Lecture #12 Complex Ions and Solubility
Stepwise exchange of NH 3 for H 2 O in M(H 2 O) 4 2+ M(H 2 O) 2 (NH 3 ) 2 2+ M(H 2 O) 4 2+ M(NH 3 ) 4 2+ M(H 2 O) 3 (NH 3 ) 2+ M(H 2 O)(NH 3 ) 3 2+
Formation Constants (K f ) for Some Complex Ions at 25ºC
The Amphoteric Behavior of Aluminum Hydroxide 3+ 0-1 (3 H3O+) (OH)- Al(H2O)6 3+ + 3 H2O Al(H2O)3(OH)3 Al(H2O)2(OH)4 - + H2O (aq) (s) (aq)
The Amphoteric Behavior of Aluminum Hydroxide
Solubiity of Amphoteric Hydroxides
Hydrated Metal Ions as Acids Most transition metal cations form insoluble hydroxides: Fe(OH)3(s) Fe 3+ (aq) + 3 OH (aq) Ksp = 1.1 10 36 Al(OH)3(s) Al 3+ (s) + 3 OH (aq) Ksp = 1.9 10 33 For some, solubility increases in strongly basic solution: Al(H 2 O) 3 (OH) 3 (s)+ OH (aq) Al(H 2 O) 2 (OH) 4 (aq)
Hydrated Metal Ions as Acids Fe(H2O)6 3+ + H2O Fe(H2O)5(OH) 2+ + H3O +
K a for Hydrated Metal Cations
1) Calculate the ph of 0.50 M CrCl3: Cr 3+ (aq) + 4 H2O (l) Cr(H2O)4 3+ (aq) Cr(H2O)3(OH) 2+ (aq) + H+ (aq) Ka= 1.0 x 10-4 R Cr(H2O)4 3+ (aq) Cr(H2O)3(OH) 2+ (aq) + H + (aq) I 0.5 0.0 0.0 C - x + x + x E 0.50-x x x Ka = [H + ][Cr(H2O)3(OH) 2+ ] [Cr(H2O)4 3+ ] (x) = 2 = 1.0 x 10 (.500-x) -4 x = 7.1 x 10-3 ph = 2.15
Solubility Equilibria
Solubility Equilibria All ionic compounds dissolve in water to some degree. Many compounds have such low solubility in water that we classify them as insoluble. What do we mean by insoluble? MnXm (s) nm m+ (aq) + mx n- (aq) We can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water.
Solubility Product The equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp. For an ionic solid MnXm, the reaction is: MnXm (s) nm m+ (aq) + mx n- (aq) The solubility product expression is: Ksp [M m+ ] n [X n- ] m
Molar Solubility Solubility (usually g/l) is the amount of solute that will dissolve in a given amount of solution at a particular temperature. Molar solubility is the number of moles of solute that will dissolve in a liter of solution. For the general reaction: MnXm (s) nm m+ (aq) + mx n- (aq) Molar Solubility = (n+m) (K sp ) (n) n (m) m
2) What is the molar solubility of Mg(OH)2 (Ksp = 5.6 x 10-12 ) Mg(OH)2(s) Mg 2+ (aq) + 2 OH (aq) Mg(OH)2(s) Mg 2+ (aq) + 2 OH (aq) (x) (2x) x = 1.1 10 4 M
K sp and Relative Solubility Molar solubility is related to Ksp. Molar solubility and Ksp cannot always be compared. In order to compare molar solubility and Ksp the compounds must have the same dissociation stoichiometry.
Solubility-Product Constants (K sp ) of Selected Ionic Compounds at 25 0 C Name, Formula Aluminum hydroxide, Al(OH) 3 Cobalt (II) carbonate, CoCO 3 Iron (II) hydroxide, Fe(OH) 2 Lead (II) fluoride, PbF 2 Lead (II) sulfate, PbSO 4 Mercury (I) iodide, Hg 2 I 2 Silver sulfide, Ag 2 S Zinc iodate, Zn(IO 3 ) 2 K sp 3 x 10-34 1.0 x 10-10 4.1 x 10-15 3.6 x 10-8 1.6 x 10-8 4.7 x 10-29 8 x 10-48 3.9 x 10-6
Relationship Between K sp and Solubility at 25 0 C No. of Ions Formula Cation:Anion K sp Solubility (M) 2 MgCO 3 1:1 3.5 x 10-8 1.9 x 10-4 2 PbSO 4 1:1 1.6 x 10-8 1.3 x 10-4 2 BaCrO 4 1:1 2.1 x 10-10 1.4 x 10-5 3 Ca(OH) 2 1:2 5.5 x 10-6 1.2 x 10-2 3 BaF 2 1:2 1.5 x 10-6 7.2 x 10-3 3 CaF 2 1:2 3.2 x 10-11 2.0 x 10-4 3 Ag 2 CrO 4 2:1 2.6 x 10-12 8.7 x 10-5
The Effect of Common Ion on Solubility Addition of a soluble salt that contains one of the ions of the insoluble salt, decreases the solubility of the insoluble salt. For example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2: PbCl2 (s) Pb 2+ (aq) + 2 Cl - (aq) Addition of Cl- shifts the equilibrium to the left
The Effect of a Common Ion on Solubility CrO 4 2- added PbCrO 4 (s) Pb 2+ (aq) + CrO 4 2- (aq) CrO 2 Equilibrium shifts to left
3) Calculate the molar solubility of Mg(OH)2 in a solution of 0.050 M MgCl2 (Ksp for Mg(OH)2 = 5.6 10 12 ) Mg(OH)2(s) Mg 2+ (aq) + 2 OH (aq) [Mg 2+ ] = 0.05 M Added Mg 2+ will shift the equilibrium to the left. [0.050][OH-] 2 (0.050)(2x) 2 x = 5.3 x 10-6 M without added Mg 2+, x = 1.1 10 4 M
The Effect of ph on Solubility (Insoluble Ionic Hydroxides) M(OH)n (s) M n+ (aq) + n OH - (aq) A high ph decreases solubility by adding OHto the equilibrium. A low ph increases solubility by removing OHfrom the equilibrium. H3O + (aq) + OH - (aq) 2 H2O (l)
The Effect of ph on Solubility Insoluble Ionic Compounds Containing Anions of Weak Acids CaF2 M2(CO3)n (s) (s) 2 Ca M n+ 2+ (aq) (aq) + 2 n FCO3 - (aq) 2- (aq) A low ph increases solubility by removing F - from the equilibrium. H3O + (aq) + F - (aq) HF (aq) + H2O (l)
4) Which of the following compounds solubility will not be affected by a low ph in solution? A) AgCl B) Sr(OH) 2 C) CaF 2 D) CuS E) SrCO 3 Cl - + H + OH - + H + F - + H + HCl H2O HF S 2- + H + HS - CO3 2- + H + HCO3 -
Precipitation Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound. One can compare the reaction quotient Q, for the current solution concentrations to the value of Ksp: If Q = Ksp, the solution is saturated, but no precipitate forms. If Q < Ksp, the solution is unsaturated, and no precipitate forms. If Q > Ksp, the solution is above saturation, a precipitate forms. (Some solutions with Q>Ksp will not precipitate unless disturbed - these are supersaturated solutions.)
5) The following two solutions are mixed: 275 ml of 0.134 M Pb(NO3)2 125 ml of 0.0339 M NaCl Will a precipitate form? Pb(NO3)2 (aq) + 2 NaCl (aq) PbCl2 (s) + 2 NaNO3 (aq) mol of [Pb 2+ ]= (0.134 mol/l) x (0.275 L) = 0.03685 mol of [Cl - ] = (0.0339 mol/l) x (0.125 L) = 0.004238 Final volume = 400 ml Final Concentrations: [Pb 2+ ] = 0.03685 mol/0.400 L = 0.0912 M [Cl - ] = 0.004238 mol/0.400 L = 0.0106 M PbCl2 (s) Pb 2+ (aq) + 2 Cl- (aq) Ksp = 1.6 x 10-5 Q = [Pb 2+ ][Cl - ] 2 = (0.0921)(0.016) 2 = 1.03 x 10-5 Q < Ksp A precipitate does not form!!
Separating Ions by Selective Precipitation Add precipitating ion Add precipitating ion Centrifuge Centrifuge
6) Consider a solution containing 0.10 M Ca 2+ and 0.02 Mg 2+. If the Mg 2+ is removed by precipitation as the hydroxide, will the Ca 2+ remain in the solution? Ca(OH)2(s) Ca 2+ (aq) + 2 OH (aq) Ksp = 4.7 10 6 Mg(OH)2(s) Mg 2+ (aq) + 2 OH (aq) Ksp = 5.6 10 12 What [OH ] is needed to ppt 0.10 M Ca 2+? Ksp = 4.7 x 10 6 = [Ca 2+ ]x[oh ] 2 = (0.100)(x) 2 x = 6.9 x 10-3 (When [OH] > 6.9 x 10-3, Q>K) How much Mg 2+ is in equilibrium with 6.9 x 10-3 M OH-? Ksp = 5.6 x 10 12 = [Mg 2+ ]x [OH ] 2 = (?)(6.9 x 10-3 ) 2? = 1.2 x 10-7 (1.2 x 10-7 )/(0.02) x 100 =.0006 % of Mg 2+ remains
7) A solution contains 3.8 10-2 M in Al 3+ and 0.29 M in NaF. If the Kf for AlF6 3- is 7 10 19, what is the final concentration of aluminum ions at equilibrium? Al 3+ (aq) + 4 F- (aq) AlF6 3- (aq) Kf = 7.0 x 10 19 R I C Al 3+ (aq) 4 F- (aq) AlF6 3- (aq) 3.8 x 10-2 0.29 0.0 -(3.8 x 10-2 -x) -6(3.8 x 10-2 -x) +(3.8 x 10-2 -x) E x 0.062 +6x (3.8 x 10-2 -x) [AlF6 3- ] Kf = [Al 3+ ][F - ] 6 = (3.8 x 10-2 -x) (3.8 x 10-2 ) x(0.062 + 6x) 6 x(0.062) 6 = 7.0 x 1019 x = 9.6 x 10-15 M