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Calculate lambda(m)^(@) for NH(4)OH give...

Calculate `lambda_(m)^(@)` for `NH_(4)OH` given that values of `lambda_(m)^(@)` for `Ba(OH)_(2)`, `BaCI_(2)` and `NH_(4)CI` as `523.28`, `280.0` and `129.8 S cm^(2) mol^(-1)` respectively

Text Solution

Verified by Experts

`lambda_(m)^(@)(NH_(4)OH) = lambda_(m)^(@) (NH_(4)^(+)) + lambda_(m)^(@) (OH^(-))`
Now,
`Lambda_(m)^(@) (NH_(4)OH) = lambda_(m)^(@) (NH_(4)^(+)) + lambda_(m)^(@)(OH^(-)) +1/2 lambda_(m)^(@) (Ba^(2+)) -1/2lambda_(m)^(@)(Ba^(2+)) + lambda_(m)^(@)(Cl^(-)) - lambda_(m)^(0)(Cl^(-))`
`Lambda_(m)^(@)(NH_(4)OH) = lambda_(m)^(@)(NH_(4)Cl) +1/2 lambda_(m)^(@) (Ba(OH))_(2) -1/2lambda_(m)^(@)(BaCl_(2))`
`=129.8 + 1/2 lambda_(m)^(@) (Ba(OH))_(2) -1/2 lambda_(m)^(@)(BaCl_(2))`
`=129.8 + 1/2(523.28 - 280.0)`
`=251.44 S cm^(2) mol^(-1)`
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