Determine the molar solubility of CuCl in a solution containing 0.040 M KCl.K_sp (CuCl)= 1.0 × 10^-6. A) 1.0 × 10^-12 M B) 4.0 × 10^-8 M C) 2.5 × 10^-5 M D) 1.0 × 10^-3 M E) 0.050 M

Question

Quizplus · Accepted Answer

The dissolution equilibrium of CuCl can be represented as: CuCl (s) ⇌ Cu+ (aq) + Cl- (aq)

The Ksp expression for this equilibrium would be: Ksp = [Cu+][Cl-]

At equilibrium, some amount of CuCl will dissolve to give Cu+ and Cl-. Let's assume that 'x' moles of CuCl dissolve.

Then, the equilibrium concentration of Cu+ and Cl- would be 'x'.

However, the Ksp expression also includes the concentration of CuCl at equilibrium. Since the initial concentration of CuCl was zero, the equilibrium concentration would be 'x'.

Therefore, we can write: Ksp = x^2

Substituting the given value of Ksp, we get: 1.0 × 10^-6 = x^2

Solving for 'x', we get: x = 1.0 × 10^-3 M

Now, we need to consider the effect of KCl on the solubility of CuCl. KCl will dissociate to give K+ and Cl- ions, which can potentially react with the Cu+ and Cl- ions from CuCl.

The common ion effect predicts that the presence of a common ion (in this case, Cl-) will decrease the solubility of a slightly soluble salt (in this case, CuCl).

To account for this effect, we need to calculate the concentration of Cl- ions in the solution.

Since the initial concentration of KCl is given as 0.040 M, we can assume that all of it dissociates to give K+ and Cl-. Thus, the concentration of Cl- ions in the solution is also 0.040 M.

This means that the total concentration of Cl- ions in the solution is 0.040 M (from KCl) + 1.0 × 10^-3 M (from CuCl) = 0.040001 M

Now, we need to calculate the ion product (Qsp) for CuCl, taking into account the common ion effect:

Qsp = [Cu+][Cl-] = (1.0 × 10^-3 M)(0.040001 M) = 4.0004 × 10^-8

Since Qsp < Ksp, we can assume that more CuCl will dissolve until Qsp = Ksp.

Let's assume that 'y' moles of CuCl dissolve.

Then, the equilibrium concentration of Cu+ and Cl- would be (x+y).

The concentration of Cl- ions (from CuCl) would be y.

The concentration of 'extra' Cl- ions (from KCl) would be (0.040001 - y).

Therefore, we can write: Ksp = (x+y)(y) = y^2 + xy

Substituting the values we calculated, we get:

1.0 × 10^-6 = y^2 + (1.0 × 10^-3)(y)

Solving for 'y', we get:

y = 2.5 × 10^-5 M

This is the molar solubility of CuCl in the presence of 0.040 M KCl.

Therefore, the best choice is C, 2.5 × 10^-5 M.

Determine the Molar Solubility of CuCl in a Solution Containing