Question

If $50$ ml of $0.2$ M $KOH$ is added to $40$ ml of $0.5$ M $HCOOH$, the $pH$ of the resulting solution is: $(K_a=1.8\times {10}^{-4})$

• A. $3.4$
• B. $7.5$
• C. $5.6$
• D. $3.75$

Answer 1

Answer: (D)

$3.75$

The relationship between the $pH$ and $p{K}_a$ of a solution containing a weak acid $HCOOH$ and its salt $HCOOK$ is as given below.

$pH=-log{k}_a+log\frac{salt}{acid}$

KOH is added to an excess of formic acid. Thus KOH will be completely converted to form a salt. Part of formic acid will remain unreacted.

The concentration of salt is $\left[salt\right]=\frac{0.2\times 50}{1000}=0.01$ M

The concentration of acid is $\left[acid\right]=\frac{0.5\times 40-0.2\times 50}{1000}=0.01$ M

Substitute values in the expression for the relationship between $pH$ and $p{K}_a$.

$pH=-log(1.8\times {10}^{-4})+log\frac{0.01}{0.01}=4-log\left(1.8\right)=3.75$