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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Answer:
Correct Answer: 74. $(4.20)$
Solution:
- $HCOOH \rightleftharpoons H^{+}+HCOO^{-}$
$\underset{1-0.75}{HCOONa} \rightleftharpoons Na^{+}+ \underset{0.75}{HCOO^{-}}$
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 . 7 5 )}{[HCOOH]} $
${\left[H^{+}\right] } =\frac{2.4 \times 10^{-4} \times 0.20}{0.75}=6.4 \times 10^{-5} $
$pH =-\log \left(6.4 \times 10^{-5}\right)=4.20$
BETA
