Determine the molar solubility of CuCl in a solution containing 0.050 mol L^-1 KCl. K_sp (CuCl) = 1.0 × 10^-6. A) 1.0 × 10^-12 mol L^-1 B) 5.0 × 10^-7 mol L^-1 C) 2.0 × 10^-5 mol L^-1 D) 1.0 × 10^-3 mol L^-1 E) 0.050 mol L^-1

Question

Quizplus · Accepted Answer

First, write the balanced equilibrium equation for the dissolution of CuCl:

CuCl(s) ⇌ Cu2+(aq) + 2Cl-(aq)

The solubility product constant (Ksp) for this reaction is given as 1.0 × 10^-6.

Using an ICE table, we can set up the following expressions:

Ksp = [Cu2+][Cl-]^2

Let x be the molar solubility of CuCl in moles per liter, then:

[Cu2+] = x
[Cl-] = 2x (from stoichiometry)

Substituting these expressions into the Ksp expression, we have:

Ksp = x(2x)^2 = 4x^3

Now, we need to account for the presence of KCl in the solution. KCl is a soluble salt that dissociates completely in water to form K+ and Cl- ions. Therefore, the equilibrium concentration of Cl- ions will be:

[Cl-] = 0.050 mol/L (from the KCl concentration given in the problem)

This means that the concentration of Cl- ions from the dissolution of CuCl must be added to this value:

[Cl-]total = [Cl-] (from KCl) + [Cl-] (from CuCl) = 0.050 + 2x

Substituting this expression for [Cl-] into the Ksp expression, we have:

1.0 × 10^-6 = x(2x)^2 = 4x^3

Solving for x, we get:

x = 2.0 × 10^-5 mol/L

Therefore, the molar solubility of CuCl in the presence of 0.050 mol/L KCl is 2.0 × 10^-5 mol/L. The closest choice is C.

Determine the Molar Solubility of CuCl in a Solution Containing