Chemical and Ionic Equilibrium - Result Question 132
####74. 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 .
$(1985,3 M)$
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Solution:
- $HCOOH \rightleftharpoons H^{+}+HCOO^{-}$
$\underset{1-0.75}{HCOONa} \rightleftharpoons Na^{+}+HCOO^{-}$
In the above buffer solution, the significant source of formate ion $\left(HCOO^{-}\right)$is $HCOONa$. Hence,
$$ \begin{aligned} K _a & =2.4 \times 10^{-4} \ & =\frac{\leftH^{+}\right}{[HCOOH]} \ {\left[H^{+}\right] } & =\frac{2.4 \times 10^{-4} \times 0.20}{0.75}=6.4 \times 10^{-5} \ pH & =-\log \left(6.4 \times 10^{-5}\right)=4.20 \end{aligned} $$
