 # PRECIPITATION REACTIONS Chapter 17 Part 2 2 Insoluble Chlorides All salts formed in this experiment are said to be INSOLUBLE and form precipitates when.

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PRECIPITATION REACTIONS Chapter 17 Part 2

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

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

5 Make a chart. Make a chart. AgCl(s) Ag + (aq) + Cl - (aq) AgCl(s) Ag + (aq) + Cl - (aq) some 0 0 some 0 0 - 1.34 x 10 -5 1.34 x 10 -5 some - 1.34 x 10 -5 Insoluble Chlorides

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

7 Lead(II) Chloride PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq) K sp = 1.9 x 10 -5

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

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

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

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

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

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

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

15 Now raise [Cl - ] to 1.0 M. Now raise [Cl - ] to 1.0 M. What is the value of [Hg 2 2+ ] at this point? What is the value of [Hg 2 2+ ] at this point?Solution [Hg 2 2+ ] = K sp / [Cl - ] 2 = K sp / (1.0) 2 = 1.1 x 10 -18 M = K sp / (1.0) 2 = 1.1 x 10 -18 M The concentration of Hg 2 2+ has been reduced by 10 16 ! The concentration of Hg 2 2+ has been reduced by 10 16 ! Precipitating an Insoluble Salt Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 Sample Problems Sample Problems

16 REVIEW PROBLEMS Write the equilibrium equation and the equilibrium constant expression for saturated solutions of: Ag 2 S and PbI 2. 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 K sp = 8.4 x 10 -28. Calculate the molar solubility of calcium fluoride given K sp = 3.9 x 10 -11. Compare the molar solubilities for CaF 2, PbCl 2, and Ag 2 CrO 4. 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. K sp = 1.7 x 10 -6.

18 Separating Metal Ions Cu 2+, Ag +, Pb 2+ K sp Values AgCl1.8 x 10 -10 PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14 K sp Values AgCl1.8 x 10 -10 PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14

19 Separating Salts by Differences in K sp A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. Which precipitates first? A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- 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 K sp for PbCrO 4 = 1.8 x 10 -14 Solution The substance whose K sp is first exceeded will precipitate first. The substance whose K sp is first exceeded will precipitate first. The ion requiring the lesser amount of CrO 4 2- precipitate first. The ion requiring the lesser amount of CrO 4 2- precipitate first. 19

20 [CrO 4 2- ] to ppt. PbCrO 4 = K sp / [Pb 2+ ] = 1.8 x 10 -14 / 0.020 = 9.0 x 10 -13 M = 1.8 x 10 -14 / 0.020 = 9.0 x 10 -13 M [CrO 4 2- ] to ppt. Ag 2 CrO 4 = K sp / [Ag + ] 2 = 9.0 x 10 -12 / (0.020) 2 = 2.3 x 10 -8 M = 9.0 x 10 -12 / (0.020) 2 = 2.3 x 10 -8 M PbCrO 4 precipitates first. PbCrO 4 precipitates first. Separating Salts by Differences in K sp A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. Which precipitates first? A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- 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 K sp for PbCrO 4 = 1.8 x 10 -14 Solution

21 A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. PbCrO 4 precipitates first. A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. PbCrO 4 precipitates first. K sp (Ag 2 CrO 4 )= 9.0 x 10 -12 K sp (PbCrO 4 ) = 1.8 x 10 -14 Separating Salts by Differences in K sp How much Pb 2+ remains in solution when Ag + begins to precipitate? How much Pb 2+ remains in solution when Ag + begins to precipitate?Solution We know that [CrO 4 2- ] = 2.3 x 10 -8 M to begin to precipitates Ag 2 CrO 4. We know that [CrO 4 2- ] = 2.3 x 10 -8 M to begin to precipitates Ag 2 CrO 4. What is the Pb 2+ concentration at this point? What is the Pb 2+ concentration at this point?

22 A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. PbCrO 4 precipitates first. A solution contains 0.020 M Ag + and Pb 2+. Add CrO 4 2- to precipitate red Ag 2 CrO 4 and yellow PbCrO 4. PbCrO 4 precipitates first. K sp (Ag 2 CrO 4 )= 9.0 x 10 -12 K sp (PbCrO 4 ) = 1.8 x 10 -14 Separating Salts by Differences in K sp How much Pb 2+ remains in solution when Ag + begins to precipitate? How much Pb 2+ remains in solution when Ag + begins to precipitate?Solution [Pb 2+ ] = K sp / [CrO 4 2- ] = 1.8 x 10 -14 / 2.3 x 10 -8 M [Pb 2+ ] = K sp / [CrO 4 2- ] = 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 Lead ion has dropped from 0.020 M to < 10 -6 M

23 Common Ion Effect Adding an Ion “Common” to an Equilibrium

24 Calculate the solubility of BaSO 4 in: (a) pure water and Calculate the solubility of BaSO 4 in: (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution (a) Solubility in pure water = [Ba 2+ ] = [SO 4 2- ] = s Solubility in pure water = [Ba 2+ ] = [SO 4 2- ] = s K sp = [Ba 2+ ] [SO 4 2- ] = s 2 s = (K sp ) 1/2 = 1.1 x 10 -5 M The Common Ion Effect

25 Calculate the solubility of BaSO 4 in: (a) pure water and Calculate the solubility of BaSO 4 in: (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution (b) The Common Ion Effect Now dissolve BaSO 4 in water already containing 0.010 M Ba 2+. Now dissolve BaSO 4 in water already containing 0.010 M Ba 2+. Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSO 4 be less than or greater than in pure water?___ Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSO 4 be less than or greater than in pure water?___ Left Less

26 [Ba 2+ ] [SO 4 2- ] [Ba 2+ ] [SO 4 2- ]initialchangeequilib. The Common Ion Effect Calculate the solubility of BaSO 4 in: (a) pure water and Calculate the solubility of BaSO 4 in: (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution (b) 0.010 0 + s + s + s + s 0.010 + s s

27 The Common Ion Effect Calculate the solubility of BaSO 4 in: Calculate the solubility of BaSO 4 in: (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution K sp = [Ba 2+ ] [SO 4 2- ] = (0.010 + s ) ( s) K sp = [Ba 2+ ] [SO 4 2- ] = (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,K sp = 1.1 x 10 -10 = (0.010)( s ) s < 1.1 x 10 -5 M (solubility in pure water), this means 0.010 + s is about equal to 0.010. Therefore,K sp = 1.1 x 10 -10 = (0.010)( s ) s = 1.1 x 10 -8 M = solubility in presence of added Ba 2+ ion.

28 Calculate the solubility of BaSO 4 in: (a) pure water and Calculate the solubility of BaSO 4 in: (a) pure water and (b) in 0.010 M Ba(NO 3 ) 2. K sp for BaSO 4 = 1.1 x 10 -10 BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Solution The Common Ion Effect Solubility in pure water = s = 1.1 x 10 -5 M Solubility in presence of added Ba 2+ = 1.1 x 10 -8 M Le Chatelier’s Principle is followed! Sample Problems Sample Problems

29 REVIEW PROBLEMS lead (II) sulfateWill a precipitate of lead (II) sulfate form when 150 ml of 0.030 M sodium sulfate is mixed with 120 mL of 0.020 M lead (II) nitrate. K sp = 1.8 x 10 -8. molar solubilitycalcium fluorideCalculate the molar solubility for calcium fluoride, K sp = 3.9 x 10 -11, in: water. 0.0025 M calcium nitrate. 0.080 M sodium fluoride. Write appropriate net-ionic equations 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 K a equation with the K sp equation. If we consider CaC 2 O 4 in the presence of strong acid, the following is the net equilibrium equation:We have discovered in Experiment 23 that salts of weak acids are generally soluble in acidic solutions. This principle is illustrated by combining the K a equation with the K sp equation. If we consider CaC 2 O 4 in the presence of strong acid, the following is the net equilibrium equation: CaC 2 O 4(s) + 2 H + H 2 C 2 O 4(aq) + Ca +2 K net = K sp. ( 1/K a 1 ). ( 1/K a 2 ) Since K a 1 and K a 2 are both less than one, K net > K sp.Since K a 1 and K a 2 are both less than one, K net > K sp. If the acid is weak enough, K net may be greater than one and products be favored. If the anion is the conjugate base of a strong acid, the K sp equation is the only equilibrium equation.If the acid is weak enough, K net may be greater than one and products be favored. If the anion is the conjugate base of a strong acid, the K sp 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 NH 3(aq) Ag(NH 3 ) 2 + (aq) + Br - K net = K sp. K f = ( 3.3 x 10 -13 ) ( 1.6 x 10 7 ) = 5.3 x 10 -6 ( 3.3 x 10 -13 ) ( 1.6 x 10 7 ) = 5.3 x 10 -6

32 Simultaneous Equilibria 1. If you add sufficient chromate ion to an aqueous suspension of PbCl 2, can PbCl 2 be converted to PbCrO 4 ? PbCl 2 Pb 2+ + 2 Cl - Pb 2+ + CrO 4 2- PbCrO 4 PbCl 2 + CrO 4 2- PbCrO 4 + 2 Cl - 1.7 x 10 -5 1/1.8 x 10 -14 9.4 x 10 8 Yes!

33 Simultaneous Equilibria 2. Can AgCl be dissolved by adding a solution of NH 3 ? Write the overall equation and determine the K value. AgCl Ag + + Cl - Ag + + 2 NH 3 Ag(NH 3 ) 2 + AgCl + 2 NH 3 Ag(NH 3 ) 2 + + Cl - 1.8 x 10 -10 1.6 x 10 7 2.9 x 10 -3 No, unless very high [NH 3 ]

34 Simultaneous Equilibria 3. Can CaC 2 O 4 be dissolved by adding a solution of HCl? Write the overall equation and determine the K value. CaC 2 O 4 Ca 2+ + C 2 O 4 2- H + + C 2 O 4 2- HC 2 O 4 - H + + HC 2 O 4 - H 2 C 2 O 4 CaC 2 O 4 + 2 H + H 2 C 2 O 4 + Ca 2+ 2.3 x 10 -9 1/6.4 x 10 -5 1/5.9 x 10 -2 No, unless very high [H + ] 6.1 x 10 -4

35 REVIEW PROBLEMS A solution contains 0.0035 M Ag + and 0.15 M Pb +2.A solution contains 0.0035 M Ag + and 0.15 M Pb +2. Which precipitates first when I - is added?Which precipitates first when I - is added? K sp AgI = 1.5 x 10 -16 K sp PbI 2 = 8.7 x 10 -9. Calculate the concentration of the first precipitated ion when the second ion begins to precipitate.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 K sp for lead chloride. 2. The K sp for Al(OH) 3 is 1.9 x 10 -33. Calculate the molar solubility of Al(OH) 3 and determine [Al 3+ ] and [OH 1- ]. 3. What is the molar solubility of BaSO 4 in a solution that contains 0.100 M Na 2 SO 4 ? (K sp for BaSO 4 = 1.1 x 10 -10 )

37 Practice Problems 4. Will precipitation occur when 50.0 ml of 0.030 M Pb(NO 3 ) 2 is added to 50.0 ml of 0.0020 M KBr? (K sp for lead bromide = 6.3 x 10 -6 ) 5. Would it be possible to separate a solution containing 0.0020 M Pb 2+ and 0.030 M Ag + by adding drops of Na 2 CO 3 solution? (K sp for lead carbonate = 1.5 x 10 -13 and K sp 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 Hg 2 Cl 2 (s) Hg 2 + (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 Lead(II) Chloride PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq) K sp = 1.9 x 10 -5 Silver Chloride AgCl(s) Ag + (aq) + Cl - (aq) K sp = 1.8 x 10 -10

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

41 K sp from Solubility 2. A saturated solution of PbBr 2 has [Pb 2+ ] = 1.05 x 10 -1 M. Calculate the K sp. PbBr 2 (s) Pb 2+ (aq) + 2 Cl - (aq) PbBr 2 (s) Pb 2+ (aq) + 2 Cl - (aq) - 0.0105 0.02100.0105 0.02100.0105 K sp = [Pb 2+ ] [Cl - ] 2 = (0.0105)(0.0210) 2 = 4.63 x 10 -3 Solid

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

43 Solubility from K sp 1. The K sp of SrCO 3 is 7.0 x10 -10. Calculate the molar solubility of SrCO 3. (s) Sr 2+ (aq) + CO 3 2- (aq) SrCO 3 (s) Sr 2+ (aq) + CO 3 2- (aq) - s ss ss Sr 2+ ]CO 3 2- ] K sp = [Sr 2+ ] [CO 3 2- ] = (s)(s) = s 2 = 7.0 x 10 -10 s = 2.6 x 10 -5 M Solid

44 Solubility from K sp 2.The K sp of Ca(OH) 2 is 7.9 x10 -6. Calculate the molar solubility of Ca(OH) 2. (s) Ca 2+ (aq) + 2 OH - (aq) Ca(OH) 2 (s) Ca 2+ (aq) + 2 OH - (aq) - s 2ss s Ca 2+ ] - ] 2 K sp = [Ca 2+ ] [OH - ] 2 = (s)(2s) 2 = 4s 3 = 7.9 x 10 -6 s = 1.3 x 10 -2 M Solid

45 Solubility from K sp 3.The K sp of Al(OH) 3 is 2.0 x 10 -33. Calculate the molar solubility of Al(OH) 3. (s) Al 3+ (aq) + 3 OH - (aq) Al(OH) 3 (s) Al 3+ (aq) + 3 OH - (aq) - s 3ss s Al 3+ ] - ] 3 K sp = [Al 3+ ] [OH - ] 3 = (s)(3s) 3 = 27s 4 = 2.0 x 10 -33 s = 2.9 x 10 -9 M Solid

46 Precipitating an Insoluble Salt Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) 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? Hg 2 Cl 2 (s) Hg 2 2+ (aq) + 2 Cl - (aq) Q = [Hg 2 2+ ] [Cl - ] 2 Q = [Hg 2 2+ ] [Cl - ] 2 [Hg 2 2+ ] = [Hg 2 2+ ] = 5.0 x 10 -6 (200./300.) = 3.3 x 10 -6 M [Cl - ] = [Cl - ] = 5.0 x 10 -8 (100./300.) = 1.7 x 10 -8 M Q = ()() 2 = 9.5 x 10 -22 Q = (3.3 x 10 -6 )(1.7 x 10 -8 ) 2 = 9.5 x 10 -22 Q < K sp No ppt Q < K sp 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? MgC 2 O 4 (s) Mg 2+ (aq) + C 2 O 4 2- (aq) MgC 2 O 4 (s) Mg 2+ (aq) + C 2 O 4 2- (aq) Q = [Mg 2+ ][C 2 O 4 2- ] Q = [Mg 2+ ][C 2 O 4 2- ] [Mg 2+ ] = 0 [Mg 2+ ] = 0.20 (100./400.) = 0.050 M [C 2 O 4 2- ] = 0 [C 2 O 4 2- ] = 0.40 (300./400.) = 0.30 M Q = (0)(0) = 1.5 x 10 -2 Q = (0.050)(0.30) = 1.5 x 10 -2 Q > K sp ppt K sp = 8.6 x 10 -5 Q > K sp 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) AgCl(s) Ag + (aq) + Cl - (aq) Q = [Ag + ][Cl - ] Q = [Ag + ][Cl - ] [Ag + ] = 0 [Ag + ] = 0.0090 (2.0/3.0) = 0.0060 M [Cl - ] = 0 M [Cl - ] = 0.00010 (1.0/3.0) = 0.000033 M Q = (0)(0) = 2.0 x 10 -7 Q = (0.0060)(0.000033) = 2.0 x 10 -7 Q > K sp ppt (K sp = 1.8 x 10 -10 ) Q > K sp ppt

49 Precipitating an Insoluble Salt What [Sr 2+ ] is required to ppt SrSO 4 in a 0.20 M Na 2 SO 4 solution? (K sp = 2.8 x 10 -7 ) (s) Sr 2+ (aq) + SO 4 2- (aq) SrSO 4 (s) Sr 2+ (aq) + SO 4 2- (aq) 0.20x K sp = [Sr 2+ ] [SO 4 2- ] 2.8 x 10 -7 = (x)(0.20) x = 1.4 x 10 -6 M = [Sr 2+ ] For ppt [Sr 2+ ] > 1.4 x 10 -6 M

50 Precipitating an Insoluble Salt How many moles of HCl are required to ppt AgCl from 100. mL 0.10 M AgNO 3 ? (K sp = 1.8 x 10 -10 ) (s) Ag + (aq) + Cl - (aq) AgCl(s) Ag + (aq) + Cl - (aq) x0.10 K sp = [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

51 Precipitating an Insoluble Salt Calculate [Cl - ] required to ppt PbCl 2 from 0.100 M Pb(NO 3 ) 2. (K sp = 1.7 x 10 -5 ) (s) Pb 2+ (aq) + 2 Cl - (aq) PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq) x0.100 K sp = [Pb 2+ ] [Cl - ] 2 1.7 x 10 -5 = (0.100)(x) 2 x = 1.3 x 10 -2 M = [Cl - ] For ppt [Cl - ] > 1.3 x 10 -2 M

52 Precipitating an Insoluble Salt If [Cl - ] is raised to 0.10 M, calculate [Pb 2+ ] (s) Pb 2+ (aq) + 2 Cl - (aq) PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq) 0.10x K sp = [Pb 2+ ] [Cl - ] 2 1.7 x 10 -5 = (x)(0.10) 2 x = 1.7 x 10 -3 M = [Pb 2+ ]

53 Precipitating an Insoluble Salt 100. mL 0.200 M silver nitrate is mixed with 100. mL 0.100 M hydrochloric acid. Calculate [Ag + ] and [Cl - ]. (K sp = 1.8 x 10 - 10 ) (s) <-- Ag + (aq) + Cl - (aq) AgCl(s) <-- Ag + (aq) + Cl - (aq) 20.0 10.0 20.0 10.0 010.0 -10.0 [Ag + ] == 0.0500 M 10.0 200.

54 (s) Ag + (aq) + Cl - (aq) AgCl(s) Ag + (aq) + Cl - (aq) - x x0.0500 xx Ag + ]Cl - ] K sp = [Ag + ] [Cl - ] 1.8 x 10 -10 = (0.0500)(x) x = 3.6 x 10 -9 M = [Cl - ] 0.0500 Precipitating an Insoluble Salt Solid

55 Common Ions 1. The K sp of SrCO 3 is 7.0 x10 -10. Calculate the molar solubility of SrCO 3 in 0.10 M Na 2 CO 3. (s) Sr 2+ (aq) + CO 3 2- (aq) SrCO 3 (s) Sr 2+ (aq) + CO 3 2- (aq) - s 0.10s ss Sr 2+ ]CO 3 2- ] K sp = [Sr 2+ ] [CO 3 2- ] = (s)(.10) = s 2 = 7.0 x 10 -10 s = 7.0 x 10 -9 M 0.10 Remember in H 2 O: s= 2.6 x 10 -5 M Solid

56 Common Ions 2. The K sp of Ca(OH) 2 is 7.9 x10 -6. Calculate the molar solubility of Ca(OH) 2 in 0.50 M NaOH. (s) Ca 2+ (aq) + 2 OH - (aq) Ca(OH) 2 (s) Ca 2+ (aq) + 2 OH - (aq) - s 0.50s 2ss Ca 2+ ] - ] 2 K sp = [Ca 2+ ] [OH - ] 2 7.9 x 10 -6 = (s)(0.50) 2 s = 3.2 x 10 -5 M 0.50 Remember in H 2 O: s= 1.3 x 10 -2 M Solid

57 Common Ions 3. The K sp of Al(OH) 3 is 2.0 x 10 -33. Calculate the molar solubility of Al(OH) 3 in 1.0 M KOH. (s) Al 3+ (aq) + 3 OH - (aq) Al(OH) 3 (s) Al 3+ (aq) + 3 OH - (aq) - s 1.0s 3ss Al 3+ ] - ] 3 K sp = [Al 3+ ] [OH - ] 3 2.0 x 10 -33 = (s)(1.0) 3 s = 2.0 x 10 -33 M 1.0 Remember in H 2 O: s= 2.9 x 10 -9 M Solid

58 Common Ions 4. Calculate the solubility of calcium chromate in 0.0050 M calcium chloride. (K sp = 7.1 x 10 -4 ) (s) Ca 2+ (aq) + CrO 4 2- (aq) CaCrO 4 (s) Ca 2+ (aq) + CrO 4 2- (aq) - s s0.0050 + s ss Ca 2+ ]CrO 4 2- ] K sp = [Ca 2+ ] [CrO 4 2- ] 7.1 x 10 -4 = (0.0050 + s)(s) s = 2.4 x 10 -2 M 0.0050Solid

59 Separating Salts by Differences in K sp 1. Separation of.10 M Ag + and.10 M Pb 2+ AgBrK sp =3.3 x 10 -13 PbBr 2 K sp =6.3 x 10 -6 Plan: Add Br - until all AgBr is ppt, but no PbBr 2 is ppt. a. Calculate [Br - ] required to ppt. b. Calculate [Ag + ] left in solution. The substance whose K sp is first exceeded will precipitate first. The substance whose K sp is first exceeded will precipitate first. The ion requiring the lesser amount of Br - precipitate first. The ion requiring the lesser amount of Br - precipitate first.

60 AgBr(s) Ag + (aq) + Br - (aq) AgBr(s) Ag + (aq) + Br - (aq) x.10 K sp = [Ag + ] [Br - ] 3.3 x 10 -13 = (.10)(x) x = 3.3 x 10 -12 = [Br - ] For ppt [Br - ] > 3.3 x 10 -12 Separating Salts by Differences in K sp

61 PbBr 2 (s) Pb 2+ (aq) + 2 Br - (aq) PbBr 2 (s) Pb 2+ (aq) + 2 Br - (aq) x.10 K sp = [Pb 2+ ] [Br - ] 2 6.3 x 10 -6 = (.10)(x) 2 x = 7.9 x 10 -3 = [Br - ] For ppt [Br - ] > 7.9 x 10 -3 Separating Salts by Differences in K sp

62 For ppt AgBr [Br - ] > 3.3 x 10 -12 For ppt PbBr 2 [Br - ] > 7.9 x 10 -3 Therefore; Start ppt of AgBr [Br - ] > 3.3 x 10 -12 Max. ppt of AgBr [Br - ] > 7.9 x 10 -3 At max. ppt of AgBr, what is the [Ag + ] left in solution? Separating Salts by Differences in K sp

63 AgBr(s) Ag + (aq) + Br - (aq) AgBr(s) Ag + (aq) + Br - (aq) 7.9 x 10 -3 x K sp = [Ag + ] [Br - ] 3.3 x 10 -18 = (x)(7.9 x 10 -3 ) x = 4.2 x 10 -11 = [Ag + ] Good Separation: [Ag + ] < 10 -5 Separating Salts by Differences in K sp

64 Separating Salts by Differences in K sp 2. Separation of.10 M CO 3 2- and.10 M C 2 O 4 2- BaCO 3 K sp =8.1 x 10 -9 BaC 2 O 4 K sp =1.1 x 10 -7 Plan: Add Ba 2+ until all BaCO 3 is ppt, but no BaC 2 O 4 is ppt. a. Calculate [Ba 2+ ] required to ppt. b. Calculate [CO 3 2- ] left in solution. The substance whose K sp is first exceeded will precipitate first. The ion requiring the lesser amount of Ba 2+ precipitate first. The ion requiring the lesser amount of Ba 2+ precipitate first.

65 BaCO 3 (s) Ba 2+ (aq) + CO 3 2- (aq) BaCO 3 (s) Ba 2+ (aq) + CO 3 2- (aq).10x Ba 2+ ]CO 3 2- K sp = [Ba 2+ ][CO 3 2- ] 8.1 x 10 -9 = (x)(.10) Ba 2+ ] x = 8.1 x 10 -8 = [Ba 2+ ] Ba 2+ ] For ppt [Ba 2+ ] > 8.1 x 10 -8 Separating Salts by Differences in K sp

66 BaC 2 O 4 (s) Ba 2+ (aq) + C 2 O 4 2- (aq) BaC 2 O 4 (s) Ba 2+ (aq) + C 2 O 4 2- (aq).10x C 2 O 4 2- K sp = [Ba 2+ ] [C 2 O 4 2- ] 1.1 x 10 -7 = (x)(.10) x = 1.1 x 10 -6 = [Ba 2+ ] For ppt [Ba 2+ ] > 1.1 x 10 -6 Separating Salts by Differences in K sp

67 BaCO 3 For ppt BaCO 3 [Ba 2+ ] > 8.1 x 10 -8 BaC 2 O 4 For ppt BaC 2 O 4 [Ba 2+ ] > 1.1 x 10 -6 Therefore; BaCO 3 Start ppt of BaCO 3 [Ba 2+ ] > 8.1 x 10 -8 BaCO 3 Max. ppt of BaCO 3 [Ba 2+ ] > 1.1 x 10 -6 BaCO 3 At max. ppt of BaCO 3, what is the [CO 3 2- ] left in solution? Separating Salts by Differences in K sp

68 BaCO 3 (s) Ba 2+ (aq) + CO 3 2- (aq) BaCO 3 (s) Ba 2+ (aq) + CO 3 2- (aq) 1.1 x 10 -6 x Ba 2+ ]CO 3 2- K sp = [Ba 2+ ][CO 3 2- ] 8.1 x 10 -9 = (1.1 x 10 -6 )(x) CO 3 2- x = 7.4 x 10 -3 = [CO 3 2- ] CO 3 2- > Poor Separation: [CO 3 2- ] > 10 -5 Separating Salts by Differences in K sp

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