Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) chloride precipitate. What was the molar concentration of chloride ions in the solution (in mol/L) ?

Question

Jerry Ritter

Chemistry

31 May, 18:57

Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) chloride precipitate. What was the molar concentration of chloride ions in the solution (in mol/L) ?

+1

Answers (1)

Know the Answer?

Master · Answer 1

[Cl⁻] = 0.016M

Explanation:

First of all, we determine the reaction:

Pb (NO₃) ₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓ + Mg (NO₃) ₂ (aq)

This is a solubility equilibrium, where you have a precipitate formed, lead (II) chloride. This salt can be dissociated as:

PbCl₂ (s) ⇄ Pb²⁺ (aq) + 2Cl⁻ (aq) Kps

Initial x

React s

Eq x - s s 2s

As this is an equilibrium, the Kps works as the constant (Solubility product):

Kps = s. (2s) ²

Kps = 4s³ = 1.7ₓ10⁻⁵

4s³ = 1.7ₓ10⁻⁵

s = ∛ (1.7ₓ10⁻⁵. 1/4)

s = 0.016 M