Question

Calculate molar conductance for `NH_(4)OH`, given that molar conductance for `Ba(OH)_(2), BaCI_(2)` and `NH_(4)CI` are `523.28, 280.0` and `129.8 ohm^(-1) cm^(2) mol^(-1)` respectively.

Answer 1

`mu_(Ba(OH)_(2))^(prop) = lambda_(Ba^(2+))^(prop) +2lambda_(OH^(-))^(prop) = 523.28` ...(i)
`mu_(BaCI_(2))^(prop) = lambda_(Ba^(2+))^(prop) +2lambda_(CI^(-))^(prop) = 280.00` ...(ii)
`mu_(NH_(4)CI)^(prop) = lambda_(NH_(4)^(+))^(prop) +lambda_(CI^(-))^(prop) = 129.80` ...(iii)
`mu_(NH_(4)OH)^(prop) = lambda_(NH_(4)^(+))^(prop) +lambda_(OH^(-))^(prop) = ?`
Eq.(iii) `+(Eq(i))/(2) -(Eq.(ii))/(2)` will gives
`lambda_(NH_(4)^(+))^(prop) +lambda_(OH^(-))^(prop) = lambda_(NH_(4)OH)^(prop) = (502.88)/(2) = 251.44 ohm^(-1) cm^(2) mol^(-1)`