Presentation on theme: "Solubility Equilibria AP Chemistry"— Presentation transcript:
1Solubility Equilibria AP Chemistry Chapter 15Solubility EquilibriaAP Chemistry
2The Solubility Product Constant, Ksp Many important ionic compounds are only slightly soluble in water and equations are written to represent the equilibrium between the compound and the ions present in a saturated aqueous solution.The solubility product constant, Ksp, is the product of the concentrations of the ions involved in a solubility equilibrium, each raised to a power equal to the stoichiometric coefficient of that ion in the chemical equation for the equilibrium.
3The Solubility Equilibrium Equation And Ksp CaF2 (s) Ca2+ (aq) + 2 F- (aq)Ksp = [Ca2+][F-]2 Ksp = 5.3x10-9As2S3 (s) 2 As3+ (aq) + 3 S2- (aq)Ksp = [As3+]2[S2-]3* Remember, solids are not in equilibrium expressions!
4Some Values For Solubility Product Constants (Ksp) At 25 oC
5Ksp And Molar Solubility The solubility product constant (Ksp) is related to the solubility of an ionic solute, but Ksp andMolar Solubility - the molarity of a solute in a saturated aqueous solutionKsp and Molar solubility are not the same thing.Calculating solubility equilibria fall into two categories:determining a value of Ksp from experimental datacalculating equilibrium concentrations when Ksp is known.
6Calculating Ksp From Molar Solubility It is found that 1.2x10-3 mol of lead (II) iodide, PbI2, dissolves in 1.0 L of aqueous solution at 25 oC. What is the Ksp at this temperature?PbI2 Pb I-2 iodide ions form, so you must multiply molarity by 2!Ksp = [Pb2+][I-]2Ksp = [1.2 x 10-3] [2(1.2 x 10-3)]2Ksp = 6.9 x 10-9
7Calculating Molar Solubility From Ksp Calculate the molar solubility of silver chromate, Ag2CrO4, in water from Ksp = 1.1x10-12 for Ag2CrO4.Ag2CrO4(s) 2Ag+ + CrO42-At equilibrium x xKsp = [Ag+]2[CrO4-2]1.1 x = (x)(2x)21.1 x = 4x3x = 6.5 x 10-5 M = [Ag2CrO4] = [CrO42-][Ag+] = 2(6.5 x 10-5)
8The Common Ion Effect In Solubility Equilibria The common ion effect also affects solubility equilibria.Le Châtelier’s principle is followed for the shift in concentration of products and reactants upon addition of either more products or more reactants to a solution.The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution.
9Solubility Equilibrium Calculation -The Common Ion Effect What is the solubility of Ag2CrO4 in 0.10 M K2CrO4?Ksp = 1.1x10-12 for Ag2CrO4.Comparison of solubility of Ag2CrO4In pure water: 6.5 x MIn 0.10 M K2CrO4: 1.7 x 10-6 MThe common ion effect!!Adding more CrO42- ions shifts the equilibrium back to the reactants, which is solid Ag2CrO4Ag2CrO4 2Ag+ + CrO42-
10Common Ion Effect Example: What will the molar solubility of CaF2 (Ksp = 4.0 x 10-11) in a 0.025M NaF solutionCaF2 Ca F and Ksp = [Ca2+][F-]2Before it dissolves[Ca2+] = 0 [F-] = 0.025After it dissolves[Ca2+] = x [F-] = xFrom NaFFrom CaF2Assume x is very small due to small Ksp4.0 x = x (0.025 – 2x)24.0 x = x(0.025)2x = 6.4 x 10-8 = molar solubility
11Determining Whether Precipitation Occurs Qsp is the ion product reaction quotient and is based on initial conditions of the reaction.Qsp can then be compared to Ksp.To predict if a precipitation occurs:- Precipitation should occur if Qsp > Ksp.- Precipitation cannot occur if Qsp < Ksp.- A solution is just saturated if Qsp = Ksp.Sometimes the concentrations of the ions are not high enough to produce a precipitate!
12Determining Whether Precipitation Occurs – An Example The concentration of calcium ion in blood plasma is M. If the concentration of oxalate ion is 1.0x10-7 M, do you expect calcium oxalate to precipitate? Ksp = 2.3x10-9.Three steps:Determine the initial concentrations of ions.Evaluate the reaction quotient Qip.Compare Qsp with Ksp.
13Determining Whether Precipitation Occurs The concentration of calcium ion in blood plasma is M. If the concentration of oxalate ion is 1.0x10-7 M, do you expect calcium oxalate, CaC2O4, to precipitate? Ksp = 2.3x10-9.Qsp = [Ca2+][C2O42-] = (0.0025)(1.0 x 10-7)Qsp = 2.5 x 10-10Qsp < Ksp therefore, no precipitate will form!!!
14Determining Whether Precipitation Occurs In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered.Example: A mL sample of M Pb(NO3)2 (aq) is mixed with mL of M NaI (aq). Should precipitation of PbI2 (s), Ksp = 7.1x10-9, occur?Calculate new concentrations in total volume of 400mls = 0.4L[Pb2+] = (0.250L)(0.0012M)/(0.400L) = 7.5 x 10-4 M[I-] = (0.150L)(0.0640M)/(0.400L) = MQsp = [Pb2+][I-]2 = (7.5 x 10-4)(0.024)2 = x 10-7Qsp > Ksp therefore a precipitate will form!
15Selective Precipitation The first precipitate to form when AgNO3(aq) is added to an aqueous solution containing Cl- and I- is yellow AgI(s).Essentially all the I- has precipitated before the precipitation of white AgCl(s) begins.AgCl (s) Ag+ (aq) + Cl- (aq) Ksp = 1.8x10-10AgI (s) Ag+ (aq) + I- (aq) Ksp = 8.5x10-17
16Selective Precipitation An Example Example: An aqueous solution that is 2.00 M in AgNO3 is slowly added from a buret to an aqueous solution that is M in Cl- and also M in I-AgCl (s) Ag+ (aq) + Cl- (aq) Ksp = 1.8x10-10AgI (s) Ag+ (aq) + I- (aq) Ksp = 8.5x10-17Which ion. Cl- or I-, is the first to precipitate from solution?(this depends on the Molar solubility not Ksp, but since both compounds give equal ions, then you can use Ksp)Is separation of the two ions by selective precipitation feasible?AgI will precipitate first, lower Ksp
17Comparing Solubilities Which salt will be the most soluble in water.CuS Ksp = 8.5 x 10-45Ag2S Ksp = 1.6 x 10-493. Bi2S3 Ksp = 1.1 x 10-73Molar solubilityx = 9.2 x 10-23Ksp = x2x = 3.4 x 10-17Ksp = (2x)2x = 4x3x = 1.0 x 10-15Ksp = (2x)2(3x)3 = 108x5Most Soluble = highest molar solubility!!!! Do not use Ksp!!
18Effect of pH on Solubility The solubility of an ionic solute may be greatly affected by pH if an acid-base reaction also occurs as the solute dissolves.In other words, some salts will not dissolve well in pure water, but will dissolve in an acid or a base.If the anion (A-) of the salt/precipitate is that of a weak acid, the salt/precipitate will dissolve more when in a strong acid (H+ ions will form HA with A-)However, if the anion of the precipitate is that of a strong acid, adding a strong acid will have no effect on the precipitate dissolving more.
19Effect of pH on Solubility How would the addition of HCl affect the solubility of PbCl2?Cl- is the conjugate base of a strong acid, thus it is a weak base.It will not react with H+ ions, so there is no effect.How would the addition of HCl affect the solubility of FeS?S2- is a strong base (conjugate base of weak acid)Thus, it will react with H+ ions to form H2S,This will shift the equilibrium to make more FeS dissolve!
20SummaryThe solubility product constant, Ksp, represents equilibrium between a slightly soluble ionic compound and its ions in a saturated aqueous solution.The common ion effect is responsible for the reduction in solubility of a slightly soluble ionic compound.The solubilities of some slightly soluble compounds depends strongly on pH.
21Qualitative Inorganic Analysis Acid-base chemistry, precipitation reactions, oxidation-reduction, and complex-ion formation all come into sharp focus in an area of analytical chemistry called classical qualitative inorganic analysis.“Qualitative” signifies that the interest is in determining what is present, not how much is present.Although classical qualitative analysis is not as widely used today as instrumental methods, it is still a good vehicle for applying all the basic concepts of equilibria in aqueous solutions.
23Cations of Group 1If aqueous HCl is added to an unknown solution of cations, and a precipitate forms, then the unknown contains one or more of these cations: Pb2+, Hg22+, or Ag+.These are the only ions to form insoluble chlorides.If there is no precipitate, then these ions must be absent from the mixture.If there is a precipitate, it is filtered off and saved for further analysis.The supernatant liquid is also saved for further analysis.
24Cation Group 1 (continued) Analyzing For Pb2+ Of the three possible ions in solution, PbCl2 is the most soluble in water.The precipitate is washed with hot water and the washings then treated with aqueous K2CrO4.If Pb2+ is present, chromate ion combines with lead ion to form a precipitate of yellow lead chromate, which is less soluble than PbCl2.If Pb2+ is absent, then the washings just become tinged yellow but no precipitate is in evidence.
25Cation Group 1 (continued) Analyzing For Ag+ Next, the undissolved precipitate is treated with aqueous ammonia.If AgCl is present, it will dissolve in this solution.If there is any remaining precipitate, it is separated from the supernatant liquid and saved for further analysis.The supernatant liquid (which contains the Ag+, if present) is then treated with aqueous nitric acid.If a precipitate reforms, then Ag+ was present in the solution, if no precipitate forms, then Ag+ was not present in the solution.
26Cation Group 1 (continued) Analyzing For Hg22+ When precipitate was treated with aqueous ammonia in the previous step, any Hg22+ underwent an oxidation-reduction reaction to form a dark gray mixture of elemental mercury and HgNH2Cl that precipitates from the solution.If this dark gray precipitate was observed, then mercury was present in the original unknown sample.If this dark gray precipitate was not observed, then mercury must have been absent from the original unknown sample.
27Group 1 Cation Precipitates left: cation goup 1 ppt: PbCl2, PbCl2, AgCl (all white)middle: product from test for Hg22+: mix of Hg (black) and HgNH2Cl (white)right: product from test for Pb2+: PbCrO4 (yellow) when K2CrO4(aq) is reacted with saturated PbCl2