Question

Learning Goal:

To understand the relationship between precipitation and the solubility product and to be able to predict whether a substance will precipitate or not.

Precipitation is the formation of an insoluble substance. For the equation AB(s)⇌A+(aq)+B−(aq), precipitation represents a shift to the left and the production of a solid. From Le Châtelier's principle, we know that when the product of the concentrations of A+ and B− gets above a certain level, the reaction will respond by shifting left to decrease the concentrations of A+ and B−. This critical level, Ksp, is a constant at a certain temperature. In this case, Q=[A+][B−]. where Q is the ion product.

When Q>Ksp, precipitation occurs until Q is equal to Ksp.

When Q≤Ksp, a precipitate does not form.

Part A - Calculate the value of Q

What is the value of Q when the solution contains 2.50×10⁻³M Mg2+ and 2.00×10−3M CO32−?

Express your answer numerically.

Part C

What concentration of the lead ion, Pb2+, must be exceeded to precipitate PbCl2 from a solution that is 1.00×10−2 M in the chloride ion, Cl−? Ksp for lead(II) chloride is 1.17×10⁻⁵ .

Express your answer with the appropriate units.

Answer 1

Part (A)

Reaction:

Mg₂CO₃ Mg²⁺ + CO₃^2-

Q = [Mg²⁺][CO₃^2-]

Q = 2.5x10^-3 x 2x10^-3

Q = 5 x 10^-6 M².

Part (B)

For precipitation to occur, Q must exceed K_sp by a marginal amount. So we calculate the concentration of Pb²⁺ for which Q will equal K_sp and any concentration of Pb²⁺ above that would cause precipitation of PbCl₂.

K_sp = [Pb²⁺][Cl^-]²

1.17x10^-5 = [Pb²⁺](1x10^-2)²

Hence [Pb²⁺] = 1.17x10^-1 M

For precipitation to occur, the concentration of Pb²⁺ must be anything greater than 1.17x10^-1 M.