Answer
$$[H_3O^+] = 5.75 \times 10^{-5} \space M$$
$$[HClO] = 0.115 \space M$$
$$pH = 4.24 $$
Work Step by Step
$$K_a = 10^{-pKa} = 10^{-7.54} = 2.88 \times 10^{-8}$$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HClO ]& [ ClO^- ]& [ H_3O^+ ]\\
Initial& 0.115 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.115 -x& x& x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ ClO^- ][ H_3O^+ ]}{[ HClO ]}$$
$$K_a = \frac{(x)(x)}{[ HClO ]_{initial} - x}$$
3. Assuming $ 0.115 \gt\gt x:$
$$K_a = \frac{x^2}{[ HClO ]_{initial}}$$
$$x = \sqrt{K_a \times [ HClO ]_{initial}} = \sqrt{ 2.88 \times 10^{-8} \times 0.115 }$$
$x = 5.75 \times 10^{-5} $
4. Test if the assumption was correct:
$$\frac{ 5.75 \times 10^{-5} }{ 0.115 } \times 100\% = 0.050 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 5.75 \times 10^{-5} $
6. $$[H_3O^+] = [ClO^-] = x = 5.75 \times 10^{-5} \space M$$
$$[HClO] = 0.115 - 5.75 \times 10^{-5} = 0.115 \space M$$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 5.75 \times 10^{-5} ) = 4.24 $$
