An aqueous solution of Na2CrO4 at 25oC is slowly added to an aqueous solution containing 0.001 M Pb(NO3)2and 0.100 M Ba(NO3)2. Which solid will precipitate first? The Ksp of BaCrO4 is 1.17 × 10−10, and Ksp of PbCrO4 is 2.80 × 10−13.

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.

the dissolution of polymer can be thermodynamically described in 2 steps.