Question

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

Answer 1

Answer:

The solubility of CuX is 1.425x10⁻⁷M

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:

$K=\frac{[Cu(CN)4]^{2} [X]}{[CN]^{4} } \\1.27x10^{-11} =\frac{x^{2} }{(0.2-x)^{4} }$

Here, solving for x:

x=1.425x10⁻⁷M=CuX

Answer 2

Answer:

Solubility= 1.08×10-12

Explanation:

Take the cube root of 1.27×10-36