Solubility Product - funnel.sfsu.edu

1 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:4242 CaSOSOCaSPaaaK +=If anhydrite is a pure solid, then aCaSO4 = in dilute solutions: aCa2+ (Ca2+) and aSO42- (SO42-).KSP [Ca2+][SO42-] = is the Solubility of anhydrite in pure water?If anhydrite dissolution is the only source of both Ca2+ andSO42-, then: [Ca2+] = [SO42-] = xx2 = = = mol/LMWanhydrite = g/molSolubility = ( mol/L)( g/mol) = g/L4242 CaSOSOCaSPaaaK +=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 = [Ca2+]eq[SO42-]eqIAP = [Ca2+]act[SO42-]act[][]SPSP actactKIAPKSOCa== +242 SaturationindexSuppose a groundwater is analyzed to contain 5x10-2 mol/L Ca2+ and7x10-3 mol/L SO42.

2 Is this water saturated with respect to anhydrite (CaSO4(s))?KSP = = (5x10-2)(7x10-3) = = = = = > 1, , IAP > KSP, so the solution is supersaturatedand anhydrite should precipitate If = 1, , IAP = KSP, the solution would be saturated (equilibrium conditions) If < 1, , IAP < KSP, the solution would be undersaturated; the mineral should dissolveCaSO4 Ca2+ + SO42-How much salt should precipitate?Returning to the previous example, , 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-] = (5x10-2 - x)(7x10-3 - x) = - ( )x + ( ) = 0 After solving for the quadratic equation:x1 = mol/L; x2 = mol/LWe choose x2 (= )because the first root(x1= ) causes [SO4] to be , mol/L of anhydrite precipitates, or:( mol/L)( g/mol) = g/Land[Ca2+] = (5x10-2) - ( ) = mol/L[SO42-] = (7x10-3) - ( ) = mol/L[Ca2+]/[SO42-] increases with precipitation of anhydriteBefore precipitation:5x10-2/7x10-3 = precipitation: = mol/L of anhydrite precipitatesPrecipitation not only reduces the concentrations of ions and actually changes the chemical composition if theremaining the initial [Ca2+]/[SO42-] > 1, the remainingSolution is enriched in Ca2+.

3 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 : 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-] = Ba2+ + SO42-,KSP = [Ba2+][SO42-] = [Ca2+][SO42-] = [Ba2+][SO42-] = [SO42-] by substituting [Ca+]:[Ba2+] [Ca2+] = ) 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-] + = [SO42-]2[SO42-] = ( + )1/2 = mol/L[Ca2+] = = mol/L[Ba2+] = = mol/LThe least soluble salt (barite, KSP=10-10), contributes a negligibleamount of sulfate to the solution.

4 The more soluble salt(gypsum, KSP= ) supresses the Solubility of the lesssoluble salt (the common-ion effect). Barite can replacegypsum because barite is less soluble than gypsum.[][] += [][] += [][][] = + ,The Solubility of gypsum is hardly affected by thepresence of barite: Solubility of barite alone:[Ba2+][SO42-] = [Ba2+]2 = [Ba2+] = mol/LSolubility of gypsum alone:[Ca2+][SO42-] = [Ca2+]2 = [Ca2+] = mol/LReplacement reactionsWe can also calculate [Ba2+]/[Ca2+] in equilibriumwith both barite and gypsum.[][] += [][] += 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+] = [Ba2+] 10-10[Ca2+] , 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 classify salts by the strength of the acid and base from which they form:1.

5 Strong acid + strong basedo not hydrolyze2. Strong acid + weak basecations + OH- = acidic soln3. Weak acid + strong baseanions + H+ = basic soln4. Weak acid + weak baserelease both cation & anionsMost common rock-forming minerals of the crust are saltsof weak acids and strong bases, , carbonates and silicatesof alkali metals (Group 1) and alkaline earths (Group 2)form these is why groundwater in carbonate aquifersis commonly basic?K?KK OHCOHOHHCOK OHHCOOHCOCO2 KCOKH2H1H2322-3H1-3223332==+ ++ ++ +2 HydrolysisDissociationA223-3A1-332K HCOHCOK HHCOCOH+ ++ + What is the pH of a solution prepared by dissolving molof K2CO3 in 1 L of water?