Question

After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2 is

Answer 1

The addition of NaCl to a saturated PbCl2 solution demonstrates the common ion effect, resulting in the decreased equilibrium concentration of Pb2+ ions in the solution, as predicted by Le Châtelier's principle.

Step-by-step explanation:

The student asks about the change in equilibrium concentration of Pb2+ ions when an excess of PbCl2 is mixed in a fixed amount of water and additional sodium chloride (NaCl) is added to the solution. This is a classic example of the common ion effect in solubility equilibria. The presence of a common ion decreases the solubility of a slightly soluble salt.

The dissolution of lead(II) chloride in water is given by the following equilibrium expression:

PbCl2 (s) ⇌ Pb2+ (aq) + 2Cl- (aq)

The initial solubility of PbCl2 would result in a certain concentration of Pb2+ and Cl- ions in solution. With the addition of NaCl, the equilibrium shifts to the left according to Le Châtelier's principle to counter the increase in Cl- ion concentration. This results in the formation of more solid PbCl2, causing a decrease in the concentration of Pb2+ in solution.

Originally, the concentration of lead(II) ions might have been higher, but with extra chloride ions added from NaCl, the solubility is suppressed and Pb2+ concentration decreases. If the chloride ion concentration is assumed to be around 0.1M due to excess NaCl, the equilibrium is further pushed towards the formation of solid PbCl2, reducing the solubility even more.

Answer 2

Ksp = 8.8x10⁻⁵

Step-by-step explanation:

Full question is:

After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?

When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:

PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻

Ksp = [Pb²⁺] [Cl⁻]²

If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):

Ksp = [X] [2X]²

Ksp = 4X³

As X is the amount of Pb²⁺ = 2.8x10⁻²M:

Ksp = 4(2.8x10⁻²)³

Ksp = 8.8x10⁻⁵