1.

The ionization constant of a weak acid is 1.6xx10^(-5) and the molar conducitvity at infinite dilutionis 380xx10^(-4)" S " m^(2) mol^(-1). If the cell constant is 0.01 m^(-1), then the conductance of 0.01 M solutionis :

Answer»

`1.52xx10^(-5)" S "`
`1.52" S "`
`1.52xx10^(-3)" S "`
`1.52xx10^(-4)" S "`

Solution :(d) For a WEAK acid `K_(a)=C alpha^(2)`
`K_(a)=C((lambda_(m)^(c ))/(lambda_(m)^(alpha)))^(2)`
`1.6xx10^(-5)=0.01((lambda_(m)^(c ))/(380xx10^(-4)))^(2)`
`16xx10^(-6)=10^(-2)((lambda_(m)^(c ))/(380xx10^(-4)))^(2)`
`(lambda_(m)^(c ))/(380xx10^(-4))=(16xx10^(-4))^(1//2)=4XX10^(-2)`
`lambda_(m)^(c )=(380xx10^(-4))xx(4xx10^(-2))`
`=152xx10^(-5)=1.52xx10^(-3)`


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