Using the table of acid-base couples, calculate ( by means of the simplified formula ) the $pH$ a solution $S$ obtained by dissolving 5.77 $g$ lactic (2-hydroxypropanoic) acid in a volumetric flask of $1\;L$.
$CH_3CHOHCOOH$ weak acid $n_{CH_3CHOHCOOH}$ $=$ $\frac{m_{CH_3CHOHCOOH}}{M_{CH_3CHOHCOOH}}$ = $\frac{5.77}{90.08}$ = 0.064$ \;mol$ $c$ $=$ $ c_{CH_3CHOHCOOH}$ = $\frac{n_{CH_3CHOHCOOH}}{V_S}$ = $\frac{0.064}{1}$ = 0.064 $\frac{mol}{L}$ $pH$ $=$ $\frac{1}{2}pK_a$ $-$ $\frac{1}{2}log\;c$ = $\frac{1}{2}$ 3.86-$\frac{1}{2}log $ 0.064 = 2.53
