Question

Calculate the solubility of CuX (Ksp=[Cu2+][X2−]=1.27×10−36) in a solution that is 0.200 M in NaCN.

I have already tried to square root the Ksp value to get the answer but it was wrong.

Answer 1

The solubility of CuX is 1.425x10⁻⁷M

Step-by-step explanation:

Given:

initial concentration of NaCN=0.2M

Ksp=1.27x10⁻³⁶

The reaction are:

CuX → Cu²⁺ + X²⁺, Ksp=1.27x10⁻³⁶

Cu²⁺ + 4CN⁻ → (Cu(CN)₄)²⁻, Kf=1x10²⁵

The overall reaction is:

CuX + 4CN⁻ → (Cu(CN)₄)²⁻ + X²⁺

The equilibrium constant is:

K=Ksp*Kf=1.27x10⁻³⁶*1x10²⁵=1.27x10⁻¹¹

CuX + 4CN⁻ → (Cu(CN)₄)²⁻ + X²⁺

I - 0.2 0 0

C - -4 +x +x

E - 0.2-4 x x

The equation for equilibrium is:

`K=([Cu(CN)4]^(2) [X])/([CN]^(4) ) \\1.27x10^(-11) =(x^(2) )/((0.2-x)^(4) )`

Here, solving for x:

x=1.425x10⁻⁷M=CuX

Answer 2

Solubility= 1.08×10-12

Step-by-step explanation:

Take the cube root of 1.27×10-36