Answer:
The one that will begin to precipitate first will be the lead chromate (PbCrO₄)
Explanation:
First of all, let's determine the equations involved:
Pb(NO₃)₂ → Pb²⁺ + 2NO₃⁻
0.001 0.001 0.002
Ba(NO₃)₂ → Ba²⁺ + 2NO₃⁻
0.100 0.100 0.200
Sodium chromate as a soluble salt, can be also dissociated in:
Na₂CrO₄ → 2Na⁺ + CrO₄²⁻
As the chromate can react to both cations of the aqueous solution, there will be formed 2 precipitates. When the saturation point is reached, which is determined by the Kps, everything that cannot be dissolved will precipitate.
The first to saturate the solution will precipitate first.
CrO₄²⁻ + Pb²⁺ ⇄ PbCrO₄
s s s² = Kps
Kps = s² ⇒ [CrO₄²⁻] . [Pb²⁺] = 2.80×10⁻¹³
[CrO₄²⁻] . 0.001 = 2.80×10⁻¹³
[CrO₄²⁻] = 2.80×10⁻¹³ / 0.001 = 2.80×10⁻¹⁰
This is the concentration for the chromate when the lead chromate starts to precpitate.
CrO₄²⁻ + Ba²⁺ ⇄ BaCrO₄
s s s² = Kps
Kps = [CrO₄²⁻] . [Ba²⁺]
1.17×10⁻¹⁰ = [CrO₄²⁻] . 0.100
[CrO₄²⁻] = 1.17×10⁻¹⁰ / 0.100 = 1.17×10⁻⁹
The first one that precipitates needs less chromate ion to start precipitating, in conclusion the one that will begin to precipitate first will be lead chromate.
