Solubility of Gypsum (CaSO4)
How much gypsum dissolves in water at 25?
We solve this task in two different ways: (i) by an analytical approach (with the “pocket calculator”) and (ii) numerically, by applying the whole gamut of chemical thermodynamics. The results will differ considerably and we will explain why.
Analytical Approach
The reaction formula and its log K value (at 25°C) is given by:^{1}^{2}^{3}
(1)  CaSO_{4} = Ca^{+2} + SO_{4}^{2}  log K = 4.58 
The calculation rests upon two equations:
(2)  law of mass action:^{4}  K ≈ K_{sp}^{*} = [Ca^{+2}] · [SO_{4}^{2}] = 10^{4.58} 
(3)  charge balance:  [Ca^{+2}] = [SO_{4}^{2}] 
Inserting Eq.(3) into Eq.(2) yields the simple quadratic dependence:
(4)  K = [SO_{4}^{2}]^{2} 
Since the dissolved amount of gypsum equals the value of [SO_{4}], we immediately get the result:
(5)  ΔGypsum  =  K^{1/2} 
(6)  ΔGypsum  =  (10^{4.58})^{1/2} M 
(7)  ΔGypsum  =  10^{2.29} M = 5.12×10^{3} M = 5.12 mM 
In words: The amount of 5.12 mmol gypsum dissolves in 1 liter of pure water. [Please note that the final result is independent of the initial inventory of gypsum.]
Numerical Approach with aqion
We start with pure water (button H2O) and switch to molar units (activate checkbox mol). To open the mineral table click on Minerals, then enter the amount of 20 mmol/L gypsum (by doubleclicking on the ‘Gypsum’ line).^{5} The corresponding screenshot is shown on the right side.
Run the calculation with Start, a first schematic overview appears. After click on next, the results are shown again in the output table. In the outermost right column you find:
pH  =  7.07 
Ca  =  15.6 mM 
SO4  =  15.6 mM 
According to this calculation, the amount of 15.4 mmol gypsum dissolves in 1 liter of pure water – that is three times as much as in the above calculation!
Here, SO4 and Ca abbreviate the total concentrations:
(8)  [Ca]_{T}  =  [Ca^{+2}] + [CaSO_{4}(aq)] + [CaHSO_{4}^{+}] + [CaOH^{+}] 
(9)  [SO4]_{T}  =  [SO_{4}^{2}] + [HSO_{4}^{}] + [CaSO_{4}(aq)] + [CaHSO_{4}^{+}] 
The corresponding equilibrium speciation is listed in the table Ions:
Ca^{+2}  10.5  mM 
CaSO_{4}(aq)  5.19  mM 
CaHSO_{4}^{+}  3.2×10^{6}  mM 
CaOH^{+}  1.2×10^{5}  mM 
SO_{4}^{2}  10.5  mM 
HSO_{4}^{}  5.1×10^{5}  mM 
In fact, there are only three major players: Ca^{+2}, SO_{4}^{2} and CaSO_{4}(aq). The last species is an aqueous complex; it should not be confused with the mineral phase gypsum, which is usually abbreviated as CaSO_{4}(s).
Résumé
Both results differ considerably:
(10a)  analytical approach (pocket calculator):  5.1 mM 
(10b)  numerical approach (chem. thermodynamics):  15.6 mM 
The reasons for the deficiency of the analytical approach are:
 neglect of complexation (especially CaSO_{4}(aq))
 neglect of activity corrections (at ionic strength I = 42 mM, γ = 0.48)
Remarks & Footnotes

The correct formula of gypsum is CaSO_{4}:2H_{2}O. To keep the notation simple, we use the short form CaSO_{4} here. ↩

Here and in the following, the notation ‘log x’ is used for the decadic logarithm ‘log_{10} x’. ↩

Here we approximate K_{sp} by the stoichiometric solubility product (Strictly speaking, this only applies for ideal solutions where ionic activities = concentrations). ↩

The initial amount of gypsum should at least exceed the solubility limit. For example, if you enter 3 mmol/L gypsum, it is far too little (the saturation index remains below zero). ↩