Answer
$[H_3O^+] = 1.265 \times 10^{- 5}M $
$[A^-] = 1.265 \times 10^{- 5}M $
$[HA] \approx 0.04M$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
-$[H_3O^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x = 0.04 - x$
For approximation, we consider: $[HA] = 0.04M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][A^-]}{ [HA]}$
$Ka = 4 \times 10^{- 9}= \frac{x * x}{ 0.04}$
$Ka = 4 \times 10^{- 9}= \frac{x^2}{ 0.04}$
$ 1.6 \times 10^{- 10} = x^2$
$x = 1.265 \times 10^{- 5}$
Percent ionization: $\frac{ 1.265 \times 10^{- 5}}{ 0.04} \times 100\% = 0.03162\%$
%ionization < 5% : Right approximation.
Therefore: $[H_3O^+] = [A^-] = x = 1.265 \times 10^{- 5}M $
And, since 'x' has a very small value (compared to the initial concentration): $[HA] \approx 0.04M$
