Answer
$$[H_3O^+] = [ClCH_2COO^-] = 0.041 \space M$$
$$[ClCH_2COOH] = 1.21 \space M$$
$$pH = 1.39 $$
Work Step by Step
$$K_a = 10^{-pKa} = 10^{-2.87} = 1.35 \times 10^{-3}$$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ ClCH_2COOH ]& [ ClCH_2COO^- ]& [ H_3O^+ ]\\
Initial& 1.25 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 1.25 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ ClCH_2COO^- ][ H_3O^+ ]}{[ ClCH_2COOH ]}$$
$$K_a = \frac{(x)(x)}{[ ClCH_2COOH ]_{initial} - x}$$
3. Assuming $ 1.25 \gt\gt x:$
$$K_a = \frac{x^2}{[ ClCH_2COOH ]_{initial}}$$
$$x = \sqrt{K_a \times [ ClCH_2COOH ]_{initial}} = \sqrt{ 1.35 \times 10^{-3} \times 1.25 }$$
$x = 0.041 $
4. Test if the assumption was correct:
$$\frac{ 0.041 }{ 1.25 } \times 100\% = 3.3 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 0.041 $
6. $$[H_3O^+] = [ClCH_2COO^-] = x = 0.041 $$
$$[ClCH_2COOH] = 1.25 - 0.041 = 1.21$$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 0.0411 ) = 1.39 $$
