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CH 17: Solubility and Complex-Ion Equilibria

CH 17: Solubility and Complex-Ion Equilibria. Renee Y. Becker CHM 1046 Valencia Community College. Solubility Equilibria. Solubility Product Constant, K sp Same as K c , K p , K w , K a , & K b Prod / reactant Coefficients are exponents, omit solids and pure liquids

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CH 17: Solubility and Complex-Ion Equilibria

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  1. CH 17: Solubility and Complex-Ion Equilibria Renee Y. Becker CHM 1046 Valencia Community College

  2. Solubility Equilibria Solubility Product Constant, Ksp Same as Kc, Kp, Kw, Ka, & Kb Prod / reactant Coefficients are exponents, omit solids and pure liquids CaF2(s) Ca2+(aq) + 2 F-(aq) Ksp = [Ca2+][F-]2

  3. Example1: Ksp Expressions Write Ksp expressions for the following a) Mg(OH)2 • SrCO3 • Ca3(AsO4)2 • Fe(OH)3

  4. Example 2: The solubility of silver bromate, AgBrO3, in water is 0.0072 g/L. Calculate Ksp

  5. Example 3: Calculate Ksp for copper(II)iodate, Cu(IO3)2. The solubillity of copper(II)iodate in water is 0.13 g/100mL

  6. Precipitation of Ionic Compounds Ion Product (IP) Same as Ksp but at some time, t, snapshot like Qc, reaction quotient CaF2(s) Ca2+ + 2 F- IP = [Ca2+][F-]2 If IP > Ksp solution is supersaturated and precipitation will occur If IP = Ksp the solution is saturated and equilibrium exists If IP< Ksp the solution is unsaturated and ppt will not occur

  7. Example 4: Will a precipitate form on mixing equal volumes of the following solutions? • 3.0 x 10-3 M BaCl2 and 2.0 x 10-3 M Na2CO3 (Ksp = 2.6 x 10-9 for BaCO3) • 1.0 x 10-5 M Ba(NO3)2 and 4.0 x 10-5 M Na2CO3

  8. Example 4: • From each of the following ion concentrations in a solution, predict whether a ppt will form in the solution A) [Ba2+] = 0.020 M [F-] = 0.015 M B) [Pb2+] = 0.035 M [Cl-] = 0.15 M

  9. Example 5: • The following solutions are mixed: 1 L of a 0.00010 M NaOH 1 L of a 0.0020 M MgSO4 Is a ppt expected, explain

  10. Measuring Ksp and Calculating Solubility from Ksp Example 6: A saturated solution of Ca3(PO4)2 has [Ca2+] = 2.01 x 10-8 M and [PO43-] = 1.6 x 10-5 M. Calculate Ksp for Ca3(PO4)2

  11. Factors that Affect Solubility 1. The Common ion effect MgF2(s) Mg2+(aq) + 2 F-(aq) If we try dissolve this in a aqueous solution of NaF the equilibrium will shift to the left. This will make MgF2 less soluble 2. Formation of Complex ions Complex ion: An ion that contains a metal cation bonded to one or more small molecules or ions, NH3, CN- or OH- AgCl(s) Ag+ + Cl- Ag+ + 2 NH3 Ag(NH3)2+ Ammonia shifts the equilibrium to the right by tying up Ag+ ion in the form of a complex ion

  12. Factors that Affect Solubility 3. The pH of the solution a) An ionic compound that contains a basic anion becomes more soluble as the acidity of the solution increases CaCO3(s) Ca2+ + CO32- H3O+ + CO32- HCO3- + H2O Net: CaCO3(s) + H3O+ Ca2+ + HCO3- + H2O Solubility of calcium carbonate increases as the pH decreases because the CO32- ions combine with protons to give HCO3- ions. As CO32- ions are removed from the solution the equilibrium shifts to the right to replenish the carbonate PH has no effect on the solubility of salts that contain anions of strong acids because these anions are not protonated by H3O+

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