Question

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)?

Answer 1

Answer:

[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