Question

Half of the formic acid solution in 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: $(\log 2=0.3010)$.

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

Answer 1

The correct option is A $3.6990$

$K_a=2\times {10}^{-4}$

$p{K}_a=-lo{g}_{10}{K}_a$

$p{K}_a=-lo{g}_{10}2\times {10}^{-4}$

$p{K}_a=-lo{g}_{10}2-lo{g}_{10}{10}^{-4}$

$p{K}_a=-0.3010+4$

$p{K}_a=3.6990$

At half neutralisation point (i.e, when half of formic acid solution is neutralised on addition of KOH solution), $pH=p{K}_a$

$\therefore pH=3.6990$