Answer
pH = 9.18
Work Step by Step
1. Calculate the Ka of the acid.
$$K_a = 10^{-pKa} = 10^{-4.77} = 1.7 \times 10^{-5}$$
2. The compound have the conjugate base of that acid:
$$K_b = \frac{10^{-14}}{1.7 \times 10^{-5}} = 5.9 \times 10^{-10}$$
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1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ C_6H_7O{_2}^- ]& [ C_6H_7O_2H ]& [ OH^- ]\\
Initial& 0.37 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.37 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_b$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_b = \frac{P_{Products}}{P_{Reactants}} = \frac{[ C_6H_7O_2H ][ OH^- ]}{[ C_6H_7O{_2}^- ]}$$
$$K_b = \frac{(x)(x)}{[ C_6H_7O{_2}^- ]_{initial} - x}$$
3. Assuming $ 0.37 \gt\gt x$:
$$K_b = \frac{x^2}{[ C_6H_7O{_2}^- ]_{initial}}$$
$$x = \sqrt{K_b \times [ C_6H_7O{_2}^- ]_{initial}} = \sqrt{ 5.9 \times 10^{-10} \times 0.37 }$$
$x = 1.5 \times 10^{-5} $
4. Test if the assumption was correct:
$$\frac{ 1.5 \times 10^{-5} }{ 0.37 } \times 100\% = 4.1 \times 10^{-3} \%$$
5. Thus, it is correct to say that $x = 1.5 \times 10^{-5} $
6. $[OH^-] = x = 1.5 \times 10^{-5} $
7. Calculate the pH:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.5 \times 10^{-5} } = 6.67 \times 10^{-10} \space M$$
$$pH = -log[H_3O^+] = -log( 6.67 \times 10^{-10} ) = 9.18 $$
