The `K_(sp)` of `AgCl` at `25^(@)C` is `1.5 xx 10^(-10)`. Find the solubility (in `gL^(-1))` in an aqueous solution containing `0.01M AgNO_(3)`.

Explanation

Now this is quite similar to finding the degree of dissociation of a weak acid or weak base in presence of its common ion (Le Chatelier 's principle).
Here `0.01 M AgNO_(3)` will give `0.01 M Ag^(+)` ions in solution before the addition of AgCl. (`AgNO_(3)` is 100% soluble in water ).
Let the solubility be c mol/L in `AgNO_(3)`.
`{:(AgCl hArr Ag^(+) + Cl^(-)),(" "c" "c):}`
`AgNO_(3) underset(0.01 M) to Ag^(+) + NO_(3)^(-)`
`[Ag^(+)] = c + 0.01 " and " [Cl^(-)] = c`
Now `K_("sp")` is a constant and is independent of presence of any ion in water (stated before)
`K_("sp") = [Ag^(+)][Cl^(-)] = (0.01 +c ) (c) = 1.50 xx 10^(-10)`
(Now assuming `0.01 + c ~ 0.01 " as " c lt lt 1` for sparingly soluble salt )
`rArr c = K_("sp")/(0.01) = 1.5 xx 10^(-8) M`
`[" Check that " c + 0.01 ~~ 0.01 M]`
or solubility ` = (1.5 xx 10^(-8)) xx 143.5`
` = 2.15 xx 10^(6) ` g/L