Question

The concentration of hydrogen ions in a $0.20M$ 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

$HCOOH\rightleftharpoons H^{+}+HCO{O}^{-}$

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