Question

The self ionization constant for pure formic acid $K=\left[HCOO{H}_2^{+}\right]\left[HCO{O}^{-}\right]$ has been estimated as ${10}^{-6}$ at room temperature. The density of formic acid is $1.15g/{cm}^3$. What percentage of formic acid molecules in pure formic acid are converted to formate ion?

• A. $0.002$%
• B. $0.004$%
• C. $0.006$%
• D. $0.008$%

Answer 1

The correct option is B $0.004$%

Answer is option $B.$

Density of formic acid$=1.15g/c{m}^3$

$\therefore$ Weight of formic acid in $1L$ solution $=1.15\times {10}^3$

$\therefore$ $\left[HCOOH\right]=\frac{1.15\times {10}^3}{46}=25M$

In case of auto ionisation,

$\left[HCOO{H}_2^{+}\right]\left[HCO{O}^{-}\right]={10}^{-6}$

$\left[HCO{O}^{-}\right]=\left[HCOO{H}_2^{+}\right]={10}^{-6}$

Now $\%$ dissociation of $HCOOH,$

$\frac{\left[HCO{O}^{-}\right]\times 100}{\left[HCOOH\right]}⟹\frac{{10}^{-3}\times 100}{25}=0.004\%$