Answer
$[H_3O^+] = 6.0 \times 10^{-3}$
$pH = 2.22$
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HCHO_2 ]& [ CHO{_2}^- ]& [ H_3O^+ ]\\
Initial& 0.200 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.200 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CHO{_2}^- ][ H^+ ]}{[ HCHO_2 ]}$$
$$K_a = \frac{(x)(x)}{[ HCHO_2 ]_{initial} - x}$$
3. Assuming $ 0.200 \gt\gt x:$
$$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 0.200 }$$
$x = 6.0 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 6.0 \times 10^{-3} }{ 0.200 } \times 100\% = 3.0 \%$$
5. Thus, it is correct to say that $x = 6.0 x 10^{-3} $
6. $$[H^+] = x = 6.0 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( \sqrt{ 1.8 \times 10^{-4} \times 0.200 } ) = 2.22 $$
