The copper(I) ion forms a chloride salt that has Ksp = 1.2 10-6. Copper(I) also forms a complex ion with Cl ‾.

Cu+(aq) + 2 Cl ‾(aq) reverse reaction arrow CuCl2‾(aq) K = 8.7 104

Calculate the solubility of copper(I) chloride in pure water. (Ignore CuCl2‾ formation)

CuCl --> Cu+ & Cl-

Ksp = [Cu+2] [Cl-]

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Cu+(aq) + 2 Cl ‾(aq) ==> CuCl2 ‾(aq)

K = [CuCl2 ‾] / [Cu+2] [Cl-]^2

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(K) (Ksp) = [Cu+2] [Cl-] times [CuCl2 ‾] / [Cu+2] [Cl-]^2

reduces to:
(K) (Ksp) = [CuCl2 ‾] / [Cl-]

(8.7 x 10^4) (1.2 x 10^-6. ) = [CuCl2 ‾] / [0.13]

(8.7 x 10^4) (1.2 x 10^-6. ) = [CuCl2 ‾] / [0.13]

1.044 e-1 = [CuCl2 ‾] / [0.13]

dissolved copper = 0.01357 Molar

your answer
with 2 sig figs is
0.014 Molar Copper in solution
I think this is right.

I think Dan's solution takes into account the formation of the CuCl2^- complex but the problem says to ignore that. If we ignore it, the only thing we have is

CuCl ==> Cu^+ + Cl^-
Ksp + (Cu^+)(Cl^-) = 1.2E-6
Substitute x = Cu^+ and x = Cl^- and solve for x.

To calculate the solubility of copper(I) chloride (CuCl) in pure water, we need to find the concentration of copper(I) ion (Cu+) in a saturated solution of CuCl.

Let's assume the solubility of CuCl is "x" mol/L. Since CuCl dissociates into Cu+ and Cl- ions in solution, the concentration of Cu+ ions will also be "x" mol/L.

The solubility product constant (Ksp) for CuCl is given as 1.2 x 10^-6. It can be represented as follows:

CuCl (s) ⇌ Cu+(aq) + Cl- (aq)

Using the stoichiometry of the reaction, we can write the expression for Ksp as:

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

Since the concentration of Cu+ ions is "x" mol/L and the concentration of Cl- ions is also "x" mol/L, we can rewrite the expression as:

Ksp = x * (x)^2
1.2 x 10^-6 = x^3

To solve for x, we need to take the cube root of both sides of the equation:

x = (1.2 x 10^-6)^(1/3)

Now, we can calculate the solubility of CuCl in pure water:

x = (1.2 x 10^-6)^(1/3)
x ≈ 8.81 x 10^-3 mol/L

Therefore, the solubility of copper(I) chloride in pure water is approximately 8.81 x 10^-3 mol/L.

To calculate the solubility of copper(I) chloride (CuCl) in pure water, we need to first determine the concentrations of Cu+ and Cl- ions in equilibrium.

Let's assume that the solubility of CuCl is 's' moles per liter. When CuCl dissociates in water, it forms Cu+ and Cl- ions in a 1:1 ratio. So, at equilibrium, the concentration of Cu+ ions will also be 's' moles per liter, and the concentration of Cl- ions will be '2s' moles per liter (due to the stoichiometry of the CuCl dissociation reaction).

Using the given equilibrium constant for the complex ion formation, we can write the K expression as follows:

K = [CuCl2‾] / [Cu+][Cl-]^2

Since we are ignoring the formation of CuCl2‾, the concentration of CuCl2‾ will be zero. Hence, the K expression simplifies to:

K = 0 / (s)(2s)^2
K = 0

This implies that the equilibrium concentration of the complex ion is zero. Therefore, we can conclude that the reaction to form CuCl2‾ does not occur significantly.

Now, let's consider the solubility product constant (Ksp) expression for the dissolution of CuCl:

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

From the equilibrium concentrations we just determined, we can substitute them into the Ksp expression:

Ksp = (s)(2s)^2
1.2 × 10^-6 = 4s^3

Solving this equation for 's', we find:

s = (1.2 × 10^-6 / 4)^(1/3) ≈ 3.68 × 10^-3 M

Therefore, the solubility of copper(I) chloride in pure water is approximately 3.68 × 10^-3 moles per liter (M).