Question

For CuCO3, Ksp is 2.5 x 10-10 M2. For CuCl42-, Kf is 5.0 x 105 M-4. What is the value of the equilibrium constant for the sum of the two reactions shown below?

CuCO3 ↔ Cu2+ + CO32- Ksp = 2.5 x 10-10 M2

Cu2+ + 4 Cl- ↔ CuCl42- Kf = 5.0 x 105 M-4

CuCO3 + 4 Cl- ↔ CuCl42- + CO32- K = ?

K = ___ M-2

Based on the previous problem, what is the maximum solubility of CuCO₃ in a 0.650 M NaCl solution?

S = ___ M

Answer 1

i) CuCO3(s) <------> Cu²⁺(aq) + CO3^2-(aq)   

K_sp =[Cu²⁺][CO3^2-] = 2.5×10^-10M²

Cu²⁺(aq) + 4Cl^-(aq) <--------> CuCl4^-(aq)

K_f = [CuCl4^-]/[Cu²⁺][Cl^-]⁴ = 5.0×10⁵M^-4

adding two equation

CuCO3(s) + 4Cl^-(aq) <---------> CuCl4^-(aq) + CO3^2-(aq)

K = [CuCl4^-][CO3^2-]/[Cl^-]⁴

K = K_sp × K_f

K = 2.5×10^-10M²× 5.0×10⁵M^-4

K = 1.25 × 10^-4M^-2

ii) CuCO3(s) + 4Cl^- (aq) <------> CuCl4^2-(aq) + CO3^2-(aq)

K = [CuCl4^2-][CO3^2-]/[Cl^-]⁴ = 1.25 ×10^-4M²

Initial concentration

[Cl^-] = 0.650

[CuCl4^-] = 0

[CO3^2-] = 0

change in concentration

[Cl^-] = -4x

[CuCl4^-] = +x

[CO3^2-] = +x

equillibrium concentration

[Cl^-] = 0.650 - 4x

[CuCl4^-] = x

[CO3^2-] = x

therefore

x²/(0.650 - 4x )⁴= 1.25×10^-4

solving for x

x = 0.004468

[CO3^2-] = 0.004468M

Therefore

Molar solubility of CuCO3 in 0.650M NaCl solution = 0.004468M