215.6 Solubility Equilibria Solubility of various compounds plays important roles.Sugar and table salt’s ability to dissolve in water help us to flavor food.Calcium sulfate is less soluble in hot water than cold water, which allows us to coat tubes in boilers.The low solubility of barium sulfate allows a safe way to X-ray the gastrointestinal tract.
3Aqueous EquilibriaAssume that when an ionic solid dissolves, that it dissociates completely.CaF2(s) → Ca2+ (aq) + 2F- (aq)There is a possibility for the ions to collide and re-form the solid phase.Ca2+ (aq) + 2F- (aq)→ CaF2(s)Finally, equilibrium is reached, and no more solid will dissolve (saturated).CaF2(s) ↔ Ca2+ (aq) + 2F- (aq)
4Solubility Product Constant Remember that solids are left out of equilibrium expressionsKsp=[Ca2+][F-]2Ksp is the solubility product constant (sometimes referred to as just the “solubility product”).*Notice that the amount of solid present nor the size of particles effect the equilibrium.
6Solubility vs. Solubility Product Solubility is an equilibrium position.Solubility product is an equilibrium constant and has ONE value for a particular solid at a particular temperature.
7Calculating Ksp from Solubility Calculate Ksp for bismuth sulfide (Bi2S3), which has a solubility of 1.0 x mol/L at 25°C.Ksp= [Bi3+]2[S2-]3=(2.0 x 10-15)2 (3.0 x 10-15)3 = 1.1 x 10-73HW odd on page 743Bi2S3(s) ↔2Bi3+(aq) +3S2- (aq)INot neededC-1.0 x = -x+2x+3xE2.0 x 10-153.0 x 10-15
8Calculating Solubility from Ksp The Ksp value from Cu(IO3)2 is 1.4 x 10-7 at 25°C. Calculate its solubility.Ksp=[Cu2+][IO3-]2=[x][2x]2=(x)(4x2)=4x31.4 x 10-7=4x3X=3.3 x 10-3 mol/L= Cu(IO3)2 solubilityHW on page 743Cu(IO3)2 (s) ↔Cu2+(aq) +2 IO3- (aq)INot neededC-x+x+2xE
9Relative Solubilities Use Ksp values to predict relative solubilities of a group of salts.Salts producing the same number of ionsEx. AgI (s), CuI (s), CaSO4 (s)Solubility is directly proportional to Ksp because each compound produces 2 ions.Largest Ksp= most solubleSmallest Ksp= least solubleSalts producing differing numbers of ions.Ex. CuS (s), Ag2S (s), Bi2S3 (s)Solubility NOT directly proportional to Ksp and you have to calculate the solubilities.
11Common Ion EffectConsider, the solubility of solid Ag2CrO4 (Ksp=9.0x10-12) in a M aqueous solution of AgNO3.What is the common ion?What is the [Ag+]0?What is the [CrO4-2]0?What is the Ksp expression?
12X=9.0 x 10-10 M= solubility of Ag2CrO4 (s) 2 Ag+ (aq) +CrO4 2- (aq)INot needed0.100 MC-x+2x+xExKsp = [Ag+]2[CrO4]2-9.0 x = ( x)2 (x)*Since Ksp is small, x is considered negligible compared to M; therefore x ≈ 0.1009.0 x = (0.100)2 (x)X=9.0 x M= solubility of Ag2CrO4 (s)[Ag+]=0.100 M and [CrO4-]=9.0 x M
13Solubility of Ag2CrO4 in pure water = 1.3 x 10-4 M Solubility of Ag2CrO4 in M AgNO3=9.0x10-10 M*Note that solubility of a solid is lowered due to the common ion effect.Example on page 743HW and on page 743.
14pH and SolubilityIf the anion (X-) is an effective base (or HX is a weak acid), the salt MX will increase solubility in an acidic solution.Effective bases include OH-, S2-, CO3-, C2O42-, and CrO42-.Salts with these anions are more soluble in acidic solutions that pure water.HW (in reference to 15.81) on page 743
15Precipitation Precipitation: the formation of a solid from solution Ion Product (Q)=defined the same as Ksp except initial concentrations are usedEx. Ca(NO3)2 (aq) is mixed with NaF (aq). What is the ion product for CaF2?Q=[Ca2+]0[F-]02
16To predict if precipitation will occur: If Q>Ksp, precipitation occurs until concentrations are reduced to satisfy Ksp.If Q<Ksp, no precipitation occurs.
17Determining Precipitation Conditions A solution is prepared by adding mL of 4.00 x 10-3 M Ce(NO3)3 to mL of 2.00 x 10-2 M KIO3. Will Ce(IO3)3 (Ksp=1.9 x 10-10) precipitate from this solution?Calculate initial conditions.Solve for Q.Q=[Ce3+]0[IO3-]03=(2.86x10-3)(5.71x10-3)3= 5.32 x 10-10Compare Q to Ksp.Q>Ksp, so Ce(IO3)3 will precipitate.Example on page 743 (Ksp values on page 718)
18Precipitation and Equilibrium Concentrations A solution is prepared by mixing mL of 1.00x10-2 M Mg(NO3)2 and mL of 1.00x10-1 M NaF. Calculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2 (Ksp=6.4x10-9).
19Find Initial Concentrations 2. Find Q3. Compare Q vs. KspQ>Ksp, so solid MgF2 will form.* The next steps are used to determine equilibrium concentrations.
20Run the reaction to completion. (BRA) Calculate the concentration of excess reactant.[F-]excess=22.0 mmol / mL = 5.50 x 10-2 MMg2+ +2 F →MgF2(s)Before rxn(150.0 mL) (1.00x10-2M)=1.50 mmol(250.0 mL) (1.00x10-1M)=25.0 mmolReaction-x-2 x+xAfter rxn=0*limiting reactant25.0-2(1.50)=22.0 mmol*excess1.50 mmol
21Determine concentrations at equilibrium (ICE). Ksp= 6.4x10-9=[Mg2+][F-]2=(x)(5.50x10-2+2x)2 ≈ (x)(5.50x10-2)2X=2.1 x 10-6 M = [Mg2+][F-]=5.50x10-2 MHW15.99 on page 743MgF2 (s) ↔Mg2+ (aq) +2F- (aq)INot needed0 M5.50 x 10-2 MC-x+x+2xE+ x5.50 x x
22Selective Precipitation Using anions that form precipitates with only one or a few metal ions in a mixture in order to separate the metal ions.Most insoluble sulfide salts can be precipitated in an acidic solutions.Soluble sulfide salts can be precipitated by making the solution slightly basic.
23-For PbI2: Ksp=1.4x10-8=[Pb2+][I-]2 1.4x10-8=(2.0x10-3) [I-]2 A solution contains 1.0x10-4 M Cu+ and 2.0x10-3 M Pb2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp=1.4x10-8) or CuI (Ksp=5.3x10-12) precipitate first? Specify the [I-] necessary to begin precipitation of each salt.-For PbI2: Ksp=1.4x10-8=[Pb2+][I-]21.4x10-8=(2.0x10-3) [I-]2[I-]=2.6 x 10-3 M is necessary to begin precipitation-For CuI: Ksp=5.3x10-12=[Cu+][I-]5.3x10-12=(1.0x10-4)[I-][I-]=5.3x10-8 M is necessary to begin precipitation*CuI will precipitate first since [I-] required is less.HW on page 744
25Groups: Insoluble Chlorides (Ag+, Pb2+, Hg22+) Sulfides insoluble in acid solution (Hg2+, Cd2+, Bi3+, Cu2+, Sn4+)Sulfides insoluble in basic solution (Co2+, Zn2+, Mn2+, Ni2+, Fe2+, Cr3+, Al3+)Insoluble Carbonates (Group 2A)Alkali Metal and ammonium ions (flame test)
26Complex Ion Equilibria Complex Ion: a charged species with a metal ion surrounded by ligands (Ag(NH3)+)The ligand donates its lone electron pair to an empty orbital of the metal ion to form a covalent bondLigand: a Lewis base (H2O, NH3, Cl-, CN-)Coordination number: the number of ligands attached to the metal ion
28Formation constants or stability constants: Metal ions add ligands one at a time in steps characterized by their own equilibrium constantsAg+ + NH3↔ Ag(NH3)+ K1=2.1x103Ag(NH3)+ + NH3↔ Ag(NH3)2+ K2=8.2x103All species (Ag+, NH3, Ag(NH3)+, Ag(NH3)2+) exist at equilibrium.HW on page 744
29Usually, [ligand] is much larger than [metal ion] and approximations are used to simplify problems. Assume both reactions go to completionAg+ + NH3↔ Ag(NH3)+ K1=2.1x103Ag(NH3)+ + NH3↔ Ag(NH3)2+ K2=8.2x103Ag+ + 2 NH3 → Ag(NH3)2+ K=K1 x K2From here, use BRA to find concentrations and the equilibrium constant.Assume the ligand amount consumed is neglibleExamples a, 105, 109 on page 744HW b, 107on page 744
30Strategies for Dissolving Water-Insoluble Ionic Solids If the anion of the solid is a good base, solubility is increased with the addition of an acidIf the anion is NOT a good base, solids can be dissolved in a solution containing a ligand to form stable complex ions with its cation.
31Aqueous Ammonia is Added to Silver Chloride (white) Aqueous Ammonia is Added to Silver Chloride (white). Silver Chloride, Insoluble in Water, Dissolves to Form Ag(NH3)2+ (aq) and Cl-(aq)