The conductivity of a saturated solution of `AgCl` at `25^(@)C` after subtracting the conductivity of water is `2.28 xx 10^(-6) S cm^(-1)`. Calculate the solubility product of AgCl at `25^(@)C` if `Lambda_(m)^(0) (AgCl)` 138.3 S cm^(2) mol^(-1)`

Options

A. `1.7 xx 10^(-10) (mol//L)^(2)`
B. `3.7 xx 10^(-10) (mol//L)^(2)`
C. `2.7 xx 10^(-10) (mol//L)^(2)`
D. `5.0 xx 10^(-10) (mol//L)^(2)`

Answer

C

Explanation

For the equilibrium
`AgCl(s) hArr Ag^(+)(aq.) +Cl^(-) (aq.)`
`K_(sp) = [Ag^(+)][CI^(-)]`
Let `"S' mol L^(-1)` be the solubility of `AgCl` in water then
`K_(sp.) = S.S = S^(2)`
If `S` moles of `AgCl` are dissolved in `1000 cm^(3) (1 L)` then volume containing `1 mol` of `AgCl` called dilution `(V)` is given as
`V = (1000)/(S)`
According to Equation (3.23)
`Lambda_(AgCL) = kappaV`
`= (2.28 xx 10^(-6)S cm^(-1))((1000)/(S)) cm^(3)mol^(-1)`
Since the sparingly soluble salts are at extremely low con-centrations even in a saturated solutions, the sitution can be treated as zero concentration or infinite dilution. Hence, the molar conductivity of a saturated solution of a sparingly soluble salt can be assumed to be equal to the molar conduc-tivity at infinite dilution i.e.
`Lambda_(m)(AgCl) = Lambda_(m)^(@)(AgCl)`
Thus
`2.28 xx 10^(-6) xx (1000)/(S) = 138.3`
or ` S = 1.648 xx 10^(-5) mol L^(-1)`
Finally
`K_(sp) = S^(2) = (1.648 xx 10^(-5) mol L^(-1))^(2)`
`= 2.7 xx 10^(-10) (mol L^(-1))^(2)`