A solution cotaning 0.10 M in Ba(NO_(3))_(2) and 0.10 M in Sr(NO_(3))_(2). If solid Na_(2)CrO_(4) is added to the solution, what is [Ba^(2+)], when SrCrO_(4) beings to precipitate? [K_(sp)(BaCrO_(4)) = 1.2 xx 10^(-10), K_(sp) (SrCrO_(4)) = 3.5 xx 10^(-5)]

A solution cotaning 0.10 M in Ba(NO_(3))_(2) and 0.10 M in Sr(NO_(3))_

1.	A solution cotaning 0.10 M in Ba(NO_(3))_(2) and 0.10 M in Sr(NO_(3))_(2). If solid Na_(2)CrO_(4) is added to the solution, what is [Ba^(2+)], when SrCrO_(4) beings to precipitate? [K_(sp)(BaCrO_(4)) = 1.2 xx 10^(-10), K_(sp) (SrCrO_(4)) = 3.5 xx 10^(-5)]
Answer» `7.4 XX 10^(-7)` `2.0 xx 10^(-7)` `6.1 xx 10^(-7)` `3.4 xx 10^(-7)` Solution :`K_(SP)(SrCrO_(4)) = [Sr^(2+)][CrO_(4)^(2-)]` `[CrO_(4)^(2-)] = (3.5 xx 10^(-5))/(0.1)= 3.5 xx 10^(-4)` `K_(sp)(BaCrO_(4)) = [Ba^(2)][CrO_(4)^(2-)]` `[CrO_(4)^(2-)]_("total") = [CrO_(4)^(2-)]` from `SrCrO_(4)` `[Ba^(2+)] = (1.2 xx 10^(-10))/(3.5 xx 10^(-4)) = 3.4 xx 10^(-7)`