+] & CH{O_2}^- \\ Initial & 0.0500 & 0 & 0 \\ Change & -x & +x & +x\\ Equil & 0.0500 -x & x & x \end{matrix}
For approximation, we are going to consider $0.0500 - x \approx 0.0500$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][CH{O_2}^-]}{ [HCHO_2]}$
$Ka = 1.8 \times 10^{- 4}= \frac{x * x}{ 0.0500}$
$Ka = 1.8 \times 10^{- 4}= \frac{x^2}{ 0.0500}$
$x^2 = 0.0500 \times 1.8 \times 10^{-4} $
$x = \sqrt { 0.0500 \times 1.8 \times 10^{-4}} = 3.0 \times 10^{-3} $
Percent ionization: $\frac{ 3.0 \times 10^{- 3}}{ 0.0500} \times 100\% = 6.0 \%$
%ionization > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 1.8 \times 10^{-4} = \frac{x^2}{0.0500-x}$
$(1.8 \times 10^{-4} \times 0.0500) - 1.8 \times 10^{-4}x = x^2$
$(1.8 \times 10^{-4} \times 0.0500) - 1.8 \times 10^{-4}x - x^2 = 0$
Bhaskara:
$x_1 = \frac{ - (- 1.8 \times 10^{-4})+ \sqrt {(1.8 \times 10^{-4})^2 - 4(-1)(1.8 \times 10^{-4} \times 0.0500)}}{2*(-1)}$
or
$x_2 = \frac{ - (- 1.8 \times 10^{-4})- \sqrt {(1.8 \times 10^{-4})^2 - 4(-1)(1.8 \times 10^{-4} \times 0.0500)}}{2*(-1)}$
$x_1 = - 3.1 \times 10^{- 3} (Negative)$
$x_2 = 2.9 \times 10^{- 3}$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
Percent ionization: $\frac{ 2.9 \times 10^{- 3}}{ 0.0500} \times 100\% = 5.8 \%$
