Question

The solubility product, Ksp, of PbCl2 is 1.6 × 10-5. What mass of KCl(s) must be dissolved in 150.00 mL of 1.15M Pb(NO3)2 to produce a solution that is just saturated in PbCl2?

Answer 1

Solubility product (K_sp)of a salt is the product of molar concentrations of constituent ions, raised to appropriate powers, in a saturated solution of the salt, at a given temperature.

Ionic product (Q_sp) of a salt is the product of molar concentrations of constituent ions, raised to appropriate powers, ​in a given solution, at a given temperature. The constituent ions in the solution may be due to dissolution of two different salts. The solution may contain two or more salts.

A given aqueous solution is said to be saturated with respect to a given salt, if,

Ionic product of the salt (Q_sp) = Solubility product of the salt (K_sp)

Solubility product of PbCl₂ is given as, K_sp = 1.6 * 10^-5

Ionic product of PbCl₂ in the solution obtained after dissolving KCl in given Pb(NO₃)₂ solution is,

Q_sp = [Pb²⁺] * [Cl^- ]²

The solution is saturated with respect to PbCl₂, when for PbCl₂,

Q_sp = K_sp    or   [Pb²⁺] * [Cl^- ]² = 1.6 * 10^-5

Hence, the solution will be saturated in PbCl₂, when,

[Pb²⁺] * [Cl^- ]² = 1.6 * 10^-5 ​ ...............(1)

Given, [Pb(NO₃)₂] = 1.15 M; Pb(NO₃)₂​ is a strong electrolyte and in aqueous solution it is completely dissociated into Pb²⁺ and NO^3- ions.

Pb(NO₃)₂ (aq) Pb²⁺ (aq) + 2 NO₃^- (aq)

So each mole of Pb(NO₃)₂ dissociated in solution leads to a mole of Pb²⁺ ions in the solution.

Hence, [Pb²⁺] = [Pb(NO₃)₂] = 1.15 M

Substituting for [Pb²⁺] in equation (1), we get,

(1.15) * [Cl^- ]² = 1.6 * 10^-5

Therefore, the solution becomes saturated when,

[Cl^- ]²​ = (1.6 * 10^-5) / 1.15 = 1.39 * 10^-5

or [Cl^- ] = 3.73 * 10^-3 M

​Thus, solution becomes saturated in PbCl₂ when, [Cl^- ] = 3.73 * 10^-3 M ..................(2)

KCl is a strong electrolyte and dissociates completely in water to form K⁺ and Cl^- ions.

KCl (aq) K⁺ (aq) + Cl^- (aq)

Hence, dissolution of each mole of KCl in water leads to formation of a mole of Cl^- ions in solution.

or, [Cl^-] = [KCl] .............(3)

So, the solution is saturated in PbCl₂ when,

[KCl] = [Cl^-] = 3.73 * 10^-3 M from results (2) and (3).

So, [KCl] = 3.73 * 10^-3 M is the concentration of KCl required in given 150 mL solution, so that the solution is saturated in PbCl₂.

Volume of the solution is given as 150 mL which is equal to 0.15 L    (since 1000 mL = 1 L)

Number of moles of KCl = (molarity of KCl in solution) * (volume of solution in litre)

Number of moles of KCl =​ (3.73 * 10^-3) * (0.15) = 5.595 * 10^-4 mol

Number of moles of KCl =​ 5.595 * 10^-4 mol

Molar mass of KCl = 74.551 g /mol

Mass of KCl = (number of moles of KCl) * (molar mass of KCl)

= ​(5.595 * 10^-4) * (74.551) = 0.04171 g

Hence, mass of KCl required to make the solution saturated in PbCl₂ is, 0.04171 g or 41.71 mg.

(since, 1 g = 1000 mg)

ANSWER: 0.04171 g or 41.71 mg