The expression to compute pH of sodium acetate solution is

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$pH = \frac{1}{2} p K_{w}^{\circ} + \frac{1}{2} p K_{a}^{\circ} ({CH}_{3} COOH) + \frac{1}{2} \log ([salt] / M)$

$pH = \frac{1}{2} p K_{w}^{\circ} + \frac{1}{2} p K_{a}^{\circ} ({CH}_{3} COOH) - \frac{1}{2} \log ([salt] / M)$

$pH = \frac{1}{2} p K_{a}^{\circ} ({CH}_{3} COOH) - \frac{1}{2} p K_{w}^{\circ} + \frac{1}{2} \log ([salt] / M)$

$pH = \frac{1}{2} p K_{w}^{\circ} - \frac{1}{2} p K_{a}^{\circ} ({CH}_{3} COOH) - \frac{1}{2} \log ([salt] / M)$

answer is C.

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Detailed Solution

$\begin{array}{l} {Ac}^{-} + H_{2} O ⇌ HAc + {OH}^{-} \\ K_{h} = \frac{K_{w}}{K_{a}} = \frac{[HAc] [{HO}^{-}]}{[{Ac}^{-}]} ≃ \frac{{[{OH}^{-}]}^{2}}{{[{Ac}^{-}]}_{0}} \end{array}$ ${[{OH}^{-}]}^{2} = \frac{K_{w}}{K_{b}} {[{Ac}^{-}]}_{0}; {(\frac{K_{w}}{{[H^{+}]}^{2}})}^{2} = \frac{K_{w}}{K_{a}} {[{Ac}^{-}]}_{0}$