The solubility of AgCl in 0.2 M NaCl solution (`K_(sp)` for `AgCl = 1.20 xx 10^(-10)`) is

Options

A. 0.2 M
B. `1.2 xx 10^(-10)` M
C. `0.2 xx 10^(-10)` M
D. `6.0 xx 10^(-10)`M

Answer

D

Explanation

`underset (a) ("AgCl") hArr underset (a) (Ag^(+)) + underset (a)(Cl^(-))`
`underset (0.2) (NaCl) hArrunderset (0.2)(Na^(+)) + underset (0.2)(Cl^(-))`
`K_(sp) AgCl = 1.20 xx10^(-10)`
`K_(sp) "AgCl" = [Ag^(+)] [Cl^(-)] = a xx [a+ 0.2] = a^(2) +0.2 a `
`a^(2)` is a very small so it is a neglected.
`K_(sp) "AgCl" = 0.2 a`
`1.20 xx 10^(-10) = 0.2 a `
`a = (1.20 xx10^(-10))/(0.20) = 6 xx 10^(-10) ` mole .