Question

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.

Answer 1

The one that will begin to precipitate first will be the lead chromate (PbCrO₄)

Step-by-step 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.