Question

A) Determine the Solubility product of calcite at 25c and 40c. At which temperature is calcite more soluble?

B)A sample of water has a calcium ion activity (aca2+) of 10^-3.04 and a carbonate ion activity of 10^-5.56. is this water sample in equlibrium with calcite?if not is calcite dissolving or precipitating? (assume STP< use information from A)

Answer 1

A)The solubility product of calcite at 25^oC is 3.36*10^-9.

Following is the enthalpy of formation data:

CaCO₃(s) ΔH^o_f = -1206.9 kJmol^-1

Ca²⁺(aq) ΔH^o_f = -542.8 kJmol^-1

CO₃^2-(aq) ΔH^o_f = -677.1 kJmol^-1

Thus to calculate K_sp at 40^oC we have the formula

ln(K₂/K₁) = -(ΔH^o/R)[(1/T₂) - (1/T₁)]

ΔH^o= -1206.9-(-1219.9) = 13*103 J/mol

R = 8.314 J/mol

T2 = 40^oC ; T1 = 25^oC

Thus the value of K_sp at 40^oC comes out to be 2.238*10^-19

Calcite is partially soluble in water. As the temperature of the solution is increased, the solubility of the calcite decreases as kinetic energy of the molecules increases and they get dissociated to more extent and leave the solution. Thus calcite is more soluble at 25^oC

B) The solubility product of the given water sample is given by,

K_sp = [a_Ca2+][a_CO32-]

As the ionic activity of calcium ion is 10^-3.04 = 9.12*10^-4 and that of carbonate ion is 10^-5.56 = 0.0275*10^-4

Thus the solubility product is

K_sp = (9.12*10^-4)(0.0275*10^-4)=2.508*10^-9

As the solubility product is decreasing from the solubility product at STP (from A), therefore calcite is not in equilibrium with water sample. Thus calcite is precipitating.

This is because of the decrease in solubility.

To determine the solubility of calcite we first balance the concentration of the ions.

CaCO₃ ------------- [Ca²⁺][CO₃^2-]

Here, [Ca²⁺] = [CO₃^2-] after dissociation.

Ksp = [Ca²⁺][CO₃^2-] = 3.36*10^-9

[Ca²⁺]² = 3.36*10^-9

[Ca²⁺] = 5.79*10^-5

Thus solubility of calcite is 5.79*10^-5/1L * 100g/mol/1mol = 5.79*10^-3 g/L