Solubility ProductThe solubility of a mineral is governed by the solubility product,the equilibrium constant for a reaction such as:

CaSO4(anhydrite) ↔ Ca2+ + SO42-

The solubility product is given by:

4

24

2

CaSO

SOCaSP a

aaK

−+

=

If anhydrite is a pure solid, then aCaSO4 = 1.

and in dilute solutions: aCa2+ ≈ (Ca2+) and aSO42- ≈ (SO42-).

KSP ≈ [Ca2+][SO42-] = 10-4.5

What is the solubility of anhydrite in pure water?If anhydrite dissolution is the only source of both Ca2+ and

SO42-, then:

[Ca2+] = [SO42-] = x

x2 = 10-4.5

x = 10-2.25 = 5.62x10-3 mol/LMWanhydrite = 136.14 g/mol

Solubility = (5.62x10-3 mol/L)(136.14 g/mol) = 0.765 g/L

4

24

2

CaSO

SOCaSP a

aaK

−+

=

Saturation indexIn a natural solution, it is not likely that [Ca2+] = [SO4

2-], forexample, because there will be more than one source of eachof these ions. In this case we use saturation indices todetermine if the water is saturated with respect to anhydrite.

KSP = 10-4.5 ≈ [Ca2+]eq[SO42-]eq

IAP = [Ca2+]act[SO42-]act

[ ] [ ]SPSP

actact

KIAP

KSOCa

==Ω−+ 24

2Saturationindex

Suppose a groundwater is analyzed to contain 5x10-2 mol/L Ca2+ and7x10-3 mol/L SO4

2-. Is this water saturated with respect to anhydrite (CaSO4(s))?

KSP = 10-4.5

IAP = (5x10-2)(7x10-3) = 3.5x10-4 = 10-3.45

Ω = 10-3.45/10-4.5 = 101.05 = 11.22

Ω > 1, i.e., IAP > KSP, so the solution is supersaturatedand anhydrite should precipitate

If Ω = 1, i.e., IAP = KSP, the solution would be saturated (equilibrium conditions)

If Ω < 1, i.e., IAP < KSP, the solution would be undersaturated; the mineral should dissolve

CaSO4 ↔ Ca2+ + SO42-

How much salt should precipitate?

Returning to the previous example, i.e., the groundwater with5x10-2 mol/L Ca2+ and 7x10-3 mol/L SO4

2-, how muchanhydrite should precipitate at equilibrium?

If x mol/L of anhydrite precipitate, then at equilibrium:[Ca2+] = 5x10-2 - x; [SO4

2-] = 7x10-3 - xand [Ca2+][SO4

2-] = 10-4.5

(5x10-2 - x)(7x10-3 - x) = 10-4.5

x2 - (5.7x10-2)x + (3.18x10-4) = 0After solving for the quadratic equation:

x1 = 5.07x10-2 mol/L; x2 = 6.26x10-3 mol/L

We choose x2 (=6.26x10-3)because the first root(x1=5.07x10-2) causes [SO4] to be negative.

So, 6.26x10-3 mol/L of anhydrite precipitates, or:

(6.26x10-3 mol/L)(136.1 g/mol) = 0.852 g/Land

[Ca2+] = (5x10-2) - (6.26x10-3) = 4.37x10-2 mol/L

[SO42-] = (7x10-3) - (6.26x10-3) = 7.4x10-4 mol/L

[Ca2+]/[SO42-] increases with precipitation of anhydrite

Before precipitation:5x10-2/7x10-3 = 7.1

After precipitation: 4.37x10-2/7.4x10-4 = 59

6.26x10-3 mol/L of anhydrite precipitates

Precipitation not only reduces the concentrations of ions and actually changes the chemical composition if theremaining solution…

Because the initial [Ca2+]/[SO4

2-] > 1, the remainingSolution is enriched in Ca2+; if [Ca2+]/[SO4

2-]i < 1, thesolution would be enriched in SO4

2-

This process occurs when saltsprecipitate when water undergoes

evaporative concentrationsuch as in a desert lake

The common-ion effect

Natural waters are very complex and we may have saturationwith respect to several phases simultaneously.

Example: What are the concentrations of all species in asolution in equilibrium with both barite and gypsum?

1) Law of mass action expressions:CaSO4·2H2O ↔ Ca2+ + SO4

2- + 2H2O,KSP = [Ca2+][SO4

2-] = 10-4.6

BaSO4 ↔ Ba2+ + SO42-,

KSP = [Ba2+][SO42-] = 10-10.0

[Ca2+][SO42-] = 10-4.6

[Ba2+][SO42-] = 10-10.0

Eliminate [SO42-] by substituting 10-4.6/[Ca+]:

[Ba2+]•10-4.6/[Ca2+] = 10-10.0

2) Species: Ca2+, Ba2+, SO42-, H+, OH-

H2O ↔ H+ + OH-

Kw = [H+][OH-] = 10-14

3) Mass-balance: [Ba2+] + [Ca2+] = [SO42-]

4) Charge-balance:2[Ba2+] + 2[Ca2+] + [H+] = 2[SO4

2-] + [OH-]

10-4.6 + 10-10.0 = [SO42-]2

[SO42-] = (10-4.6 + 10-10.0)1/2 = 10-2.3 mol/L

[Ca2+] = 10-4.6/10-2.3 = 10-2.3 mol/L[Ba2+] = 10-10.0/10-2.3 = 10-7.7 mol/L

The least soluble salt (barite, KSP=10-10), contributes a negligibleamount of sulfate to the solution. The more soluble salt(gypsum, KSP=10-4.6) supresses the solubility of the lesssoluble salt (the common-ion effect). Barite can replacegypsum because barite is less soluble than gypsum.

[ ] [ ]−−+ = 2

4

6.42 10SO

Ca [ ] [ ]−−+ = 2

4

0.102 10SO

Ba

[ ] [ ] [ ]−−

−

−

−=

+

242

4

0.10

24

6.4 1010 SOSOSO

,

The solubility of gypsum is hardly affected by thepresence of barite:

Solubility of barite alone:[Ba2+][SO4

2-] = 10-10.0

[Ba2+]2 = 10-10.0

[Ba2+] = 10-5.0 mol/LSolubility of gypsum alone:

[Ca2+][SO42-] = 10-4.6

[Ca2+]2 = 10-2.3

[Ca2+] = 10-2.3 mol/L

Replacement reactionsWe can also calculate [Ba2+]/[Ca2+] in equilibrium

with both barite and gypsum.

[ ] [ ]−−+ = 2

4

6.42 10SO

Ca [ ] [ ]−−+ = 2

4

0.102 10SO

Ba

What would happen if a solution with [Ba2+]/[Ca2+] = 10-3

([Ca2+] = 1,000[Ba2+]) came into contact with a gypsum-bearing rock?

Barite will precipitate (taking Ba2+ out of the solution) andgypsum will dissolve until [Ba2+]/[Ca2+] = 10-5.4.

[Ba2+] 10-10

[Ca2+] 10-4.6= = 10-5.4, or [Ca2+] = 250,000[Ba2+]

HydrolysisThe interaction between water and one or both ions of a saltthat results in the formation of the parental acid or base, or both.We classify salts by the strength of the acid and base from which they form:

1. Strong acid + strong base do not hydrolyze2. Strong acid + weak base cations + OH- = acidic soln3. Weak acid + strong base anions + H+ = basic soln4. Weak acid + weak base release both cation & anions

Most common rock-forming minerals of the crust are saltsof weak acids and strong bases, e.g., carbonates and silicatesof alkali metals (Group 1) and alkaline earths (Group 2)form these salts…this is why groundwater in carbonate aquifersis commonly basic

?K

?K

K OHCOHOHHCO

K OHHCOOHCO

CO2KCOK

H2

H1

H2322-3

H1-32

23

332

=

=

+↔+

+↔+

+→

−

−−

−+ 2

Hydrolysis

Dissociation

A223

-3

A1-332

K HCOHCO

K HHCOCOH+−

+

+↔

+↔

What is the pH of a solution prepared by dissolving 0.1 molof K2CO3 in 1 L of water?