Equivalent conductance of `BaCI_(2),H_(2)SO_(4)` & `HCI` at infinite are `A_(oo)^(1),A_(oo)^(2)& A_(oo)^(3)` conductance of `BaSO_(4)` solution is:

Options

A. `A_(oo)^(1) +A_(oo)^(2)-2 A_(oo)^(3)`
B. `A_(oo)^(2) +A_(oo)^(2)-2 A_(oo)^(3)`
C. `A_(oo)^(1) +A_(oo)^(2) +A_(oo)^(3)`
D. `A_(oo)^(1) +A_(oo)^(2)-A_(oo)^(3)`

Answer

C

Explanation

`A_(oo)^(1) =^^_(BaCI_(2))^(oo) = (1)/(2) ^^_(eq)^(oo) Ba^(2+) +^^_(eqCI^(-))^(oo)` .....(i)
`A_(oo)^(2) = ^^_(H_(2)SO_(4))^(oo) = ^^_(eqH^(+))^(oo) +(1)/(2)^^_(eq^(SO_(4)^(-2)))^(oo)` .....(ii)
`A_(oo)^(3) = ^^_(HCI)^(oo) = ^^_(H^(+))^(oo) +^^_(CI^(-))^(oo)` ....(iii)
`^^_(eq^(BaCI_(2)))^(oo) =(1)/(2) ^^_(eq^(Ba^(2+)))^(oo) +(1)/(2) ^^_(eq^(SO_(4)^(2-)))^(oo)` ....(iv)
eq. (4) =eq (i) +eq (ii) -eq(iii)
`= A_(oo)^(1) +A_(oo)^(2) - A_(oo)^(3)`