Solubility Products


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Discusses the chemical of slightly soluble compounds. Ksp and factors affecting solubility are included as well as solved problems.
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Solubility Products

  1. 1. Copyright Sautter 2015
  2. 2. SOLUBILITY EQUILIBRIUM • Solubility refers to the ability of a substance to dissolve. In the study of solubility equilibrium we generally deal with low solubility compounds (those which dissolve poorly) • Low solubility compounds are classified as precipitates. A set of solubility rules indicate which combination of ions generally form precipitates. These rules allow us to identify low solubility substances and potential precipitates. The concentrations of the ions present however are the final determiners of precipitate formation. 2
  3. 3. SOLUBILITY RULES • (1) Inorganic acids (those not consisting of carbon, hydrogen and oxygen) are soluble. Low molecular weight organic acids are soluble. • (2) Compound containing alkali metals and ammonium (NH4 +) cations are soluble. • (3) Compounds containing nitrates (NO3-), acetates (C2H3O2 -) and perchlorates (ClO4 -) are soluble. • (4) Compounds containing chlorides (Cl-) are soluble except for AgCl, PbCl2 and Hg2Cl2. Compounds containing bromides (Br-) and iodides (I-) have solubility similar to chlorides. • (5) Sulfate containing compounds are soluble except for PbSO4 and BaSO4. CaSO4 and Ag2SO4 are slightly soluble. • (6) Hydroxides are insoluble (precipitates) except Alkali Metal (I) and heavier Alkaline Earth Metal (II) hydroxides after Mg. • (7) Carbonates (CO3 -2), phosphates (PO4 -3) and arsenates (AsO4 -3) are insoluble except for Alkali Metal (I) and Alkaline Earth Metal (II) compounds • (8) Sulfides (S-2) are insoluble except for Alkali Metal (I) and Alkaline Earth Metal (II) compounds. 3
  4. 4. ALL LOW SOLUBILITY SYSTEMS ARE EQUILIBRIUM SYSTEMS RATE OF DISSOLVING = RATE OF CRYSTALIZATION RATE OF FORWARD PROCESS = RATE OF REVERSE PROCESS The amount of substance dissolved remains constant The system appears macroscopically static. 4
  5. 5. Ksp AND SOLUBILITY • Since solubility is an equilibrium phenomena, equilibrium concepts can readily be applied. • If the system is at equilibrium, the solution must be saturated (no more solid can be dissolved). Equilibrium concepts cannot be applied to unsaturated solutions (more solid can still be dissolved) since equilibrium has not yet been reached. • For a typical saturated solution of silver chloride, AgCl(s)  Ag+ (aq) + Cl- (aq) the equilibrium expression is: K = [Ag+] x [Cl-] / 1 or simply [Ag+] x [Cl-] Since AgCl is and solid it is not included in the equilibrium expression! • This K value is called the Ksp . “sp” stands for solubility product. The product of the concentrations of the soluble ions.5
  6. 6. CALCULATING Ksp • Problem: A saturated solution of BaSO4 contains 0.0025 grams in one liter. Calculate the Ksp for barium sulfate. • Solution: Since Ksp requires concentrations is moles per liter we must first calculate the number of moles present. Moles = 0.0025 gram / 233 grams per mole = 1.1 x 10-5 M • BaSO4(s)  Ba+2 (aq) + SO4 -2 (aq) • Each dissolved BaSO4 forms one Ba+2 ion and one SO4 -2 ion, therefore: • [BaSO4] = 1.1 x 10-5 M, [Ba+2] = [SO4 -2] = 1.1 x10-5 M • Ksp = [Ba+2] x [SO4 -2] = (1.1 x 10-5)2 = 1.2 x10 -10 6
  8. 8. Calculating Concentrations From Ksp • Problem: Find the concentration of dissolved Al(OH)3 in a saturated solution. • Solution: Al(OH)3 (s)  Al+3 (aq) + 3 OH- (aq) • Ksp = [Al+3] x [OH-]3 , Ksp = 1.9 x 10-33 • Each dissolved Al(OH)3 gives one Al+3 and three OH- ions therefore if [Al+3] = X, [OH-] = 3X and • Ksp = X (3X)3 = 27 X4 = 3.0 x 10-34 • X = (3.0 x 10-34 / 27)1/4 = 1.8 x 10-9 M 8
  9. 9. THE COMMON ION EFFECT • The common ion effect is an application of LeChatelier’s Principle. Recall that equilibrium shifts are caused be stresses applied to the system. • If additional products are added to an equilibrium system that system will shift towards the reactants. In a solubility equilibrium adding one of the dissolved ion components will cause a shift left and the solubility of the compound will decrease. • Reducing the concentration of one of the dissolved ions will cause a shift to the right thereby allowing the compound to dissolve to a greater extend. 9
  10. 10. THE COMMON ION EFFECT Add NaCl to the system System shifts left!10
  11. 11. THE COMMON ION EFFECT • Problem: Calculate the solubility of silver chloride in pure water and in a 0.10 M NaCl solution. Compare the results. • Solution: AgCl(s)  Ag+ (aq) + Cl- (aq) • Ksp = [Ag+] x [Cl-] = 1.8 x 10-10 • Each dissolved AgCl gives one Ag+ and one Cl- ion therefore if [Ag+] = X, [Cl-] = X • Ksp = X2 = 1.8 x 10-10, X = (1.8 x 10-10)1/2 = 1.3 x 10-5 M • In 0.10 M NaCl the concentration of Cl- = 0.10 M and • Ksp = [Ag+] (0.10)* = 1.8 x 10-10, [Ag+] = 1.8 x 10-10 / 0.10 • ]Ag+] = 1.8 x 10-9 M (the presence of the common Cl- ion in solution has shifted the equilibrium hard to the left and made the silver chloride much less soluble than in pure water • *Note: the Cl- from the AgCl in the common ion system is so small as compared to the 0.10 M Cl- from the NaCl that is was neglected.11
  12. 12. WILL A PRECIPITATE FORM? • In order to decide whether a precipitate will form when two solutions are mixed we must first calculate the value of Q (reaction quotient). Q is calculated with the same format as Ksp (an ion concentration product) • If the value of Q exceeds the Ksp value, a precipitate will form. • If the value of Q is less than Ksp, a precipitate will not be formed. • If Q = Ksp then the solution is a saturated solution. 12
  14. 14. 0.0001 M AgNO3 0.00002 M NaCl 100 ml 200 ml WILLA PRECIPITATE FORM? 14
  15. 15. WILL A PRECIPITATE FORM? • Problem: If 100 ml of 0.0001 M AgNO3 is mixed with 200 ml of 0.0002 M NaCl, will a precipitate form? • Solution: The potential precipitate is AgCl. (See the solubility rules). • Q = [Ag+] x [Cl-], when the two solutions are mixed, the concentrations of both the Ag+ ion and the Cl- ion will change due to dilution of one solution by the other. • M1V1= M2V2 (the dilution formula) • For Ag+ , 0.0001 (100) = [Ag+] (100 + 200) • [Ag+] = 3.33 x 10-5 M • For Cl- , 0.0002 (100) = [Cl-] (100 + 200) • [Cl-] = 6.67 x 10-5 M • Q = (3.33 x 10-5) (6.67 x 10-5) = 2.22 x 10-9, Ksp = 1.8 x 10-10 • Q > Ksp therefore a precipitate will form 15
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