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At 25 ^(@)C , the molar conductance of 0...

At `25 ^(@)C` , the molar conductance of 0.007 M hydrofluoric acid is 150 mho `cm^(2)` `mol^(-1)` and its `Lambda_(m)^(@)` = 500 mho `cm^(2)` `mol^(-1)` . The value of the dissociation constant of the acid at the given concentration at `25^(@)C` is

A

`7 xx 10^(-4)` M

B

`7 xx 10^(-5) M`

C

`9 xx 10^(-3) M`

D

`9 xx 10^(-4) M`

Text Solution

Verified by Experts

The correct Answer is:
D
`Lambda_(m) = 150 mho cm^(2) mol^(-1)`
`Lambda_(m)^(@) = 500 mho cm^(2) mol^(-1)`
Degree of dissociation `(alpha) = (Lambda_m)/(Lambda_m^@) = (150)/(500) =0.3`
`" "HF hArr H^(+) + F^(-)`
`Final : C - Calpha " "C alpha " "Calpha`
`therefore K_(a)= (C_(alpha).C_(alpha))/(C-C_(alpha)) = (Calpha^(2))/(1 - alpha)`
`K_(a) = (7 xx10^(-3) xx (0.3)^(2))/(1-0.3)=(7xx 10^(-3)xx9 xx 10^(-2))/(0.7)`
`rArr K_(a) = (7xx9xx 10^(-5))/(7xx10^(-1)) = 9 xx10^(-4) M`
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