Solid AgNO_(3) is added to a solution which 0.1 M in Cl^(-) and 0.1 M in CrO_(4)^(2-). K_(sp) values for AgCl and Ag_(2)CrO_(4) are 1.7 xx 10^(-10) and 1.9 xx 10^(-12) respectively. The concentration of Cl^(-) when Ag_(2)CrO_(4) starts precitating will be

Solid AgNO_(3) is added to a solution which 0.1 M in Cl^(-) and 0.1 M

1.	Solid AgNO_(3) is added to a solution which 0.1 M in Cl^(-) and 0.1 M in CrO_(4)^(2-). K_(sp) values for AgCl and Ag_(2)CrO_(4) are 1.7 xx 10^(-10) and 1.9 xx 10^(-12) respectively. The concentration of Cl^(-) when Ag_(2)CrO_(4) starts precitating will be
Answer» `3.9 xx 10^(-5) M` `2.9 xx 10^(-5) M` `3.9 xx 10^(-3) M` `1.9 xx 10^(-2) M` SOLUTION :`:' K_(sp) (AgCl) = [AG^(+)][Cl^(-)]` To precipitate `AgCl`, `:. [Ag^(+)] = (1.7 xx 10^(-10))/(0.1) = 1.7 xx 10^(-9)` `:' K_(sp) (Ag_(2)CrO_(4)) = [Ag^(+)]^(2)[CrO_(4)^(2-)]` To precipitate `Ag_(2)CrO_(4)` `:. [Ag^(+)] = SQRT((1.9 xx 10^(-12))/(0.1)) = 4.34 xx 10^(-6) M` Since `[Ag^(+)]` required to precinpitate `AgCl` is less than that required to precipitate `Ag_(2)CrO_(4), AgCl` will precipitate first. Now since `[Ag^(+)] = 4.34 xx 10^(-6)` when `Ag_(2)CrO_(4)` starts can be CALCULATED as `K_(sp)[AgCl] = [Ag^(+)] [Cl^(-)]` or `1.7 xx 10^(-10) = (4.34 xx 10^(-6))[Cl^(-)]` or `[Cl^(-)] = 3.9 xx 10^(-5) M`.