Using the table of acid-base couples, calculate ( by means of the simplified formula ) the $pH$ a solution $S$ obtained by dissolving 4.67 $g$ acetic (ethanoic) acid in a volumetric flask of $1\;L$.
$CH_3COOH$ weak acid $n_{CH_3COOH}$ $=$ $\frac{m_{CH_3COOH}}{M_{CH_3COOH}}$ = $\frac{4.67}{60.05}$ = 0.078$ \;mol$ $c$ $=$ $ c_{CH_3COOH}$ = $\frac{n_{CH_3COOH}}{V_S}$ = $\frac{0.078}{1}$ = 0.078 $\frac{mol}{L}$ $pH$ $=$ $\frac{1}{2}pK_a$ $-$ $\frac{1}{2}log\;c$ = $\frac{1}{2}$ 4.75-$\frac{1}{2}log $ 0.078 = 2.93
