the solubility product constant k sp n.
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The Solubility Product Constant, K sp

The Solubility Product Constant, K sp

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The Solubility Product Constant, K sp

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  1. The Solubility Product Constant, Ksp • Many 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 equilibrium system is a heterogeneous system. • 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.

  2. The Solubility Equilibrium Equation And Ksp CaF2 (s)  Ca2+ (aq) + 2 F- (aq) Ksp = [Ca2+][F-]2 As2S3 (s) 2 As3+ (aq) + 3 S2- (aq) Ksp = [As3+]2[S2-]3 • * Remember, solids are not in equilibrium expressions!

  3. Some Values For Solubility Product Constants (Ksp) At 25 oC

  4. Ksp And Molar Solubility • The solubility product constant (Ksp) is related to the solubility of an ionic solute, but Ksp and Molar Solubility are not the same thing. • Molar Solubility - the molarity (concentration) of a solute in a saturated aqueous solution • Calculating solubility equilibria fall into two categories: • determining a value of Ksp from experimental data • calculating equilibrium concentrations when Ksp is known.

  5. Calculating 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? • 2 iodide ions form, so you must multiply molarity by 2! • PbI2 Pb2+ + 2I- • Ksp = [Pb2+][I-]2 • Ksp = [1.2 x 10-3] [2(1.2 x 10-3)]2 • Ksp = 6.9 x 10-9

  6. Calculating 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 2x x Ksp = [Ag+]2[CrO4-2] 1.1 x 10-12 = (x)(2x)2 1.1 x 10-12 = 4x3 x = 6.5 x 10-5 M = [Ag2CrO4] = [CrO42-] [Ag+] = 2(6.5 x 10-5)

  7. Try These • Page 464 #1-4

  8. The 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.

  9. Solubility 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 Ag2CrO4 In pure water: 6.5 x 10-5 M In 0.10 M K2CrO4: 1.7 x 10-6 M The common ion effect!! Adding more CrO42- ions shifts the equilibrium back to the reactants, which is solid Ag2CrO4 Ag2CrO4 2Ag+ + CrO42-

  10. Common Ion Effect Example: What will the molar solubility of CaF2 (Ksp = 4.0 x 10-11) in a 0.025M NaF solution CaF2 Ca2+ + 2F- and Ksp = [Ca2+][F-]2 Before it dissolves [Ca2+] = 0 [F-] = 0.025 After it dissolves [Ca2+] = x [F-] = 0.025 + 2x From NaF From CaF2 Assume x is very small due to small Ksp 4.0 x 10-11 = x (0.025 – 2x)2 4.0 x 10-11 = x(0.025)2 x = 6.4 x 10-8 = molar solubility

  11. Try These • Page 470 #1-3

  12. Determining 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!

  13. Determining Whether Precipitation Occurs – An Example The concentration of calcium ion in blood plasma is 0.0025 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.

  14. Determining Whether Precipitation Occurs The concentration of calcium ion in blood plasma is 0.0025 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 10-10 • Qsp < Ksp therefore, no precipitate will form!!!

  15. Determining Whether Precipitation Occurs In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered. Example: A 250.0 mL sample of 0.0012 M Pb(NO3)2 (aq) is mixed with 150.0 mL of 0.0640 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) = 0.024 M Qsp = [Pb2+][I-]2 = (7.5 x 10-4)(0.024)2 = 4.32 x 10-7 Qsp > Ksp therefore a precipitate will form!

  16. Try These • Page 468 #1-4

  17. Comparing Solubilities Molar solubility • Which salt will be the most soluble in water. • CuS Ksp = 8.5 x 10-45 • Ag2S Ksp = 1.6 x 10-49 3. Bi2S3 Ksp = 1.1 x 10-73 x = 9.2 x 10-23 Ksp = x2 x = 3.4 x 10-17 Ksp = (2x)2x = 4x3 x = 1.0 x 10-15 Ksp = (2x)2(3x)3 = 108x5 • Most Soluble = highest molar solubility!!!! Do not use Ksp!!

  18. Effect 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.

  19. Effect 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!

  20. Summary • The 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.

  21. Qualitative 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.

  22. Cations of Group 1 • If 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.

  23. Cation 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.

  24. Cation 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.

  25. Cation 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.

  26. Group 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