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The freezing point of a 0.08 molal solut...

The freezing point of a 0.08 molal solution of `NaHSO^(4)` is `-0.372^(@)C`. Calculate the dissociation constant for the reaction.

`K_(f)` for water =`1.86 K m^(-1)`

Text Solution

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`{:(NaHSO_(4) , to, Na^(+),+, HSO_(4)^(-)),(0.08,,0,,0),(,,(0.08),,(0.08)):}`
Suppose the degree of dissocilation of `HSO_(4)^(@)` is `alpha`
`{:(HSO_(4)^(-) ,hArr H^(+),+SO_(4)^(-2)),("At Eqm" C(1-alpha), Calpha, C alpha):}`
`rArr [H^(+)]=0.08 alpha`
`[SO_(4)^(-2)]=0.08 alpha`
`[HSO_(4)]=0.08(1-alpha)`
total ions present `=0.08+0.08(1-alpha)+0.08 alpha+0.08 alpha`
`=0.16+0.08 alpha`
`=(0.16+0.08 alpha)/(0.08)=(2+alpha)`
`DeltaT_(f)=K_(f)xx mxxi`
`0.372=(2+alpha)xx1.86xx0.08`
`rArr alpha=0.5`
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