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The freezing point depression of a 0.109...

The freezing point depression of a `0.109M` aq. Solution of formic acid is `-0.21^(@)C`. Calculate the equilibrium constant for the reaction,
`HCOOH (aq) hArr H^(+)(aq) +HCOO^(Theta)(aq)`
`K_(f)` for water `= 1.86 kg mol^(-1)K`

Text Solution

Verified by Experts

The correct Answer is:
`1.44 xx 10^(-4)`
`DeltaT_(f) = (1 - alpha) k_(f) xx m`
`0.21 = (1+alpha) xx 1.86 xx 0.109`
`1 +alpha = 1.0358`
`k_(a) = (Calpha^(2))/(1-alpha) = (1.109(0.0358)^(2))/(1-0.0358) = 1.44 xx 10^(-4)`
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