Chemical and Ionic Equilibrium 2 Question 76
74. The concentration of hydrogen ions in a $0.20 \mathrm{M}$ solution of formic acid is $6.4 \times 10^{-3} \mathrm{~mol} / \mathrm{L}$. 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 $\mathrm{pH}$ 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 .
$(1985,3 \mathrm{M})$
Show Answer
Solution:
- $\mathrm{HCOOH} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HCOO}^{-}$
$\underset{1-0.75}{\mathrm{HCOONa}} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{HCOO}^{-}$
In the above buffer solution, the significant source of formate ion $\left(\mathrm{HCOO}^{-}\right)$is $\mathrm{HCOONa}$. Hence,
$$ \begin{aligned} K_{a} & =2.4 \times 10^{-4} \ & =\frac{\left\mathrm{H}^{+}\right}{[\mathrm{HCOOH}]} \ {\left[\mathrm{H}^{+}\right] } & =\frac{2.4 \times 10^{-4} \times 0.20}{0.75}=6.4 \times 10^{-5} \ \mathrm{pH} & =-\log \left(6.4 \times 10^{-5}\right)=4.20 \end{aligned} $$
