Answer
$$pH = 10.23$$
Work Step by Step
- Identify the acid and base:
$Na^+$ does not act as an acid;
$OCl^-$ acts as a weak base.
$$K_b (OCl^-) = \frac{10^{-14}}{K_a(HOCl)} = \frac{10^{-14}}{2.9 \times 10^{-8}} = 3.4 \times 10^{-7}$$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ OCl^- ]& [ HOCl ]& [ OH^- ]\\
Initial& 0.089 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.089 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_b$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_b = \frac{P_{Products}}{P_{Reactants}} = \frac{[ HOCl ][ OH^- ]}{[ OCl^- ]}$$
$$K_b = \frac{(x)(x)}{[ OCl^- ]_{initial} - x}$$
3. Assuming $ 0.089 \gt\gt x$:
$$K_b = \frac{x^2}{[ OCl^- ]_{initial}}$$
$$x = \sqrt{K_b \times [ OCl^- ]_{initial}} = \sqrt{ 3.4 \times 10^{-7} \times 0.089 }$$
$x = 1.7 \times 10^{-4} $
4. Test if the assumption was correct:
$$\frac{ 1.7 \times 10^{-4} }{ 0.089 } \times 100\% = 0.19 \%$$
5. Thus, it is correct to say that $x = 1.7 \times 10^{-4} $
6. $[OH^-] = x = 1.7 \times 10^{-4} $
7. Calculate the pH:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.7 \times 10^{-4} } = 5.9 \times 10^{-11} \space M$$
$$pH = -log[H_3O^+] = -log( 5.9 \times 10^{-11} ) = 10.23 $$
