`K_(sp)` of AgCl is `2.8 xx 10^(-10)` at `25^(@)C`. Calculate solubility of AgCl in.
a. Pure water b. `0.1M AgNO_(3)`
c. `0.1M KCl` or `0.1M NaCl`

Explanation

a. In pure water
Let solubility of `AgC1` be `S mol L^(-1)`
For `AgC1 (s) hArr Ag^(o+) (aq) + C1^(Θ) (aq)`
`K_(sp) = [Ag^(o+)] [C1^(Θ)] = S xxS`
`:. S = sqrt((K_(sp)))=sqrt((2.8 xx 10^(-10))) = 1.673 xx 10^(-5)mol L^(-1)`
b. IN `0.1M AgNO_(3)`
`{:(AgC1(s)hArr,Ag^(o+)+,C1^( Θ),,),(,S,S,,):}`
`AgNO_(3) rarr Ag^(o+) + NO_(3)^(Θ)`
`:. K_(sp) = [Ag^(o+)] [C1^(Θ)] = (0.1 +S) (S)`
`(S lt lt lt 0.1`, presence of common ion decreases solubility)
`:. S (0.1) = 2.8 xx 10^(-10) :. S = 2.8 xx 10^(-9)M`.
c. In `0.1 M KC1`
`{:(AgC1(s)hArr,Ag^(o+)+,C1^( Θ),,),(,S,S,,):}`
`{:(KCIrarr,K^(o+),+CI^(Θ),,),(,0.1,0.1,,):}`
`K_(sp) = [Ag^(o+)] [C1^(Θ)] = (S) (S + 0.1) (S lt lt 0.1M)`
`:. S (0.1) = 2.8 xx 10^(-10)`
`S = 2.8 xx 10^(9) M`
Alternative method for (b) and (c).
For (b) and (c), `S_("new") = (K_(sp))/((C)^(n)) = (2.8 xx 10^(-10))/((0.1)^(1))`
`= 2.8 xx 10^(-9)M`
where `C =conc`. of the common ion added
`n =` number of common ion in `AgC1`
In (b) `Ag^(o+)` and in (c) `C1^(Θ)` are the common ion.