The self ionisation constant for pure `HCOOH, K = [HCOO overset(o+)H_(2)] [HCOO^(Θ)]` is `10^(-6)` at room temperature. What percentage of `HCOOH` molecules are converted to `HCOO^(Θ)` ions. The density of `HCOOH` iws `1.22 g cm^(-3)`.

Options

A. `0.002%`
B. `0.004%`
C. `0.006%`
D. `0.008%`

Answer

B

Explanation

Given density of formic acid `= 1.22g//cm^(3)`
`:.` Weight of acid in 1 litre solution
`= 1.22xx10^(3)g`
Thus, `[HCOOH]= (1.22xx10^(3))/(46)=26.5M`
Since in case of auto-ionisation
`[HCOOH_(2)^(+)]=[HCOO^(-)]`
and `[HCOO^(-)][HCOOH_(2)^(+)]=10^(-6)`
`[HCOO^(-)]=10^(-3)`
Now % dissociation of HCOOH
`=([HCOO^(-)]xx100)/([HCOOH])=(10^(-3))/(26.5)xx100`
`= 0.004%`