Answer
The pH of that solution is equal to 9.07
Work Step by Step
1. $Na^+$ does not ionize, so we can ignore it. $C_2H_3{O_2}^-$ is the conjugate base of a weak acid; thus, it is a weak base.
2. Calculate its $K_b$:
$$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$$
3. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ C_2H_3O{_2}^- ]& [ HC_2H_3O_2 ]& [ OH^- ]\\
Initial& 0.250 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.250 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
4. Write the expression for $K_b$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_b = \frac{[Products]}{[Reactants]} = \frac{[ HC_2H_3O_2 ][ H^+ ]}{[ C_2H_3O{_2}^- ]}$$
$$K_b = \frac{(x)(x)}{[ C_2H_3O{_2}^- ]_{initial} - x}$$
5. Assuming $ 0.250 \gt\gt x$:
$$K_b = \frac{x^2}{[ C_2H_3O{_2}^- ]_{initial}}$$
$$x = \sqrt{K_b \times [ C_2H_3O{_2}^- ]_{initial}} = \sqrt{ 5.6 \times 10^{-10} \times 0.250 }$$
$x = 1.183 \times 10^{-5} $
6. Test if the assumption was correct:
$$\frac{ 1.183 \times 10^{-5} }{ 0.250 } \times 100\% = 4.7 \times 10^{-3} \%$$
7. Thus, it is correct to say that $x = 1.183 \times 10^{-5} $
8. $[OH^-] = x = 1.183 \times 10^{-5} $
9. Calculate the pH:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.183 \times 10^{-5} } = 8.453 \times 10^{-10} \space M$$
$$pH = -log[H_3O^+] = -log( 8.453 \times 10^{-10} ) = 9.07 $$
