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Chapter 17 - Solubility

Chapter 17 - Solubility. Online HW 17 is due before Friday pm, 5/10/2013 Exam #2: Will cover chapters 17 & 18. Problems pg 733 - 740. Chapter 17 1 through 8, 11, 12, 13, 15, 17, 21, 22, 23, 25, 26, 27, 29, 31, 33, 35, 39, 41, 45, 53, 57.

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Chapter 17 - Solubility

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  1. Chapter 17 - Solubility Online HW 17 is due before Friday pm, 5/10/2013 Exam #2: Will cover chapters 17 & 18.

  2. Problems pg 733 - 740 Chapter 17 1 through 8, 11, 12, 13, 15, 17, 21, 22, 23, 25, 26, 27, 29, 31, 33, 35, 39, 41, 45, 53, 57

  3. I. Introduction - Solubility of Ionic Compounds • Know these 3 rules for the solubility of ionic compounds: (see page 131, Table 4.1 for expanded solubility rules – useful for online HW) 1. All group 1 compounds (except H), all NH4+, all NO3-, most C2H3O2- compounds are SOLUBLE. 2. All Cl-, Br-, I- compounds are SOLUBLE except Ag+, Hg2+2, Pb+2 3. All S-2 & OH- compds are INSOLUBLE except group 1, NH4+, Ca-Ba • This chapter deals with quantitative aspects of ionic compounds that are insoluble or slightly soluble.

  4. I. Introduction • Many chemicals are slightly soluble in water, and their solubility can be represented with an equilibrium expression. Note that solids drop out of the expression since the M of the solid within it’s phase does not change. The equilibrium constant is sometimes called the Solubility Product or Ksp. • Example: 1CaC2O4 (s)1Ca2+(aq) + 1C2O42-(aq) Ksp = 2.3x10-9 = [Ca2+]1[C2O42-]1 O O [ -O-C - C-O- ] Ca++

  5. I. Introduction Examples Continued • PbI2 (s)1Pb2+ + 2I- 6.5x10-9 = [Pb2+]1[I-]2 Note: 1) PbI2 is a yellow, toxic solid - used as a paint pigment. 2) The smaller the value for Ksp, the less soluble the solid. 3) Table 17.1 (pg 713) & A-15 are tables of Ksp values. • Ag2S (s)2Ag+ + 1S2- 6x10-50 = [Ag+]2[S2-]1 Note: Some black stains on pieces of silver may be Ag2S, a very insoluble compound.

  6. II. Calculations A. Ksp • Ksp can be calculated from solubility data. Example: • Calculate Ksp of Silver Carbonate if it’s solubility is 1.2x10-4M. 1Ag2CO3(s)2Ag+ + 1CO32- Ksp = [Ag+]2[CO32-]1 If 1.2x10-4M of Ag2CO3 dissolves at equilibrium, then determine the M of Ag+ & CO32-: [Ag+] = 2 x 1.2x10-4 = 2.4x10-4M&[CO32-] = 1.2x10-4M Ksp = [2.4x10-4]2[1.2x10-4]1 = 6.9x10-12 - Note: 1) equilibrium concentrations go into the expression 2) the square exponent [Ag+]2 is always present.

  7. II. Calculations B. SolubilityExample #1 • Calculate the M solubility of Barium Sulfate if the Ksp = 1.1x10-10 Let x = M of BaSO4 that dissolves: BaSO4 (s) (-----) Ba2+ + SO42- x dissolves x x 1.1x10-10 = [Ba2+][SO42-] 1.1x10-10 = [Ba2+][SO42-] = [x][x] = x2 x = √1.1x10-10 = 1.0x10-5M= Solubility of BaSO4

  8. II. Calculations B. Solubility Example #2 • Calculate the M solubility of silver sulfate if Ksp = 1.4x10-5 Let x = M of silver sulfate that dissolves 1Ag2SO4 (s) (-----) 2Ag+ + 1SO42- 1.4x10-5 = [Ag+]2[SO42-]1 x dissolves 2x x 1.4x10-5 = [Ag+]2[SO42-] = [2x]2[x] = 4x3 4x3 = 1.4x10-5 x = (1.4x10-5/4)1/3= (3.5x10-6)0.333=1.5x10-2M 1.5x10-2M= M of Ag2SO4dissolves = Solubility

  9. II. Calculations C. Solubility – Common Ion • The presence of a common ion from another source will lower the solubility (Le Chatelier’s principle). • Calculate the solubility of BaSO4 in the presence of 1.5 M Na2SO4. Let x = M of BaSO4 that dissolves. Ksp= 1.1x10-10 BaSO4(s) (-----) Ba2+ + SO42- x dissolves x x + 1.5 ≈ 1.5 (1.5/1.1x10-10 = 1.4x1010 >>100) 1.1x10-10 = [Ba2+][SO42-] = [x][1.5] x = (1.1x10-10)/1.5 = 7.3x10-11M (solubility) Note: solubility = 1.0x10-5M without 1.5 M SO42-(7000000 times less soluble in 1.5 M SO4-2)

  10. II. Calculations D. Precipitation • We can calculate whether a precipitate (ppt) will form. • Let Q = Ion Product (Same as K except use initial concentrations; see chapter 14) If Q is less (<) than K, then NO ppt will form If Q = K, then at equilibrium & precipitation JUST starts If Q is greater (>) than K, then YES, there will be a ppt

  11. II. Calculations D. Precipitation Example 1.0 mg of Na2CrO4 is added to 225 mL of 1.5x10-4 M AgNO3. Will a ppt form? Ag2CrO4 (s)2 Ag+ + 1 CrO42-Ksp = 1.1x10-12 1.1x10-12 = (Ag+)2(CrO42-)1 - Problem Solving Method: 1. get INITIAL M’s of all products; 2. plug into Equilibrium Expression; 3. Calculate Q & compare to Ksp. 1.Init M Na2CrO4: 1.0x10-3 g Na2CrO4 x (1 mol/162g)/0.225 L = 2.74x10-5MCrO42- Init M Ag+:1.5x10-4 M(note: we do not multiply by 2 - why?) 2. Q = (Ag+)2(CrO42-)1 = (1.5x10-4)2(2.74x10-5)1 = 6.2x10-13 3.Q (6.2x10-13) < Ksp (1.1x10-12) Since Q is less than Ksp, there is no precipitation

  12. III. Other Influences on Solubility A. pH - If the anion or cation of a slightly soluble compound is influenced by acidity, then the compound solubility is influenced by pH pH Example 1: 1) H2S (----) 2H+ + S-2Ka = small 2) ZnS(s) (----) Zn2+ + S2- Ksp = small - Will addition of H+increase , decrease, no effect on the solubility of ZnS ? - Use both equilibria. a) Which way will added HCl [H+] shift rxn #1? b) Which way will rxn #2 then be shifted? Use Le Chatelier’s Principle: Increase in H+ will force rxn #1 to the left; take S-2away from rxn #2 to form H2S and increase solubility of ZnS. - Note: May be able to separate metal cations by controlling [S2-] through pH. Least soluble metal sulfide will ppt at smallest sulfide concentation leaving more soluble sulfides in solution. FeS, Ksp=10-18 & HgS, Ksp=10-52Which will ppt first?

  13. III. Other Influences on Solubility A. pH & B. Complex Ions pH Example 2: Al(OH)3(s) + OH- (----) Al(OH)4-(soluble)Kf= Large - Le Chatlier’s principle: excess OH- will force rxn to right dissolving Al(OH)3 - Note: Excess HCl will also cause Al(OH)3(s) to dissolve. Why? B. Complex Ions - Many transition metals can form complex ions which will influence solubility. Examples: a) AgCl(s) (----) Ag+ + Cl-Ksp= 2x10-10 b) Ag+ + 2NH3 (----) Ag(NH3)2+ Kf = 1.7x107 - LeChatlier’s principle: NH3 will force rxn b to the right dissolving the AgCl in rxn a. Example: a) Pb+2 + 2Cl- PbCl2(s) b) PbCl2(s) + 2Cl- PbCl4-2Kf = Large - Large excess of Cl- will force reaction b to right thus dissolving PbCl2(s) from rxn a.

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