Question

The concentration of hydrogen ions in a $0.20\, M$ solution of formic acid is $6.4 \times 10^{-3}\, mol / L$. To this solution, sodium formate is added so as to adjust the concentration of sodium formate to one mole per litre. What will be the $pH$ of this solution? The dissociation constant of formic acid is $2.4 \times 10^{-4}$ and the degree of dissociation of sodium formate is $0.75$.

Answer 1

$HCOOH \rightleftharpoons H ^{+}+ HCOO ^{-}$

In the above buffer solution, the significant source of formate ion $\left( HCOO ^{-}\right)$is $HCOONa$. Hence,
$K_{a}=2.4 \times 10^{-4} $
$=\frac{\left[H^{+}\right](0.75)}{[H C O O H]} $
${\left[H^{+}\right]=\frac{2.4 \times 10^{-4} \times 0.20}{0.75}}$
$=6.4 \times 10^{-5} $
$pH =-\log 6.4 \times 10^{-5}=4.20$