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:4242CaSOSOCaSPaaaK−+=If anhydrite is a pure solid, then aCaSO4 = 1.and in dilute solutions: aCa2+ ≈ (Ca2+) and aSO42- ≈ (SO42-).KSP ≈ [Ca2+][SO42-] = 10-4.5What is the solubility of anhydrite in pure water?If anhydrite dissolution is the only source of both Ca2+ andSO42-, then: [Ca2+] = [SO42-] = xx2 = 10-4.5x = 10-2.25 = 5.62x10-3 mol/LMWanhydrite = 136.14 g/molSolubility = (5.62x10-3 mol/L)(136.14 g/mol) = 0.765 g/L4242CaSOSOCaSPaaaK−+=Saturation indexIn a natural solution, it is not likely that [Ca2+] = [SO42-], 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-]eqIAP = [Ca2+]act[SO42-]act[ ] [ ]SPSPactactKIAPKSOCa==Ω−+ 242SaturationindexSuppose a groundwater is analyzed to contain 5x10-2 mol/L Ca2+ and7x10-3 mol/L SO42-. Is this water saturated with respect to anhydrite (CaSO4(s))?KSP = 10-4.5IAP = (5x10-2)(7x10-3) = 3.5x10-4 = 10-3.45Ω = 10-3.45/10-4.5 = 101.05 = 11.22Ω > 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 dissolveCaSO4 ↔ 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 SO42-, how muchanhydrite should precipitate at equilibrium?If x mol/L of anhydrite precipitate, then at equilibrium:[Ca2+] = 5x10-2 - x; [SO42-] = 7x10-3 - xand [Ca2+][SO42-] = 10-4.5(5x10-2 - x)(7x10-3 - x) = 10-4.5x2 - (5.7x10-2)x + (3.18x10-4) = 0After solving for the quadratic equation:x1 = 5.07x10-2 mol/L; x2 = 6.26x10-3 mol/LWe 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 anhydriteBefore precipitation:5x10-2/7x10-3 = 7.1After precipitation: 4.37x10-2/7.4x10-4 = 596.26x10-3 mol/L of anhydrite precipitatesPrecipitation not only reduces the concentrations of ions and actually changes the chemical composition if theremaining solution…Because the initial [Ca2+]/[SO42-] > 1, the remainingSolution is enriched in Ca2+; if [Ca2+]/[SO42-]i < 1, thesolution would be enriched in SO42- This process occurs when saltsprecipitate when water undergoesevaporative concentrationsuch as in a desert lakeThe common-ion effectNatural 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+ + SO42- + 2H2O,KSP = [Ca2+][SO42-] = 10-4.6BaSO4 ↔ Ba2+ + SO42-,KSP = [Ba2+][SO42-] = 10-10.0[Ca2+][SO42-] = 10-4.6[Ba2+][SO42-] = 10-10.0Eliminate [SO42-] by substituting 10-4.6/[Ca+]:[Ba2+]•10-4.6/[Ca2+] = 10-10.02) Species: Ca2+, Ba2+, SO42-, H+, OH-H2O ↔ H+ + OH-Kw = [H+][OH-] = 10-143) Mass-balance: [Ba2+] + [Ca2+] = [SO42-]4) Charge-balance:2[Ba2+] + 2[Ca2+] + [H+] = 2[SO42-] + [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/LThe 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.[ ][ ]−−+=246.4210SOCa[ ][ ]−−+=240.10210SOBa[ ] [ ][ ]−−−−−=+24240.10246.41010SOSOSO,The solubility of gypsum is hardly affected by thepresence of barite:Solubility of barite alone:[Ba2+][SO42-] = 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/LReplacement reactionsWe can also calculate [Ba2+]/[Ca2+] in equilibriumwith both barite and gypsum.[ ][ ]−−+=246.4210SOCa[ ][ ]−−+=240.10210SOBaWhat 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 & anionsMost 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?KK OHCOHOHHCOK OHHCOOHCOCO2KCOKH2H1H2322-3H1-3223332==+↔++↔++→−−−−+ 2HydrolysisDissociationA223-3A1-332K HCOHCOK HHCOCOH+−++↔+↔What is the pH of a solution prepared by dissolving 0.1 molof K2CO3 in 1 L of