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Conductivity of 0.00241M acetic acid is ...

Conductivity of 0.00241M acetic acid is `7.896 xx 10^(-5) S cm^(-1)`. Calculate its molar condcutivity. If `Lambda_(m)^(@)` acetic acid is 390.5 S `cm^(2) "mol"^(-1)`, what is its dissociation constant?

Text Solution

Verified by Experts

(i) According to the question, the given values are
1 S `cm^(=1) = 100 S m^(-1) ` or `( 1 S cm^(-1))/( 100 S m^(-1))= 1`
( unit conversion factor)

So, the graph between `Lambda_(m)` and `C^(1//2)` is shown along side.

(ii) Given `kappa = 7.896 XX 10^(-5) S cm^(-1), M = 0.00241 "mol" L^(-1) , Lambda_(m) = ( kappa xx 1000)/( M )`
`= ( 7.896 xx 10^(-5) xx 1000) // 0.00241 S cm^(2) "mol"^(-1) = ( 7.896 xx 10^(-2) ) // ( 2.41 xx 10^(-3) ) S cm^(2) "mol"^(-1)`
`= 78.96 // 2.41 S cm^(2) "mol"^(-1) = 32.76 S cm^(2) "mol"^(-1)`
Since, degree of dissociation `alpha = ( Lambda_(m))/( Lambda_(m)^(@))`, `alpha = 32.76 // 390 .5 = 0.084`
The dissociation constant, `K = ( C alpha^(2))/( ( 1- alpha))`
Since, `alpha` is small , the expression is simply written as `: K = C alpha^(2)`
Here, C =M= Concentration in mol `L^(-1)` . Therefore, `K = 0.00241 xx ( 0.084)^(2) = 1.7 xx 10^(-5) ` mol `L^(-1)`
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