IOM 2017, Теория — Задача 5. Equilibrium between the copper minerals

Solutions

1 Calculations

$n(\mathrm{CuO})=0.02\ \mathrm{mol}$, $n(\mathrm{CO_2})=0.01\ \mathrm{mol}$, $n(\mathrm{H_2O})=0.01\ \mathrm{mol}$

Each substance – 0.5 points, each quantity – 0.5 point

Formula of M: $\mathrm{Cu_2(OH)_2CO_3}$ — 2 points

Reaction equations:

$\mathrm{Cu_2(OH)_2CO_3\rightarrow2CuO+H_2O+CO_2}$ — 1 point

$\mathrm{P_2O_5+H_2O\rightarrow2HPO_3}$ — 0.5 points

$\mathrm{CaO+CO_2\rightarrow CaCO_3}$ — 0.5 points

$\mathrm{4CuO\rightarrow2Cu_2O+O_2}$ — 1 point [8 т.]

2 Calculations

Suppose, azurite is a basic copper carbonate $\mathrm{Cu_x(OH)_y(CO_3)_z}$ — 2 points

mass balance:

charge balance:

$2x=y+2z$ — 4 points

The simplest solution in natural numbers is:$x=3,y=2,z=2$.

Formula of A: $\mathrm{Cu_3(OH)_2(CO_3)_2}$ — 4 points [10 т.]

3 Calculations

$\mathrm{3Cu_2(OH)_2CO_3+CO_2\rightarrow2Cu_3(OH)_2(CO_3)_2+H_2O}(l)$ — 2 points

$\Delta_\mathrm rG^\circ(298\ \mathrm K)=2*(-1389)+(-237)-(-394)-3*(-883)=28\ \mathrm{kJ/mol}$ — 6 points

$p_{\mathrm{CO_2}}=1/K_\mathrm p=\exp[28000/8.314/298]=81000\ \mathrm{bar}$ — 8 points [16 т.]

4 Calculations and estimates

From the previous answer it follows that at low pressure malachite is more stable than azurite. So, we need to decide which mineral is more stable tenorite(CuO) or malachite — 4 points

Consider transformation of tenorite to malachite:

$\mathrm{2CuO+H_2O}(l)+\mathrm{CO_2\rightarrow Cu_2(OH)_2CO_3}$ — 2 points

$\Delta_\mathrm rG^\circ(298\ \mathrm K)=-883-2*(-130)-(-237)-(-394)=8.0\ \mathrm{kJ/mol}$ — 4 points

$K_\mathrm p=1/p(\mathrm{CO_2})=0.040$ — 4 points

The partial pressure of CO₂ in the atmosphere: $p=4\cdot10^{-4}\ \mathrm{bar}$ — 2 points

$Q=1/(4\cdot10^{-4})=2500>K_\mathrm p\Rightarrow$ tenorite is more stable — 4 points

The most stable mineral – tenorite

(Answer without calculations – 0 points) [20 т.]

5 Calculations

a) $K_\mathrm{sp}(\mathrm M)=[\mathrm{Cu^{2+}}]^2[\mathrm{OH^-}]^2[\mathrm{CO_3^{2-}}]$

$K_\mathrm{sp}(\mathrm A)=[\mathrm{Cu^{2+}}]^3[\mathrm{OH^-}]^2[\mathrm{CO_3^{2-}}]^2$ — 4 points(2×2)

Determine the concentrations of ions in the solution:

$[\mathrm{Cu^{2+}}]=0.05/20.05=0.0025\ \mathrm M$ (dilution of the solution can be neglected) — 2 points

Hydrogen carbonate is an ampholyte, so we can use a well-known approximate formula for H⁺ concentration: $[\mathrm{H^+}]=(K_1K_2)^{1/2}=4.74\cdot10^{-9}\ \mathrm M$ — 8 points

$[\mathrm{OH^-}]=10^{-14}/[\mathrm{H^+}]=2.11\cdot10^{-6}\ \mathrm M$ — 2 points

Neglecting the dilution of the solution calculate the carbonate concentration $[\mathrm{CO_3^{2-}}]$ at a given pH:

$[\mathrm{CO_3^{2-}}]=c_0K_1K_2/([\mathrm{H^+}]^2+K_1[\mathrm{H^+}]+K_1K_2)=1.03\cdot10^{-3}\ \mathrm M$ — 8 points

Compare with $K_\mathrm{sp}$:

malachite: $(0.025)^2(2.11\cdot10^{-6})^2(1.03\cdot10^{-3})=2.9\cdot10^{-20}\gg K_\mathrm{sp}(\mathrm M)\Rightarrow$ precipitates — 2 points

azurite: $(0.025)^3(2.11\cdot10^{-6})^2(1.03\cdot10^{-3})^2=7.3\cdot10^{-26}\gg K_\mathrm{sp}(\mathrm A)\Rightarrow$ precipitates — 2 points

b) The freshly precipitated substance contains a mixture of basic carbonates, because both solubility products are much less than the product of concentrations. Upon long standing, the formation of the most thermodynamically stable mineral is possible, i.e., malachite.

Both answers can be correct with the corresponding justification. — 6 points

a) The precipitate is formed

b) Answer(yes(formula),no): No [34 т.]