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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?