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The self ionisation constant for pure fo...

The self ionisation constant for pure formic acid `K=[HCOOH_(2)^(+)][HCOO^(-)]` 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 fomic acid are converted to formation ion?

Text Solution

Verified by Experts

The correct Answer is:
4
Given that `d_(HCOOH)=1.22g//cm^(3)`
`:.` Mass of formic acid in `1L` solution `=1.22xx10^(3)g`
or `[HCOOH]=(1.22xx10^(3))/(46)=26.5M`
`HCOOH` auto ionizes as
`2HCOOHhArr[HCOOH_(2)]^(+)+HCOO^(-)`
and thus `[HCOOH_(2)^(+)]+[HCOO^(-)]`
and `[HCOO^(-)][HCOOH_(2)^(+)]=10^(-6)implies[HCOO^(-)]=10^(-3)`
or `%` dissociation `(HCOOH)=([HCOO^(-)]xx100)/([HCOOH])=(10^(-3))/(26.5)xx100`
`=0.004%=4xx10^(-3)`
`=x=4`
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