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 water (assume density is 1 kg/L).

Answer 1

According to the question

We know that

CuCO3 (s) ⇌ Cu2+ (aq) + CO32-(aq)

Ksp = 1.4 *10^-10 =[Cu2+(aq)][CO32-(aq)]-----(2)

CO2 (g) ⇌ CO2 (aq)

K1= 3.4 *10^-2 =[CO2(aq)]o/PCO2(g) ------(2)

CO2(aq) + H2O(l) ⇌ HCO3-(aq) + H+(aq)

K2 = 4.7 *10^-7 = [HCO3-][H+]/[CO2(aq)] ------(3)

HCO3- (aq) ⇌ CO32-(aq) + H+(aq)

K3 = 5.1 *10^-11 =[CO32-(aq)][H+(aq)]/[HCO3- (aq)]--------(4)

pH of the solution = 5.745

pH = -log[H+]

That is;

[H+] = 10^-5.745 = 1.8*10^-6 M

From the equation (2)

[CO2(aq)]o = PCO2(g) * 3.4 *10^-2 [Given PCO2(g) = 1 bar]

= 1*3.4 *10^-2 = 3.4 *10^-2 M

So,

From (3) and (4) we get

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

[CO32-(aq)] = K3[HCO3-]/[H+] = 2.83*10^-5 [HCO3-]-------(6)

[CO2(aq)]o = [CO2(aq)] + [HCO3-] + [CO32-]

3.4 *10^-2 M = 3.827 [HCO3-] + [HCO3-] + 2.83*10-5 [HCO3-]

[HCO3-] = 3.4 *10-2 M/(3.827+1+2.83*10^-5]

Then,

[HCO3-] = 3.4 *10^-2 M/4.827 = 0.704 M

Putting this in (6), we get

[CO32-(aq)] = 2.83*10^-5 * 0.704 M = 1.99*10^-5 M

Then,

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

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

Then,

No. of moles of Cu2+ in 1.54 L = 0.704*10^-5 mol/L * 1.54 L = 1.083*10^-5 mol

Weight of copper in 1.54 L water = no. of moles *molar mass of Cu

Then,

= 1.083*10^-5 mol x 63.5 g/mol

= 68.8*10^-5 g

Hence,

Weight of copper in 1.54 L water = 68.8*10^-5 g