Copper(I) chloride has Ksp = 1.7 × 10-7. Calculate the molar solubility of copper(I) chloride in

(a) 0.0858M HCl solution
(b) 0.419 M HCl solution
(c) 0.653 M CaCl2 solution

To calculate the molar solubility of copper(I) chloride in different solutions, we first need to write the balanced chemical equation for the dissolution of copper(I) chloride in water.

The balanced chemical equation for the dissolution of copper(I) chloride (CuCl) in water is:

CuCl(s) ⇌ Cu+(aq) + Cl-(aq)

(a) In 0.0858 M HCl solution:

Since HCl is a strong acid and completely dissociates in water, we can ignore its contribution in the Ksp expression. Therefore, the equilibrium expression for the dissolution of copper(I) chloride is:

Ksp = [Cu+][Cl-]

Let's assume that the concentration of Cu+ and Cl- ions formed at equilibrium is 'x'. Therefore, we can write:

Ksp = x * x = x^2

Therefore, x = sqrt(Ksp)

Substituting the given value of Ksp = 1.7 × 10^(-7), we get:

x = sqrt(1.7 × 10^(-7)) ≈ 1.31 × 10^(-4) M

Therefore, the molar solubility of copper(I) chloride in a 0.0858 M HCl solution is approximately 1.31 × 10^(-4) M.

(b) In 0.419 M HCl solution:

Using the same approach as in part (a), we can calculate the molar solubility of copper(I) chloride in a 0.419 M HCl solution by substituting the given value of Ksp = 1.7 × 10^(-7) into the equation:

x = sqrt(1.7 × 10^(-7)) ≈ 1.31 × 10^(-4) M

Therefore, the molar solubility of copper(I) chloride in a 0.419 M HCl solution is also approximately 1.31 × 10^(-4) M.

(c) In 0.653 M CaCl2 solution:

Since CaCl2 is a different compound that does not directly affect the solubility of copper(I) chloride, we do not include it in the Ksp expression. Therefore, the equilibrium expression remains the same:

Ksp = [Cu+][Cl-]

Using the same approach as before, we can calculate the molar solubility of copper(I) chloride in a 0.653 M CaCl2 solution:

x = sqrt(1.7 × 10^(-7)) ≈ 1.31 × 10^(-4) M

Therefore, the molar solubility of copper(I) chloride in a 0.653 M CaCl2 solution is also approximately 1.31 × 10^(-4) M.

In conclusion, the molar solubility of copper(I) chloride is the same in all three solutions: approximately 1.31 × 10^(-4) M.

What do you not understand about the problem? This is Le Chatelier's Principle in action.