If to 100 mL of 0.1 M acetic acid solution, 0.82 g of sodium acetate is added, the pH of the solution (assume no dilution factor) becomes

$M\left({\mathrm{CH}}_3\mathrm{COONa}\right)=82\mathrm{g}{\mathrm{mol}}^{-1}$

$n\left({\mathrm{CH}}_3\mathrm{COONa}\right)=m/M=0.82\mathrm{g}/82\mathrm{g}{\mathrm{mol}}^{-1}=0.01\mathrm{mol}$

$c\left({\mathrm{CH}}_3\mathrm{COONa}\right)=n/V=0.01\mathrm{mol}/0.1\mathrm{L}=0.1\mathrm{M}$

$\mathrm{pH}=\mathrm{p}{K}_{\mathrm{a}}+\log \left[\text{ salt }\right]/\left[\text{ acid }\right]=-\log \left(1.8\times {10}^{-5}\right)+\log (0.1/0.1)=4.75$