BaSO4(s) Ba2+(aq) + SO42–(aq) The Solubility Product Constant, Ksp • Many important ionic compounds are only slightly soluble in water (we used to call them “insoluble” – Chapter 4). • An equation can represent the equilibrium between the compound and the ions present in a saturated aqueous solution: • Solubility product constant, Ksp: the equilibrium constant expression for the dissolving of a slightly soluble solid. Ksp= [ Ba2+ ][ SO42–]
Example 16.1 Write a solubility product constant expression for equilibrium in a saturated aqueous solution of the slightly soluble salts (a) iron(III) phosphate, FePO4, and (b) chromium(III) hydroxide, Cr(OH)3.
Ksp and Molar Solubility • Ksp is an equilibrium constant • Molar solubility is the number of moles of compound that will dissolve per liter of solution. • Molar solubility is related to the value of Ksp, but molar solubility and Ksp are not the same thing. • In fact, “smaller Ksp” doesn’t always mean “lower molar solubility.” • Solubility depends on both Ksp and the form of the equilibrium constant expression.
Example 16.2 At 20 °C, a saturated aqueous solution of silver carbonate contains 32 mg of Ag2CO3 per liter of solution. Calculate Kspfor Ag2CO3 at 20 °C. The balanced equation is Ag2CO3(s) 2 Ag+(aq) + CO32–(aq) Ksp= ? Example 16.3 From the Ksp value for silver sulfate, calculate its molar solubility at 25 °C. Ag2SO4(s) 2 Ag+(aq) + SO42–(aq) Ksp= 1.4 x 10–5 at 25 °C
Example 16.4 A Conceptual Example Without doing detailed calculations, but using data from Table 16.1, establish the order of increasing solubility of these silver halides in water: AgCl, AgBr, AgI.
The Common Ion Effectin Solubility Equilibria • The common ion effect affects solubility equilibria as it does other aqueous equilibria. • The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution.
Common Ion Effect Illustrated When Na2SO4(aq) is added to the saturated solution of Ag2SO4 … … [Ag+] attains a new, lower equilibrium concentration as Ag+ reacts with SO42–to produce Ag2SO4. Calculate the molar solubility of Ag2SO4 in 1.00 M Na2SO4(aq).
Will Precipitation Occur? Is It Complete? • Qip can then be compared to Ksp. • Precipitation should occur if Qip > Ksp. • Precipitation cannot occur if Qip < Ksp. • A solution is just saturated if Qip = Ksp. • In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered. • Qip is the ion product reaction quotient and is based on initial conditions of the reaction. Qip andQc: new look, same great taste!
Example 16.6 If 1.00 mg of Na2CrO4 is added to 225 mL of 0.00015 M AgNO3, will a precipitate form? Ag2CrO4(s) 2 Ag+(aq) + CrO42–(aq) Ksp= 1.1 x 10–12 Example 16.8 If 0.100 L of 0.0015 M MgCl2 and 0.200 L of 0.025 M NaF are mixed, should a precipitate of MgF2 form? MgF2(s) Mg2+(aq) + 2 F–(aq) Ksp= 3.7 x 10–8
Simple explanation problem… Pictured here is the result of adding a few drops of concentrated KI(aq) to a dilute solution of Pb(NO3)2. What is the solid that first appears? Explain why it then disappears.
To Determine Whether Precipitation Is Complete • A slightly soluble solid does not precipitate totally from solution … • … but we generally consider precipitation to be “complete” if about 99.9% of the target ion is precipitated (0.1% or less left in solution). • Three conditions generally favor completeness of precipitation: • A very small value of Ksp. • A high initial concentration of the target ion. • A concentration of common ion that greatly exceeds that of the target ion.
Example 16.9 To a solution with [Ca2+] = 0.0050 M, we add sufficient solid ammonium oxalate, (NH4)2C2O4(s), to make the initial [C2O42–] = 0.0051 M. Will precipitation of Ca2+ as CaC2O4(s) be complete? CaC2O4(s) Ca2+(aq) + C2O42–(aq) Ksp= 2.7 x 10–9
CaF2(s) Ca2+(aq) + 2 F–(aq) AgCl(s) Ag+(aq) + Cl–(aq) Effect of pH on Solubility—fly in the Ointment • If the anion of a precipitate is that of a weak acid, the precipitate will dissolve somewhat when the pH is lowered: Added H+ reacts with, and removes, F–; LeChâtelier’s principle says more F– forms. • If, however, the anion of the precipitate is that of a strong acid, lowering the pH will have no effect on the precipitate. H+ does not consume Cl– ; acid does not affect the equilibrium.
Example 16.11 What is the molar solubility of Mg(OH)2(s) in a buffer solution having [OH–] = 1.0 x 10–5 M, that is, pH = 9.00? Mg(OH)2(s) Mg2+(aq) + 2 OH–(aq) Ksp= 1.8 x 10–11 Another one of those “explain” problems Without doing detailed calculations, determine in which of the following solutions Mg(OH)2(s) is most soluble: (a) 1.00 M NH3 (b) 1.00 M NH3 /1.00 M NH4+ (c) 1.00 M NH4Cl.
Neutralization Reactions • At the equivalence point in an acid–base titration, the acid and base have been brought together in precise stoichiometric proportions. • The endpoint is the point in the titration at which the indicator changes color. • Ideally, the indicator is selected so that the endpoint and the equivalence point are very close together. • The endpoint and the equivalence point for a neutralization titration can be best matched by plotting a titration curve, a graph of pH versus volume of titrant.
Titration Curve, Strong Acid with Strong Base Bromphenol blue, bromthymol blue, and phenolphthalein all change color at very nearly 20.0 mL At about what volume would we see a color change if we used methyl violet as the indicator?
Example 15.20 Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH: H3O+ + Cl– + Na+ + OH– Na+ + Cl– + 2 H2O (a) before the addition of any NaOH (b) after the addition of 10.00 mL of 0.500 M NaOH (c) after the additionof 20.00 mL of 0.500 M NaOH (d) after the addition of 20.20 mL of 0.500 M NaOH
Titration Curve, Weak Acid with Strong Base The equivalence-point pH is NOT 7.00 here. Why not?? Bromphenol blue was ok for the strong acid/strong base titration, but it changes color far too early to be useful here.
Sample problem for WA/SB titration • The titration of 100.0 mL of 0.016 M HOCl (Ka = 3.5 x 10-8 with 0.0400 M NaOH. How many mL of 0.040 M NaOH are required to reach the equivalence point? • Calc the pH after addition of 10.0 mL NaOH • Halfway to equivalence point • At equivalence point.