The expression to compute pH of sodium acetate solution is

$\mathrm{pH}=\frac{1}{2}\mathrm{p}{K}_{\mathrm{w}}^{\circ }+\frac{1}{2}\mathrm{p}{K}_{\mathrm{a}}^{\circ }\left({\mathrm{CH}}_3\mathrm{COOH}\right)+\frac{1}{2}\log \left(\right[\mathrm{salt}]/\mathrm{M})$

$\mathrm{pH}=\frac{1}{2}\mathrm{p}{K}_{\mathrm{w}}^{\circ }+\frac{1}{2}\mathrm{p}{K}_{\mathrm{a}}^{\circ }\left({\mathrm{CH}}_3\mathrm{COOH}\right)-\frac{1}{2}\log \left(\right[\mathrm{salt}]/\mathrm{M})$

$\mathrm{pH}=\frac{1}{2}\mathrm{p}{K}_{\mathrm{a}}^{\circ }\left({\mathrm{CH}}_3\mathrm{COOH}\right)-\frac{1}{2}\mathrm{p}{K}_{\mathrm{w}}^{\circ }+\frac{1}{2}\log \left(\right[\text{ salt }]/\mathrm{M})$

$\mathrm{pH}=\frac{1}{2}\mathrm{p}{K}_{\mathrm{w}}^{\circ }-\frac{1}{2}\mathrm{p}{K}_{\mathrm{a}}^{\circ }\left({\mathrm{CH}}_3\mathrm{COOH}\right)-\frac{1}{2}\log \left(\right[\mathrm{salt}]/\mathrm{M})$