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The equivalent conductance of 0.1 N of H...

The equivalent conductance of 0.1 N of `H_(3)PO_(4)` at `18^(f@)C` is 96.5 `Omega^(-1) cm^(2) eq^(-1)` if `wedge_(HCI)^(@)=378.3 , wedge_(NaCI)^(@)=109 , wedge_(NaH_(2)PO_(4))^(@)=70 Omega^(-1) cm^(-2)eq^(-1)` respectively calculate the degree of dissoication and dissoication constant for the reaction
`H_(3)PO_(4)rarrH_(2)PO_(4)`

A

`alpha=0.668 ,K =6.63 xx10^(-2)`

B

`alpha=0.389 ,K =5.53 xx10^(-2)`

C

`alpha=0.2844 ,K =1.13 xx10^(-2)`

D

`alpha=0.2587 ,K =2.56 xx10^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
C

`alpha=0.2844 ,K =1.13 xx10^(-2)`
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