The solubility product of AgCl is `4.0 xx 10^(-10)` at 298 K . The solubility of AgCl in 0.04 M Ca `Cl_(2)` will be

Options

A. `2.0 xx 10^(-5)`M
B. `1.0 xx 10^(-4)` M
C. `5.0 xx 10^(-9)` M
D. `2.2 xx 10^(-4)`M

Answer

C

Explanation

`{:(,CaCl_(2),hArr,Ca^(2+),+,2Cl^(-)),("Concen",0.04,,0.04,,0.08),("For AgCl",,,,,):}`
`AgCl hArr Ag^(+) +Cl^(-)`
`K_(sp) = [Ag^(+)] [Cl^(-)] `
`4.0 xx 10^(-10) = [S] [S+0.08] `
`4.0 xx 10^(-10) = S^(2) S xx 0.08 [S^(2) lt lt lt 1] `
`4.0 xx 10^(-10) = S xx 0.08`
`S = (4.0 xx10^(-10))/(0.08) = 0.5 xx10^(-8) = 5 xx 10^(-9)`