Question

Half of the formic acid solution is neutralised on addition of a $KOH$ solution to it. If $K_a\left(HCOOH\right)=2\times {10}^{-4}$, then $pH$ of the solution is:

$(log2=0.3010)$

• A. $3.6990$
• B. $10.3010$
• C. $3.85$
• D. $4.3010$

Answer 1

The correct option is A $3.6990$

$HCOOH+KOH\to HCOOK+H_{2}O$

Let initial concentration of $HCOOH$ be $C,$ then after reaction it becomes $C/2$ and $c/2$ concentration of $HCOOK$ also produced$.$ But since $HCO{O}^{-}$ is conjugate base of $HCOOH,$ a buffer solution is formed$.$So, we apply the buffer formula

$pH=p{K}_a+\log \left(\frac{salt}{acid}\right)$

$=p{K}_a=-\log \left(2\times {10}^{-4}\right)$

$=$ $3.6990$

$=3.7$

Hence$,$ option $\left(A\right)$ is correct