The self ionisation constant for pure formic acid `K=[HCOOH_(2)^(+)][HCOO^(-)]` has been estimated as `10^(-6)` at room temperature .The density of fo

Question

The self ionisation constant for pure formic acid `K=[HCOOH_(2)^(+)][HCOO^(-)]` has been estimated as `10^(-6)` at room temperature .The density of 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?
A. `0.002%`
B. `0.004%`
C. `0.006%`
D. `0.008%`

Monjur Alahi

7 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Given density of formic acid =`1.15g//cm^(3)`
:.Weight of formic acid in `1` litre solution =`1.15xx10^(3)g`
thus `[HCOOH]=(1.15xx10^(3))/(46)=25M`
since in case of auto ionisation
`[HCOOH_(2)^(+)]=[HCOO^(-)]` and `[HCOO^(-)][HCOOH_(2)^(+)]=10^(-6)rArr [HCOO^(-)]=10^(-3)`
Now `%` dissociation of `HCOOH=([HCOO^(-)]xx100)/([HCOOH])=(10^(-3))/(25)xx100=0.004%`