Answer
$[H_3O^+] = 2.3 \times 10^{-4}$
$pH = 3.64$
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.
-$[H_3O^+] = [H{CO_3}^-] = x$
-$[H_2CO_3] = [H_2CO_3]_{initial} - x = 0.125 - x$
For approximation, we consider: $[H_2CO_3] = 0.125M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][H{CO_3}^-]}{ [H_2CO_3]}$
$Ka = 4.3 \times 10^{- 7}= \frac{x * x}{ 0.125}$
$Ka = 4.3 \times 10^{- 7}= \frac{x^2}{ 0.125}$
$ 5.375 \times 10^{- 8} = x^2$
$x = 2.318 \times 10^{- 4}$
Percent dissociation: $\frac{ 2.318 \times 10^{- 4}}{ 0.125} \times 100\% = 0.1855\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [H{CO_3}^-] = x = 2.3 \times 10^{- 4}M $
And, since 'x' has a very small value (compared to the initial concentration): $[H_2CO_3] \approx 0.125M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 2.3 \times 10^{- 4})$
$pH = 3.64$
- We don't need to consider the second dissociation $(K_{a2} = 5.6 \times 10^{-11})$, because it is negligible if compared to the hydronium concentration given by the first one.
