Question

Consider the following system in equilibrium: CuCO3 (s) ⇌ Cu2+ (aq) + CO3 2- (aq) Ksp = 1.4 x10-10 CO2 (g) ⇌ CO2 (aq) K1 = 3.4 x10-2 CO2 (aq) + H2O (l) ⇌ HCO3 - (aq) + H+ (aq) K2 = 4.7 x10-7 HCO3 - (aq) ⇌ CO3 2- (aq) + H+ (aq) K3 = 5.1 x10-11 At equilibrium, the pH of the solution is equal to 5.745 when the system is saturated with CuCO3. In standard conditions (pressure of 1 bar), calculate how many grams of copper are contained in 1.54 L of wate

Answer 1

Given,

CuCO₃ (s) ⇌ Cu²⁺ (aq) + CO₃^2-(aq)

K_sp = 1.4 x10^-10 =[Cu²⁺(aq)][CO₃^2-(aq)] ..................(1)

CO₂ (g) ⇌ CO₂ (aq)

K1= 3.4 x10^-2 =[CO₂(aq)]_o/P_CO2(g) .......................(2)

CO₂(aq) + H2O(l) ⇌ HCO₃^-(aq) + H⁺(aq)

K₂ = 4.7 x10^-7 = [HCO₃^-][H⁺]/[CO₂(aq)] .....................(3)

HCO₃^- (aq) ⇌ CO₃^2-(aq) + H⁺(aq)

K₃ = 5.1 x10^-11 =[CO₃^2-(aq)][H⁺(aq)]/[HCO₃^- (aq)] ......................(4)

pH of the solution = 5.745

pH = -log[H⁺]

i.e. [H⁺] = 10^-5.745 = 1.8x10^-6 M

From the equation (2)

[CO₂(aq)]_o = P_CO2(g) x 3.4 x10^-2 [Given P_CO2(g) = 1 bar]

= 1x3.4 x10^-2 = 3.4 x10^-2 M

From (3) and (4) we get

[CO₂(aq)] = [HCO₃^-][H⁺]/K₂ = 3.827 [HCO₃^-] ...........................(5)

[CO₃^2-(aq)] = K₃[HCO₃^-]/[H⁺] = 2.83x10^-5 [HCO₃^-] ...........................(6)

[CO₂(aq)]_o = [CO₂(aq)] + [HCO₃^-] + [CO₃^2-]

3.4 x10^-2 M = 3.827 [HCO₃^-] + [HCO₃^-] + 2.83x10^-5 [HCO₃^-]

[HCO₃^-] = 3.4 x10^-2 M/(3.827+1+2.83x10^-5]

[HCO₃^-] = 3.4 x10^-2 M/4.827 = 0.704 M

Putting this in (6), we get

[CO₃^2-(aq)] = 2.83x10^-5 x 0.704 M = 1.99x10^-5 M

Now using the equation (1), we get the concentration of [Cu²⁺]

[Cu²⁺] = 1.4 x10^-10/1.99x10^-5 M = 0.704x10^-5 M

No. of moles of Cu²⁺ in 1.54 L = 0.704x10^-5 mol/L x 1.54 L = 1.083x10^-5 mol

weight of copper in 1.54 L water = no. of moles x molar mass of Cu

= 1.083x10^-5 mol x 63.5 g/mol

= 68.8x10^-5 g

Hence, weight of copper in 1.54 L water = 68.8x10^-5 g