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PRECIPITATION REACTIONS Chapter 17 Part 2

PRECIPITATION REACTIONS Chapter 17 Part 2. Insoluble Chlorides. All salts formed in this experiment are said to be INSOLUBLE and form precipitates when mixing moderately concentrated solutions of the metal ion with chloride ions. Insoluble Chlorides.

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PRECIPITATION REACTIONS Chapter 17 Part 2

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  1. PRECIPITATION REACTIONSChapter 17 Part 2

  2. InsolubleChlorides All salts formed in this experiment are said to be INSOLUBLE and form precipitates when mixing moderately concentrated solutions of the metal ion with chloride ions.

  3. InsolubleChlorides Although all salts formed in this experiment are said to be insoluble, they do dissolve to some SLIGHT extent. AgCl(s) Ag+(aq) + Cl-(aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED.

  4. InsolubleChlorides AgCl(s) <--> Ag+(aq) + Cl-(aq) When the solution is SATURATED, experiment shows that [Ag+] = 1.34 x 10-5 M. This is equivalent to the SOLUBILITYof AgCl. What is [Cl-]? This is also equivalent to the AgCl solubility.

  5. InsolubleChlorides Make a chart. AgCl(s) <--> Ag+(aq) + Cl-(aq) some 0 0 - 1.34 x 10-5 1.34 x 10-5 1.34 x 10-5 some - 1.34 x 10-5 1.34 x 10-5 1.34 x 10-5

  6. InsolubleChlorides Ksp = [Ag+] [Cl-] = (1.34 x 10-5)(1.34 x 10-5) = 1.80 x 10-10 Ksp = solubility product constant See Table 18.2 and Appendix J 18A & 18B

  7. Lead(II) Chloride PbCl2(s) <--> Pb2+(aq) + 2 Cl-(aq) Ksp = 1.9 x 10-5

  8. Solubility of Lead(II) Iodide Consider PbI2 dissolving in water PbI2(s) Pb2+(aq) + 2 I-(aq) Calculate Ksp if solubility =0.00130 M Solution Solubility = [Pb2+] = 1.30 x 10-3 M [I-] = _____________ ? 2(1.30 x 10-3 M)

  9. Solubility of Lead(II) Iodide Consider PbI2 dissolving in water PbI2(s) Pb2+(aq) + 2 I-(aq) Calculate Ksp if solubility =0.00130 M Solution 1. Solubility = [Pb2+] = 1.30 x 10-3 M [I-] = 2 x [Pb2+] = 2.60 x 10-3 M

  10. Solubility of Lead(II) Iodide Consider PbI2 dissolving in water PbI2(s) Pb2+(aq) + 2 I-(aq) Calculate Ksp if solubility = 0.00130 M Solution 1. Solubility = [Pb2+] = 1.30 x 10-3 M [I-] = 2 x [Pb2+] = 2.60 x 10-3 M 2. Ksp = [Pb2+] [I-]2 = [Pb2+] {2 • [Pb2+]}2 = 4 [Pb2+]3

  11. Solubility of Lead(II) Iodide Consider PbI2 dissolving in water PbI2(s) Pb2+(aq) + 2 I-(aq) Calculate Ksp if solubility = 0.00130 M Solution 2. Ksp = 4 [Pb2+]3 = 4 (solubility)3 Ksp = 4 (1.30 x 10-3)3 = 8.8 x 10-9 Sample Problems

  12. Precipitating an Insoluble Salt Hg2Cl2(s) <--> Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-] 2 If [Hg22+] = 0.010 M, what [Cl-] is required to just begin the precipitation of Hg2Cl2? What is the maximum [Cl-] that can be in solution with 0.010 M Hg22+ without forming Hg2Cl2?

  13. Precipitating an Insoluble Salt Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 Recognize that Ksp = product of maximum ion concentrations. Precipitation begins when product of ion concentrations EXCEEDS the Ksp.

  14. K sp - -18 [ Cl ] = = 1.1 x 10 M 4(0.010) Precipitating an Insoluble Salt Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [2Cl-]2 Solution [Cl-] that can exist when [Hg22+] = 0.010 M, If this concentration of Cl- is just exceeded, Hg2Cl2 begins to precipitate.

  15. Precipitating an Insoluble Salt Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 Now raise [Cl-] to 1.0 M. What is the value of [Hg22+] at this point? Solution [Hg22+] = Ksp / [Cl-]2 = Ksp / (1.0)2 = 1.1 x 10-18 M The concentration of Hg22+ has been reduced by 1016 ! Sample Problems

  16. REVIEW PROBLEMS • Write the equilibrium equation and the equilibrium constant expression for saturated solutions of: Ag2S and PbI2. • The molar solubility of barium carbonate is 9.0 x 10-5 M. Calculate the solubility product constant. • The molar solubility of barium fluoride is 7.5 x 10-3 M. Calculate the solubility product constant.

  17. REVIEW PROBLEMS • Calculate the molar solubility of galena, PbS, given Ksp= 8.4 x 10-28. • Calculate the molar solubility of calcium fluoride given Ksp= 3.9 x 10-11. • Compare the molar solubilities forCaF2, PbCl2, and Ag2CrO4. • A solution is found to be 0.0060 M in barium ion and 0.019 M in fluoride ion.Is the system in equilibrium? If not what will occur as equilibrium is reached. Ksp = 1.7 x 10 -6.

  18. Separating Metal Ions Cu2+, Ag+, Pb2+ Ksp Values AgCl 1.8 x 10-10 PbCl2 1.7 x 10-5 PbCrO4 1.8 x 10-14

  19. Separating Salts by Differences in Ksp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first? Ksp for Ag2CrO4 = 9.0 x 10-12 Ksp for PbCrO4 = 1.8 x 10-14 Solution The substance whose Ksp is first exceeded will precipitate first. The ion requiring the lesser amount of CrO42- precipitate first. 19

  20. Separating Salts by Differences in Ksp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. Which precipitates first? Ksp for Ag2CrO4 = 9.0 x 10-12 Ksp for PbCrO4 = 1.8 x 10-14 Solution [CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+] = 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M [CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2 = 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M PbCrO4 precipitates first.

  21. Separating Salts by Differences in Ksp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. PbCrO4 precipitates first. Ksp (Ag2CrO4)= 9.0 x 10-12 Ksp (PbCrO4) = 1.8 x 10-14 How much Pb2+ remains in solution when Ag+ begins to precipitate? Solution We know that [CrO42-] = 2.3 x 10-8 M to begin to precipitates Ag2CrO4. What is the Pb2+ concentration at this point?

  22. Separating Salts by Differences in Ksp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate red Ag2CrO4 and yellow PbCrO4. PbCrO4 precipitates first. Ksp (Ag2CrO4)= 9.0 x 10-12 Ksp (PbCrO4) = 1.8 x 10-14 How much Pb2+ remains in solution when Ag+ begins to precipitate? Solution [Pb2+] = Ksp / [CrO42-] = 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

  23. Common Ion EffectAdding an Ion “Common” to an Equilibrium

  24. The Common Ion Effect Calculate the solubility of BaSO4 in: (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) Ba2+(aq) + SO42-(aq) Solution (a) Solubility in pure water = [Ba2+] = [SO42-] = s Ksp = [Ba2+] [SO42-] = s2 s = (Ksp)1/2 = 1.1 x 10-5 M

  25. The Common Ion Effect Calculate the solubility of BaSO4 in: (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) Ba2+(aq) + SO42-(aq) Solution (b) Now dissolve BaSO4 in water already containing 0.010 M Ba2+. Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSO4 be less than or greater than in pure water?___ Left Less

  26. The Common Ion Effect Calculate the solubility of BaSO4 in: (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) Ba2+(aq) + SO42-(aq) Solution (b) [Ba2+] [SO42-] initial change equilib. 0.010 0 + s + s 0.010 + s s

  27. The Common Ion Effect Calculate the solubility of BaSO4 in: (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) Ba2+(aq) + SO42-(aq) Solution Ksp = [Ba2+] [SO42-] = (0.010 + s) (s) s < 1.1 x 10-5 M (solubility in pure water), this means 0.010 + s is about equal to 0.010. Therefore, Ksp = 1.1 x 10-10 = (0.010)(s) s = 1.1 x 10-8 M = solubility in presence of added Ba2+ ion.

  28. The Common Ion Effect Calculate the solubility of BaSO4 in: (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) Ba2+(aq) + SO42-(aq) Solution Solubility in pure water = s = 1.1 x 10-5 M Solubility in presence of added Ba2+ = 1.1 x 10-8 M Le Chatelier’s Principle is followed! Sample Problems

  29. REVIEW PROBLEMS • Will a precipitate of lead (II) sulfate form when 150 ml of 0.030 M sodium sulfateis mixed with 120 mL of 0.020 M lead (II) nitrate. Ksp = 1.8 x 10 -8. • Calculate the molar solubility for calcium fluoride, Ksp= 3.9 x 10 -11, in: water. 0.0025 M calcium nitrate. 0.080 M sodium fluoride. Write appropriate net-ionic equations.

  30. SOLUBILITY AND pH • We have discovered in Experiment 23 that salts of weak acids are generally soluble in acidic solutions. This principle is illustrated by combining the Ka equation with the Ksp equation. If we consider CaC2O4 in the presence of strong acid, the following is the net equilibrium equation: CaC2O4(s) + 2 H+ <======> H2C2O4(aq) + Ca+2 Knet = Ksp. ( 1/Ka 1 ) . ( 1/Ka 2 ) • Since Ka 1 and Ka 2 are both less than one, Knet > Ksp. • If the acid is weak enough, Knet may be greater than one and products be favored. If the anion is the conjugate base of a strong acid, the Ksp equation is the only equilibrium equation.

  31. SOLUBILITY AND COMPLEX IONS • If the metal cation can form a complex ion with the other species present, a new net equilibrium will exist. The process is similar to that in the previous slide. • If silver bromide is treated with ammonia solution, some of the solid dissolves and the complex ion is formed. AgBr(s) + 2 NH3(aq) <=====> Ag(NH3)2+(aq) + Br- Knet = Ksp. Kf = ( 3.3 x 10-13 ) ( 1.6 x 107) = 5.3 x 10-6

  32. Simultaneous Equilibria 1. If you add sufficient chromate ion to an aqueous suspension of PbCl2, can PbCl2 be converted to PbCrO4? PbCl2 <--> Pb2+ + 2 Cl- Pb2+ + CrO42- <--> PbCrO4 1.7 x 10-5 1/1.8 x 10-14 9.4 x 108 PbCl2 + CrO42- <--> PbCrO4 + 2 Cl- Yes!

  33. Simultaneous Equilibria 2. Can AgCl be dissolved by adding a solution of NH3? Write the overall equation and determine the K value. AgCl <--> Ag+ + Cl- Ag+ + 2 NH3 <--> Ag(NH3)2+ 1.8 x 10-10 1.6 x 107 2.9 x 10-3 AgCl + 2 NH3 <--> Ag(NH3)2++ Cl- No, unless very high [NH3]

  34. Simultaneous Equilibria 3. Can CaC2O4 be dissolved by adding a solution of HCl? Write the overall equation and determine the K value. CaC2O4 <--> Ca2+ + C2O42- H+ + C2O42- <--> HC2O4- H+ + HC2O4- <--> H2C2O4 2.3 x 10-9 1/6.4 x 10-5 1/5.9 x 10-2 6.1 x 10-4 CaC2O4 + 2 H+ <--> H2C2O4 + Ca2+ No, unless very high [H+]

  35. REVIEW PROBLEMS • A solution contains 0.0035 M Ag+ and 0.15 M Pb+2. • Which precipitates first when I- is added? Ksp AgI = 1.5 x 10 -16 Ksp PbI2 = 8.7 x 10 -9. • Calculate the concentration of the first precipitated ion when the second ion begins to precipitate. • Write the equation for silver bromide changing to silver iodide with the addition of iodide ion. Calculate K for this reaction. Solubility product constants for silver bromide and silver iodide are 3.3 x 10 -13 and 1.5 x 10-16 respectively.

  36. Practice Problems 1. A saturated solution of lead chloride contains 4.50 g of lead chloride per liter. Calculate the Ksp for lead chloride. 2. The Ksp for Al(OH)3 is 1.9 x 10-33. Calculate the molar solubility of Al(OH)3 and determine [Al3+] and [OH1-]. 3. What is the molar solubility of BaSO4 in a solution that contains 0.100 M Na2SO4? (Ksp for BaSO4 = 1.1 x 10-10)

  37. Practice Problems 4. Will precipitation occur when 50.0 ml of 0.030 M Pb(NO3)2 is added to 50.0 ml of 0.0020 M KBr? (Ksp for lead bromide = 6.3 x 10-6) 5. Would it be possible to separate a solution containing 0.0020 M Pb2+ and 0.030 M Ag+ by adding drops of Na2CO3 solution? (Ksp for lead carbonate = 1.5 x 10-13 and Ksp for silver carbonate = 8.2 x 10-12) 6. Can CuBr be dissolved by adding a solution of NaCl? Write the overall equation and determine the K value

  38. Practice Problems Answers 1. 1.7 x 10-5 2. 2.9 x 10-9 M, 2.9 x 10-9 M, 8.7 x 10-9 M 3. 1.1 x 10-9 M 4. no 5. yes 6. No, unless [Cl-] is very large, K = 5.3 x 10-4 The End

  39. Mercury(I) Chloride Hg2Cl2(s) <--> Hg2+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 Lead(II) Chloride PbCl2(s) <--> Pb2+(aq) + 2 Cl-(aq) Ksp = 1.9 x 10-5 Silver Chloride AgCl(s) <--> Ag+(aq) + Cl-(aq) Ksp = 1.8 x 10-10

  40. Ksp from Solubility 1. A saturated solution of CuCl has a gram solubility of 0.05643 g/L. Calculate the Ksp. (0.05643g/L)(1 mole/99.0g) = 0.000570 M CuCl(s) <--> Cu+(aq) + Cl-(aq) Ksp = [Cu+] [Cl-] = (0.000570)(0.000570) = 3.25 x 10-7 Solid - 0.000570 0.000570 0.000570 0.000570 0.000570 Solid

  41. Ksp from Solubility 2. A saturated solution of PbBr2 has [Pb2+] = 1.05 x 10-1 M. Calculate the Ksp. PbBr2(s) <--> Pb2+(aq) + 2 Cl-(aq) Ksp = [Pb2+] [Cl-]2 = (0.0105)(0.0210)2 = 4.63 x 10-3 Solid - 0.0105 0.0105 0.0210 Solid 0.0105 0.0210

  42. Ksp from Solubility 3. A saturated solution of Ag2CrO4 has [Ag+] = 1.6 x 10-4 M. Calculate the Ksp. Ag2CrO4(s) <--> 2 Ag+(aq) + CrO42-(aq) Ksp = [Ag+]2 [CrO42-] = (1.6 x 10-4)2(8.0 x 10-5) = 2.0 x 10-12 Solid - 8.0 x 10-5 1.6 x 10-4 8.0 x 10-5 Solid 1.6 x 10-4 8.0 x 10-5

  43. Solubility from Ksp 1. The Ksp of SrCO3 is 7.0 x10-10. Calculate the molar solubility of SrCO3. SrCO3(s) <--> Sr2+(aq) + CO32-(aq) Ksp = [Sr2+] [CO32-] = (s)(s)= s2 = 7.0 x 10-10 s = 2.6 x 10-5 M Solid - s s s Solid s s

  44. Solubility from Ksp 2.The Ksp of Ca(OH)2 is 7.9 x10-6. Calculate the molar solubility of Ca(OH)2. Ca(OH)2(s) <--> Ca2+(aq) + 2 OH-(aq) Ksp = [Ca2+] [OH-]2 = (s)(2s)2 = 4s3 = 7.9 x 10-6 s = 1.3 x 10-2 M Solid - s s 2s Solid s 2s

  45. Solubility from Ksp 3.The Ksp of Al(OH)3 is 2.0 x 10-33. Calculate the molar solubility of Al(OH)3. Al(OH)3(s) <--> Al3+(aq) + 3 OH-(aq) Ksp = [Al3+] [OH-]3 = (s)(3s)3 = 27s4 = 2.0 x 10-33 s = 2.9 x 10-9 M Solid - s s 3s Solid s 3s

  46. Precipitating an Insoluble Salt Will mixing 200. mL 5.0 x 10-6 M mercury(I) nitrate and 100. mL 5.0 x 10-8 M sodium chloride cause a precipitate to form? Hg2Cl2(s) <--> Hg22+(aq) + 2 Cl-(aq) Q = [Hg22+] [Cl-] 2 [Hg22+] = 5.0 x 10-6 (200./300.) = 3.3 x 10-6 M [Cl-] = 5.0 x 10-8 (100./300.) = 1.7 x 10-8 M Q = (3.3 x 10-6)(1.7 x 10-8) 2 = 9.5 x 10-22 Q < Ksp No ppt

  47. Precipitating an Insoluble Salt Will mixing 100. mL 0.20 M magnesium nitrate and 300. mL 0.40 M sodium oxalate cause a precipitate to form? MgC2O4(s) <--> Mg2+(aq) + C2O42-(aq) Q = [Mg2+][C2O42-] [Mg2+] = 0.20 (100./400.) = 0.050 M [C2O42-] = 0.40 (300./400.) = 0.30 M Q = (0.050)(0.30) = 1.5 x 10-2 Ksp = 8.6 x 10-5 Q > Ksp ppt

  48. Precipitating an Insoluble Salt Will mixing 1.0 L 0.00010 M sodium chloride and 2.0 L 0.0090 M silver nitrate cause a precipitate to form? AgCl(s) <--> Ag+(aq) + Cl-(aq) Q = [Ag+][Cl-] [Ag+] = 0.0090 (2.0/3.0) = 0.0060 M [Cl-] = 0.00010 (1.0/3.0) = 0.000033 M Q = (0.0060)(0.000033) = 2.0 x 10-7 (Ksp = 1.8 x 10-10) Q > Ksp ppt

  49. Precipitating an Insoluble Salt What [Sr2+] is required to ppt SrSO4 in a 0.20 M Na2SO4 solution? (Ksp = 2.8 x 10-7) SrSO4(s) <--> Sr2+(aq) + SO42-(aq) Ksp = [Sr2+] [SO42-] 2.8 x 10-7 = (x)(0.20) x = 1.4 x 10-6 M = [Sr2+] For ppt [Sr2+] > 1.4 x 10-6 M x 0.20

  50. Precipitating an Insoluble Salt How many moles of HCl are required to ppt AgCl from 100. mL 0.10 M AgNO3? (Ksp = 1.8 x 10-10) AgCl(s) <--> Ag+(aq) + Cl-(aq) Ksp = [Ag+] [Cl-] 1.8 x 10-10 = (0.10)(x) x = 1.8 x 10-9 M = [Cl-] 1.8 x 10-9 mole/L)(0.100 L) = 1.8 x 10-10 mole For ppt mole HCl > 1.8 x 10-10 0.10 x

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