Question

Suppose 231.8 mgmg of PbCl2PbCl2 was added to 15.0 mLmL of water in a flask, and the solution was allowed to reach equilibrium at 20.0 oCoC. Some solute remained at the bottom of the flask after equilibrium, and the solution was filtered to collect the remaining PbCl2PbCl2, which had a mass of 74 mgmg . What is the solubility of PbCl2PbCl2 (in g/Lg/L)?

Answer 1

Answer:

ksp = 2.2 x ⁻⁴

Explanation:

The equilibrium here is:

PbCl₂ (s)     ⇄ Pb²⁺ + 2 Cl⁻

we can recognize it as a product solubilty equilibrium once we are told that some undissolved PbCl₂ remained.

The equilibrium constant, Ksp is given by the equation

Ksp = [Pb²⁺][Cl⁻]²

where [Pb²⁺] and [Cl⁻]² are the concentrations (M) of Pb²⁺ and Cl⁻ in solution.

we have the mass of solid PbCl₂ placed in solution, so we can determine the number of moles it represents, and if  we  substract the moles of undissolved PbCl₂ we will know the moles of Pb²⁺ and Cl⁻ which went into solution.

From there we can calculate the molarity (M= moles/L solution) and finally plug the values into our expression for Ksp to answer this question.

molar mas PbCl₂ = 278.1 g/mol

1 milligram = 1 x 10⁻³ g

mol PbCl₂ initially = 231.8 x 10⁻³ g / 278.1 mol = 8.3 x 10⁻⁴ mol

Volume solution = 15 mL x 1L / 1000 mL = 0.015 L

mol undissolved PbCl₂ = 74 x 10⁻³ g / 278.1 g/mol = 2.7 x 10⁻⁴ mol

mol PbCl dissolved =   8.3 x 10⁻⁴ mol -  2.7 x 10⁻⁴ mol = 5.7 x 10⁻⁴ mol

Concentration of Pb²⁺ in solution = 5.7 x 10⁻⁴ mol / 0.015 L = 3.8 x 10⁻² M

Concentration of Cl⁻ in solution = 2 x 3.8 x 10⁻² M = 7.6 x 10⁻² M

(Note from the formula we we get 2 mol Cl⁻ per mol PbCl₂)

Plugging these values into the expression for Ksp we have

Ksp = 3.8 x 10⁻² x (7.6 x 10⁻²)² = 2.2 x 10⁻⁴