Page III-15-1 / Chapter Fifteen Lecture Notes Flashback - Aqueous Salts! If one ion from the Soluble Compd. list is present in a compound, the compound is water soluble. PRECIPITATION REACTIONS Chapter 15 Ba(NO3)2? soluble BaCl2? soluble Chemistry 223 Professor Michael Russell Solubility of a Salt Ag Pb Hg2 Consider NaCl dissolving in water NaCl(s) qe Na BaSO4? insoluble Cl- Solubility of NaCl exceeded when solid precipitate does not dissolve All salts formed in this experiment are said to be INSOLUBLE They form when mixing moderately concentrated solutions of the metal ion with chloride ions. Ag Pb Hg2 Although all salts formed in this experiment are said to be insoluble, they do dissolve to some SLIGHT extent. AgCl(s) qe Ag Cl- When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED. Ag Pb Hg2 AgCl(s) qe Ag Cl- When solution is SATURATED, expt. shows that [Ag = 1.67 x 10-5 M. This is equivalent to the SOLUBILITY of AgCl. What is [Cl-? This is also equivalent to the AgCl solubility. Page III-15-1 / Chapter Fifteen Lecture Notes
Page III-15-2 / Chapter Fifteen Lecture Notes Ag Pb Hg2 Ag Pb Hg2 AgCl(s) qe Ag Cl- AgCl(s) qe Ag Cl- Saturated solution has [Ag = [Cl- = 1.67 x 10-5 M Kc = [Ag [Cl- = 2.79 x 10-10 Use this to calculate Kc Because this is the product of "solubilities", we call it Kc = [Ag [Cl- Ksp = solubility product constant = (1.67 x 10-5)(1.67 x 10-5) See: Solubility Guide = 2.79 x 10-10 Lead(II) Chloride PbCl2(s) qe Pb 2 Cl- Ksp = 1.9 x 10-5 Soluble Compounds Solubility of Lead(II) Iodide Solubility of Lead(II) Iodide Consider PbI2 dissolving in water Consider PbI2 dissolving in water PbI2(s) qe Pb 2 I- PbI2(s) qe Pb 2 I- Calculate Ksp if solubility = 0.00130 M Calculate Ksp if solubility = 0.00130 M 1. Solubility refers to how many moles of solid dissolve per L 1. Solubility = [Pb = 1.30 x 10-3 M [I- = 2 x [Pb = 2.60 x 10-3 M Solubility = [Pb = 1.30 x 10-3 M [I- = 2 x [Pb = 2.60 x 10-3 M 2. Ksp = [Pb [I-2 = [Pb {2 [Pb}2 = 4 [Pb3 Page III-15-2 / Chapter Fifteen Lecture Notes
Page III-15-3 / Chapter Fifteen Lecture Notes Solubility of Lead(II) Iodide Consider PbI 2 dissolving in water PbI 2 (s) qe Pb 2 I - Calculate K sp if solubility = 0.00130 M 2. K sp = 4 [Pb 3 = 4 (solubility) 3 K sp = 4 (1.30 x 10-3 ) 3 = 8.79 x 10-9 Notice that solubility of PbI 2 (x) and K sp related here by: K sp = 4x 3 Solubility and K sp Relations # cations # anions K sp and solubility (x) Examples 1 1 K sp = x2 NaCl, SrO, KClO 3 1 2 K sp = 4x 3 PbI 2, Mg(OH) 2 2 1 K sp = 4x 3 Na 2O, (NH 4 ) 2 SO 3 3 1 K sp = 27x 4 Li 3P, (NH 4 ) 3 PO 4 1 3 K sp = 27x 4 AlBr 3, Cr(NO 3 ) 3 2 3 K sp = 108x 5 Fe 2O 3, Al 2 (SO 4 ) 3 3 2 K sp = 108x 5 Ti 3 As 2, Mg 3 (PO 4 ) 2 See: Solubility Guide Solubility and K sp Relations Example: What is the solubility of copper(ii) phosphate if K sp = 1.4*10-37? Answer: Formula = Cu 3 (PO 4 ) 2 3 cations & 2 anions, so K sp = 108x 5 x = (1.4*10-37 /108) (1/5) = 1.7*10-8 M Example: What is K sp for magnesium carbonate if the solubility at 25 C is 2.6*10-3 M? Answer: Formula = MgCO 3, 1 cation & 1 anion, so K sp = x 2 K sp = (2.6*10-3 ) 2 = 6.8*10-6 Precipitating an Insoluble Salt Hg 2 Cl 2 (s) qe Hg 2 2 Cl - K sp = 1.1 x 10-18 = [Hg 2 [Cl - 2 If [Hg 2 = 0.010 M, what [Cl - is req'd to just begin the precipitation of Hg 2 Cl 2? That is, what is the maximum [Cl - that can be in solution with 0.010 M Hg 2 without forming Hg 2 Cl 2? Precipitating an Insoluble Salt Hg 2 Cl 2 (s) qe Hg 2 2 Cl - K sp = 1.1 x 10-18 = [Hg 2 [Cl - 2 Recognize that K sp = product of maximum ion concs. Precip. begins when product of ion concs. EXCEEDS the K sp. Precipitating an Insoluble Salt Hg 2 Cl 2 (s) qe Hg 2 2 Cl - K sp = 1.1 x 10-18 = [Hg 2 [Cl - 2 [Cl - that can exist when [Hg 2 = 0.010 M: [Cl - = If this conc. of Cl - is just exceeded, Hg 2 Cl 2 begins to precipitate. K sp 0.010 = 1.0 x 10-8 M Page III-15-3 / Chapter Fifteen Lecture Notes
Precipitating an Insoluble Salt Hg 2 Cl 2 (s) qe Hg 2 2 Cl - K sp = 1.1 x 10-18 Now raise [Cl - to 1.0 M when [Hg 2 = 0.010 M. What is the value of [Hg 2 at this point? [Hg 2 = K sp / [Cl - 2 = K sp / (1.0) 2 = 1.1 x 10-18 M The concentration of Hg 2 has been reduced by 10 16! Page III-15-4 / Chapter Fifteen Lecture Notes Adding an ion "common" to an equilibrium causes the equilibrium to shift back to reactant. Common Ion Effect Adding an Ion "Common" to an Equilibrium K sp for BaSO 4 = 1.1 x 10-10 = [Ba [SO 4 BaSO 4 (s) qe Ba SO 4 Solubility in pure water = [Ba = [SO 4 = x K sp = [Ba [SO 4 = x 2 x = (K sp ) 1/2 = 1.0 x 10-5 M Note 1:1 ratio of cation to anion: K sp = x 2 K sp for BaSO 4 = 1.1 x 10-10 = [Ba [SO 4 BaSO 4 (s) qe Ba SO 4 So... Solubility in pure water = 1.0 x 10-5 mol/l. Now dissolve BaSO 4 in water already containing 0.010 M Ba. Which way will the "common ion" shift the equilibrium? Will solubility of BaSO 4 be less than or greater than in pure water? K sp for BaSO 4 = 1.1 x 10-10 = [Ba [SO 4 BaSO 4 (s) qe Ba SO 4 [Ba [SO 4 initial 0.010 0 change y y equilib. 0.010 y y Page III-15-4 / Chapter Fifteen Lecture Notes
Page III-15-5 / Chapter Fifteen Lecture Notes K sp for BaSO 4 = 1.1 x 10-10 = [Ba [SO 4 BaSO 4 (s) qe Ba SO 4 K sp = [Ba [SO 4 = (0.010 y) (y) Because y < 1.0 x 10-5 M (= x, the solubility in pure water), this means 0.010 y is about equal to 0.010. Therefore, K sp = 1.1 x 10-10 = (0.010)(y) y = 1.1 x 10-8 M = solubility in presence of added Ba ion. K sp for BaSO 4 = 1.1 x 10-10 = [Ba [SO 4 BaSO 4 (s) qe Ba SO 4 Solubility in pure water = x = 1.0 x 10-5 M Solubility in presence of added Ba = 1.1 x 10-8 M Le Chatelier's Principle is followed! See: Solubility Guide DISTRACTIONS!!! Separating Metal Ions Cu, Ag, Pb K sp Values AgCl 1.8 x 10-10 PbCl 2 1.7 x 10-5 PbCrO 4 1.8 x 10-14 Separating Salts by Differences in K sp A solution contains 0.020 M Ag and 0.020 M Pb. Add CrO 4 to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. Which precipitates first? K sp for Ag 2 CrO 4 = 9.0 x 10-12 = [Ag 2 [CrO 4 K sp for PbCrO 4 = 1.8 x 10-14 = [Pb [CrO 4 The substance whose K sp is first exceeded precipitates first. The ion requiring the lesser amount of CrO 4 ppts. first. Separating Salts by Differences in K sp Page III-15-5 / Chapter Fifteen Lecture Notes A solution contains 0.020 M Ag and 0.020 M Pb. Add CrO 4 to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. Which precipitates first? K sp for Ag 2 CrO 4 = 9.0 x 10-12 = [Ag 2 [CrO 4 K sp for PbCrO 4 = 1.8 x 10-14 = [Pb [CrO 4 - Calculate [CrO 4 required by each ion [CrO 4 to ppt. PbCrO 4 = K sp / [Pb = 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M [CrO 4 to ppt. Ag 2 CrO 4 = K sp / [Ag 2 = 9.0 x 10-12 / (0.020) 2 = 2.3 x 10-8 M PbCrO 4 precipitates first.
Page III-15-6 / Chapter Fifteen Lecture Notes Separating Salts by Differences in Ksp Separating Salts by Differences in Ksp A solution contains 0.020 M Ag and 0.020 M Pb. Add CrO4 to precipitate red Ag2CrO4 and yellow PbCrO4. PbCrO4 ppts. first. Ksp (Ag2CrO4)= 9.0 x 10-12 = [Ag2 [CrO4 A solution contains 0.020 M Ag and 0.020 M Pb. Add CrO4 to precipitate red Ag2CrO4 and yellow PbCrO4. Ksp (Ag2CrO4)= 9.0 x 10-12 = [Ag2 [CrO4 Ksp (PbCrO4) = 1.8 x 10-14 = [Pb [CrO4 How much Pb remains in solution when Ag begins to precipitate? Ksp (PbCrO4) = 1.8 x 10-14 = [Pb [CrO4 How much Pb remains in solution when Ag begins to precipitate? We know that [CrO4 = 2.3 x 10-8 M to begin to precipitate Ag2CrO4. What is the Pb conc. at this point? [Pb = Ksp / [CrO4 = 1.8 x 10-14 / 2.3 x 10-8 M = 7.8 x 10-7 M Lead ion has dropped from 0.020 M to < 10-6 M Formation Constants Formation Constants (Kf) at 25 C Complex Ions are systems with Lewis bases connected around the (Lewis) acidic metal center. Examples: Zn(NH3)4, Ag(CN)1 Can write a Formation Constant, Kf Ag 2 CN-1 qe Ag(CN)2-1, Example: Ag 2 CN-1 qe Ag(CN)1 ions are reactants, complex ion is product, usually written as net ionic reactions and [Ag(CN) 1 2 Kf = = 5.6 *1018 1 2 [Ag [CN Kf values usually quite large (product-favored) and product is always the complex ion End of Chapter 15 Formation Constants Complex Ions can be helpful when dissolving solids. Ex: AgCl(s) and Ag(NH3) AgCl(s) qe Ag Ag Cl- 2 NH3 qe Ag(NH3)2 Ksp Kf -------------------------------------AgCl(s) 2 NH3 qe Ag(NH3) ClKnet = Ksp * Kf See: Chapter Fifteen Study Guide Chapter Fifteen Concept Guide Types of Equilibrium Constants Solubility Guide Page III-15-6 / Chapter Fifteen Lecture Notes