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

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 CaF 2(s)  Ca 2+ (aq) + 2 F - (aq) K sp = [Ca 2+ ][F - ] 2 2

Example1: K sp Expressions Write K sp expressions for the following a) Mg(OH) 2 b)SrCO 3 c)Ca 3 (AsO 4 ) 2 d)Fe(OH) 3 3

Example 2: The solubility of silver bromate, AgBrO 3, in water is g/L. Calculate K sp 4

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

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

Example 4: Will a precipitate form on mixing equal volumes of the following solutions? a)3.0 x M BaCl 2 and 2.0 x M Na 2 CO 3 (K sp = 2.6 x for BaCO 3 ) b)1.0 x M Ba(NO 3 ) 2 and 4.0 x M Na 2 CO 3 7

From each of the following ion concentrations in a solution, predict whether a ppt will form in the solution A) [Ba 2+ ] = M [F - ] = M B) [Pb 2+ ] = M [Cl - ] = 0.15 M 8 Example 4:

Example 5: The following solutions are mixed: 1 L of a M NaOH 1 L of a M MgSO 4 Is a ppt expected, explain 9

Measuring K sp and Calculating Solubility from K sp Example 6: A saturated solution of Ca 3 (PO 4 ) 2 has [Ca 2+ ] = 2.01 x M and [PO 4 3- ] = 1.6 x M. Calculate K sp for Ca 3 (PO 4 ) 2 10

Factors that Affect Solubility 1.The Common ion effect MgF 2(s)  Mg 2+ (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 MgF 2 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, NH 3, CN - or OH - AgCl (s)  Ag + + Cl - Ag NH 3  Ag(NH 3 ) 2 + Ammonia shifts the equilibrium to the right by tying up Ag + ion in the form of a complex ion 11

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 CaCO 3(s)  Ca 2+ + CO 3 2- H 3 O + + CO 3 2-  HCO H 2 O Net: CaCO 3(s) + H 3 O +  Ca 2+ + HCO H 2 O Solubility of calcium carbonate increases as the pH decreases because the CO 3 2- ions combine with protons to give HCO 3 - ions. As CO 3 2- 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 H 3 O + 12