Answer
The pH of this solution is equal to 3.83
Work Step by Step
The second proton from $H_2CO_3$ has a negligible contribution to the pH.
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ H_2CO_3 ]& [ HC{O_3}^- ]& [ H_3O^+ ]\\
Initial& 0.050 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.050 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ HC{O_3}^- ][ H_3O^+ ]}{[ H_2CO_3 ]}$$
$$K_a = \frac{(x)(x)}{[ H_2CO_3 ]_{initial} - x}$$
3. Assuming $ 0.050 \gt\gt x:$
$$K_a = \frac{x^2}{[ H_2CO_3 ]_{initial}}$$
$$x = \sqrt{K_a \times [ H_2CO_3 ]_{initial}} = \sqrt{ 4.3 \times 10^{-7} \times 0.050 }$$
$x = 1.466 \times 10^{-4} $
4. Test if the assumption was correct:
$$\frac{ 1.466 \times 10^{-4} }{ 0.050 } \times 100\% = 0.3 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 1.466 x 10^{-4} $
6. $$[H_3O^+] = x = 1.466 \times 10^{-4} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 1.466 \times 10^{-4} ) = 3.83 $$
