Question

From the following molar conductivities at infinite dilution `:`
`wedge_(m)^(@)` for `Ba(OH)_(2)=457.6Omega^(1)cm^(2)mol^(-1)`
`wedge_(m)^(@)` for `BaCl_(2)=240.6Omega^(-1) cm^(2)mol^(-1)`
`wedge_(m)^(@)` for `NH_(4)Cl=129.8 Omega^(-1) cm^(2) mol^(-1)`
Calculate `wedge_(m)^(@)` for `NH_(4)OH`.

Answer 1

`wedge_(m[Ba(OH)_(2)])^(@)=lambda_(Ba^(2+))^(@)+2 lambda_(overset(c-)(O)H)^(@)=457.6 S cm^(2)mol^(-1)`
`wedge_(m(BaCl_(2)))^(@)=lambda_(Ba^(2+))^(@)+2 lambda_(Cl^(c-))^(@)=240.6 S cm^(2)mol^(-1)`
`wedge_(m(NH_(4)Cl))^(@)=lambda_(NH_(4)^(o+))^(@)+ lambda_(Cl^(c-))^(@)=129.8 S cm^(2)mol^(-1)`
`wedge_(m(NH_(4)OH))^(@)=wedge_(m(NH_(4)Cl))^(@)+(1)/(2)wedge_(m[Ba(OH)_(2)])^(@)-(1)/(2)wedge_(m(BaCl_(2)))^(@)`
`=129.8+(1)/(2)(457.6)-(1)/(2)(240.6)`
`=238.3S cm^(2)mol^(-1)`