Using the table of acid-base couples, calculate ( by means of the simplified formula ) the $pH$ a solution $S$ obtained by dissolving 3.11 $g$ formic (methanoic) acid in a volumetric flask of $1\;L$.
$HCOOH$ weak acid $n_{HCOOH}$ $=$ $\frac{m_{HCOOH}}{M_{HCOOH}}$ = $\frac{3.11}{46.03}$ = 0.068$ \;mol$ $c$ $=$ $ c_{HCOOH}$ = $\frac{n_{HCOOH}}{V_S}$ = $\frac{0.068}{1}$ = 0.068 $\frac{mol}{L}$ $pH$ $=$ $\frac{1}{2}pK_a$ $-$ $\frac{1}{2}log\;c$ = $\frac{1}{2}$ 3.75-$\frac{1}{2}log $ 0.068 = 2.46
