John C. Kotz State University of New York, College at Oneonta John C. Kotz Paul M. Treichel John Townsend Chapter 18 Principles.

John C. Kotz State University of New York, College at Oneonta John C. Kotz Paul M. Treichel John Townsend http://academic.cengage.com/kotz Chapter 18 Principles of Reactivity: Other Aspects of Aqueous Equilibria

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5 © 2009 Brooks/Cole - Cengage Acid-Base Reactions Strong acid + strong baseStrong acid + strong base HCl + NaOH f HCl + NaOH f Strong acid + weak baseStrong acid + weak base HCl + NH 3 f HCl + NH 3 f Weak acid + strong baseWeak acid + strong base HOAc + NaOH f HOAc + NaOH f Weak acid + weak baseWeak acid + weak base HOAc + NH 3 f HOAc + NH 3 f What is relative pH before, during, & after reaction? Need to study: a) Common ion effect and buffers b) Titrations

6 © 2009 Brooks/Cole - Cengage QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O e NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O e NH 4 + (aq) + OH - (aq) Here we are adding NH 4 +, an ion COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the left (1), right (2), no change (3). The pH will go up (1), down (2), no change (3). NH 4 + is an acid! The Common Ion Effect Section 18.1

7 © 2009 Brooks/Cole - Cengage Let us first calculate the pH of a 0.25 M NH 3 solution. [NH 3 ][NH 4 + ][OH - ] initial0.2500 change-x+x+x equilib0.25 - xx x QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O e NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O e NH 4 + (aq) + OH - (aq) pH of Aqueous NH 3

8 © 2009 Brooks/Cole - Cengage QUESTION: What is the effect on the pH of adding NH 4 Cl to 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O e NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O e NH 4 + (aq) + OH - (aq) pH of Aqueous NH 3 Assuming x is << 0.25, we have [OH - ] = x = [K b (0.25)] 1/2 = 0.0021 M This gives pOH = 2.67 and so pH = 14.00 - 2.67 = 11.33 for 0.25 M NH 3 = 11.33 for 0.25 M NH 3

9 © 2009 Brooks/Cole - Cengage Problem: What is the pH of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O e NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O e NH 4 + (aq) + OH - (aq) We expect that the pH will decline on adding NH 4 Cl. Let’s test that! [NH 3 ][NH 4 + ][OH - ] initialchangeequilib pH of NH 3 /NH 4 + Mixture

10 © 2009 Brooks/Cole - Cengage Problem: What is the pH of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O e NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O e NH 4 + (aq) + OH - (aq) We expect that the pH will decline on adding NH 4 Cl. Let’s test that! [NH 3 ][NH 4 + ][OH - ] initial0.250.100 change-x+x+x equilib0.25 - x0.10 + x x pH of NH 3 /NH 4 + Mixture

11 © 2009 Brooks/Cole - Cengage Problem: What is the pH of a solution with 0.10 M NH 4 Cl and 0.25 M NH 3 (aq)? NH 3 (aq) + H 2 O e NH 4 + (aq) + OH - (aq) NH 3 (aq) + H 2 O e NH 4 + (aq) + OH - (aq) pH of NH 3 /NH 4 + Mixture Assuming x is very small, [OH - ] = x = (0.25 / 0.10)(K b ) = 4.5 x 10 -5 M [OH - ] = x = (0.25 / 0.10)(K b ) = 4.5 x 10 -5 M This gives pOH = 4.35 and pH = 9.65 pH drops from 11.33 to 9.65 on adding a common ion

13 © 2009 Brooks/Cole - Cengage A buffer solution is a special case of the common ion effect. The function of a buffer is to resist changes in the pH of a solution. Buffer Composition Weak Acid+Conj. Base HOAc+OAc - H 2 PO 4 - +HPO 4 2- NH 4 + +NH 3 Buffer Solutions

14 © 2009 Brooks/Cole - Cengage Consider HOAc/OAc - to see how buffers work ACID USES UP ADDED OH - We know that OAc - + H 2 O e HOAc + OH - OAc - + H 2 O e HOAc + OH - has K b = 5.6 x 10 -10 Therefore, the reverse reaction of the WEAK ACID with added OH - has K reverse = 1/ K b = 1.8 x 10 9 K reverse is VERY LARGE, so HOAc completely snarfs up OH - !!!! Buffer Solutions

15 © 2009 Brooks/Cole - Cengage Consider HOAc/OAc - to see how buffers work CONJ. BASE USES UP ADDED H + HOAc + H 2 O e OAc - + H 3 O + has K a = 1.8 x 10 -5 Therefore, the reverse reaction of the WEAK BASE with added H + has K reverse = 1/ K a = 5.6 x 10 4 K reverse is VERY LARGE, so OAc - completely snarfs up H + ! Buffer Solutions

16 © 2009 Brooks/Cole - Cengage Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M? HOAc + H 2 O e OAc - + H 3 O + K a = 1.8 x 10 -5 Buffer Solutions 0.700 M HOAc has pH = 2.45 The pH of the buffer will have 1.pH < 2.45 2.pH > 2.45 3.pH = 2.45 0.700 M HOAc has pH = 2.45 The pH of the buffer will have 1.pH < 2.45 2.pH > 2.45 3.pH = 2.45

17 © 2009 Brooks/Cole - Cengage [HOAc][OAc - ][H 3 O + ] [HOAc][OAc - ][H 3 O + ]initialchangeequilib Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M? HOAc + H 2 O e OAc - + H 3 O + K a = 1.8 x 10 -5 Buffer Solutions 0.7000.6000 -x+x+x 0.700 - x 0.600 + x x

18 © 2009 Brooks/Cole - Cengage [HOAc][OAc - ][H 3 O + ] equilib0.700 - x0.600 + xx Assuming that x << 0.700 and 0.600, we have Problem: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M? HOAc + H 2 O e OAc - + H 3 O + K a = 1.8 x 10 -5 Buffer Solutions [H 3 O + ] = 2.1 x 10 -5 and pH = 4.68

19 © 2009 Brooks/Cole - Cengage Notice that the expression for calculating the H + conc. of the buffer is Buffer Solutions Notice that the H 3 O + or OH - concs. depend on (1) K and (2) the ratio of acid and base concs.

20 © 2009 Brooks/Cole - Cengage Henderson-Hasselbalch Equation Take the negative log of both sides of this equation The pH is determined largely by the pK a of the acid and then adjusted by the ratio of acid and conjugate base.

21 © 2009 Brooks/Cole - Cengage Adding an Acid to a Buffer Problem: What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (before HCl, pH = 7.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M (pH = 4.68) Solution to Part (a) Calc. [HCl] after adding 1.00 mL of HCl to 1.00 L of water C 1 V 1 = C 2 V 2 C 1 V 1 = C 2 V 2 C 2 = 1.00 x 10 -3 M = [H 3 O + ] C 2 = 1.00 x 10 -3 M = [H 3 O + ] pH = 3.00 pH = 3.00

22 © 2009 Brooks/Cole - Cengage Adding an Acid to a Buffer Solution to Part (b) Step 1 — do the stoichiometry H 3 O + (from HCl) + OAc - (from buffer) f HOAc (from buffer) The reaction occurs completely because K is very large. What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (after HCl, pH = 3.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M (pH before = 4.68)

23 © 2009 Brooks/Cole - Cengage Adding an Acid to a Buffer Solution to Part (b): Step 1—Stoichiometry [H 3 O + ] +[OAc - ] [HOAc] [H 3 O + ] +[OAc - ] [HOAc] Before rxn Change After rxn What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (pH = 3.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M (pH = 4.68) 0.00100 mol 0.600 mol 0.700 mol -0.00100-0.00100 +0.00100 0 0.599 mol 0.701 mol

24 © 2009 Brooks/Cole - Cengage Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H 2 O e H 3 O + + OAc - HOAc + H 2 O e H 3 O + + OAc - [HOAc] [H 3 O + ] [OAc - ] [HOAc] [H 3 O + ] [OAc - ] Before rxn (M) Change (M) After rxn (M) What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (pH = 3.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc - ] = 0.600 M (pH = 4.68) 0.701 mol/L 0 0.599 mol/L -x +x +x 0.599 + x x0.701-x

25 © 2009 Brooks/Cole - Cengage Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H 2 O e H 3 O + + OAc - HOAc + H 2 O e H 3 O + + OAc - [HOAc] [H 3 O + ] [OAc - ] [HOAc] [H 3 O + ] [OAc - ] After rxn0.701-x0.599+xx Because [H 3 O + ] = 2.1 x 10 -5 M BEFORE adding HCl, we again neglect x relative to 0.701 and 0.599. What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (pH = 3.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68)

26 © 2009 Brooks/Cole - Cengage Adding an Acid to a Buffer Solution to Part (b): Step 2—Equilibrium HOAc + H 2 O e H 3 O + + OAc - HOAc + H 2 O e H 3 O + + OAc - What is the pH when 1.00 mL of 1.00 M HCl is added to a)1.00 L of pure water (pH = 3.00) b)1.00 L of buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M (pH = 4.68) [H 3 O + ] = 2.1 x 10 -5 M f pH = 4.68 The pH has not changed on adding HCl to the buffer!

27 © 2009 Brooks/Cole - Cengage Preparing a Buffer You want to buffer a solution at pH = 4.30. This means [H 3 O + ] = 10 -pH = 5.0 x 10 -5 M It is best to choose an acid such that [H 3 O + ] is about equal to K a (or pH ≈ pK a ). —then you get the exact [H 3 O + ] by adjusting the ratio of acid to conjugate base.

28 © 2009 Brooks/Cole - Cengage You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M POSSIBLE ACIDSK a POSSIBLE ACIDSK a HSO 4 - / SO 4 2- 1.2 x 10 -2 HSO 4 - / SO 4 2- 1.2 x 10 -2 HOAc / OAc - 1.8 x 10 -5 HOAc / OAc - 1.8 x 10 -5 HCN / CN - 4.0 x 10 -10 HCN / CN - 4.0 x 10 -10 Best choice is acetic acid / acetate. Preparing a Buffer

29 © 2009 Brooks/Cole - Cengage You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x 10 -5 M Therefore, if you use 0.100 mol of NaOAc and 0.278 mol of HOAc, you will have pH = 4.30. Solve for [HOAc]/[OAc - ] ratio Solve for [HOAc]/[OAc - ] ratio Preparing a Buffer

30 © 2009 Brooks/Cole - Cengage A final point — CONCENTRATION of the acid and conjugate base are not important. It is the RATIO OF THE NUMBER OF MOLES of each. Result: diluting a buffer solution does not change its pH Preparing a Buffer

31 © 2009 Brooks/Cole - Cengage Commercial Buffers The solid acid and conjugate base in the packet are mixed with water to give the specified pH. Note that the quantity of water does not affect the pH of the buffer.

37 © 2009 Brooks/Cole - Cengage QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. pH at equivalence point? pH of solution of benzoic acid, a weak acid Benzoic acid + NaOH pH at half-way point?

38 © 2009 Brooks/Cole - Cengage Acid-Base Titration Section 18.3 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? HBz + NaOH Na + + Bz - + H 2 O HBz + NaOH f Na + + Bz - + H 2 O C 6 H 5 CO 2 H = HBz Benzoate ion = Bz -

39 © 2009 Brooks/Cole - Cengage Acid-Base Titration Section 18.3 The pH of the final solution will be 1.Less than 7 2.Equal to 7 3.Greater than 7 The pH of the final solution will be 1.Less than 7 2.Equal to 7 3.Greater than 7 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? HBz + NaOH Na + + Bz - + H 2 O HBz + NaOH f Na + + Bz - + H 2 O

40 © 2009 Brooks/Cole - Cengage Acid-Base Titrations The product of the titration of benzoic acid is the benzoate ion, Bz -. Bz - is the conjugate base of a weak acid. Therefore, final solution is basic. Bz - + H 2 O e HBz + OH - Bz - + H 2 O e HBz + OH - K b = 1.6 x 10 -10 K b = 1.6 x 10 -10 + + e

42 © 2009 Brooks/Cole - Cengage Acid-Base Reactions Strategy — find the conc. of the conjugate base Bz - in the solution AFTER the titration, then calculate pH. This is a two-step problem 1.stoichiometry of acid-base reaction 2.equilibrium calculation QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?

43 © 2009 Brooks/Cole - Cengage Acid-Base Reactions STOICHIOMETRY PORTION 1. Calc. moles of NaOH req’d (0.100 L HBz)(0.025 M) = 0.0025 mol HBz This requires 0.0025 mol NaOH 2.Calc. volume of NaOH req’d 0.0025 mol (1 L / 0.100 mol) = 0.025 L 25 mL of NaOH req’d QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?

44 © 2009 Brooks/Cole - Cengage Acid-Base Reactions STOICHIOMETRY PORTION 25 mL of NaOH req’d 3. Moles of Bz - produced = moles HBz = 0.0025 mol 4. Calc. conc. of Bz - There are 0.0025 mol of Bz - in a TOTAL SOLUTION VOLUME of QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution? 125 mL [Bz - ] = 0.0025 mol / 0.125 L = 0.020 M

45 © 2009 Brooks/Cole - Cengage Acid-Base Reactions Equivalence Point Most important species in solution is benzoate ion, Bz -, the weak conjugate base of benzoic acid, HBz. Bz - + H 2 O e HBz + OH - K b = 1.6 x 10 -10 [Bz - ] [HBz][OH - ] [Bz - ] [HBz][OH - ] initial0.020 00 change- x+x+x equilib0.020 - xxx QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH at equivalence point?

46 © 2009 Brooks/Cole - Cengage Acid-Base Reactions Equivalence Point Most important species in solution is benzoate ion, Bz -, the weak conjugate base of benzoic acid, HBz. Bz - + H 2 O e HBz + OH - K b = 1.6 x 10 -10 QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH at equivalence point? x = [OH - ] = 1.8 x 10 -6 pOH = 5.75 and pH = 8.25 pOH = 5.75 and pH = 8.25

47 © 2009 Brooks/Cole - Cengage QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH at half-way point? pH at half-way point? 1.< 7 2.= 7 3.> 7 pH at half-way point? 1.< 7 2.= 7 3.> 7 Equivalence point pH = 8.25 Equivalence point pH = 8.25

48 © 2009 Brooks/Cole - Cengage QUESTION: You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH at half-way point? pH at half- way point Equivalence point pH = 8.25 Equivalence point pH = 8.25

49 © 2009 Brooks/Cole - Cengage Acid-Base Reactions HBz + H 2 O e H 3 O + + Bz - K a = 6.3 x 10 -5 You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH. What is the pH at the half-way point? What is the pH at the half-way point? At the half-way point, [HBz] = [Bz - ] Therefore, [H 3 O + ] = K a = 6.3 x 10 -5 pH = 4.20 = pK a of the acid Both HBz and Bz - are present. This is a BUFFER! Both HBz and Bz - are present. This is a BUFFER!

53 © 2009 Brooks/Cole - Cengage Titration of a 1. Strong acid with strong base? 2. Weak acid with strong base? 3. Strong base with weak acid? 4. Weak base with strong acid? 5. Weak base with weak acid 6. Weak acid with weak base? pH Volume of titrating reagent added -->

59 © 2009 Brooks/Cole - Cengage Types of Chemical Reactions EXCHANGE REACTIONS: AB + CD f AD + CBEXCHANGE REACTIONS: AB + CD f AD + CB –Acid-base: CH 3 CO 2 H + NaOH f NaCH 3 CO 2 + H 2 O –Gas forming: CaCO 3 + 2 HCl f CaCl 2 + CO 2 (g) + H 2 O –Precipitation: Pb(NO 3 ) 2 + 2 KI f PbI 2 (s) + 2 KNO 3 OXIDATION REDUCTIONOXIDATION REDUCTION –4 Fe + 3 O 2 f 2 Fe 2 O 3 Apply equilibrium principles to acid-base and precipitation reactions.Apply equilibrium principles to acid-base and precipitation reactions.

60 © 2009 Brooks/Cole - Cengage Analysis of Silver Group All salts formed in this experiment are said to be INSOLUBLE and form when mixing moderately concentrated solutions of the metal ion with chloride ions. PLAY MOVIE

61 © 2009 Brooks/Cole - Cengage Analysis of Silver Group Although all salts formed in this experiment are said to be insoluble, they do dissolve to some SLIGHT extent. AgCl(s) e Ag + (aq) + Cl - (aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED.

62 © 2009 Brooks/Cole - Cengage Analysis of Silver Group AgCl(s) e Ag + (aq) + Cl - (aq) 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 - ]? [Cl - ] = [Ag + ] = 1.67 x 10 -5 M

63 © 2009 Brooks/Cole - Cengage Analysis of Silver Group AgCl(s) e Ag + (aq) + Cl - (aq) Saturated solution has [Ag + ] = [Cl - ] = 1.67 x 10 -5 M Use this to calculate K c K c = [Ag + ] [Cl - ] = (1.67 x 10 -5 )(1.67 x 10 -5 ) = (1.67 x 10 -5 )(1.67 x 10 -5 ) = 2.79 x 10 -10 = 2.79 x 10 -10

64 © 2009 Brooks/Cole - Cengage Analysis of Silver Group AgCl(s) e Ag + (aq) + Cl - (aq) K c = [Ag + ] [Cl - ] = 2.79 x 10 -10 Because this is the product of “solubilities”, we call it K sp = solubility product constant See Table 18.2 and Appendix J See Table 18.2 and Appendix J

67 © 2009 Brooks/Cole - Cengage Solution 1. Solubility = [Pb 2+ ] = 1.30 x 10 -3 M [I - ] = ? [I - ] = ? [I - ] = 2 x [Pb 2+ ] = 2.60 x 10 -3 M Solubility of Lead(II) Iodide Consider PbI 2 dissolving in water PbI 2 (s) e Pb 2+ (aq) + 2 I - (aq) Calculate K sp if solubility = 0.00130 M

68 © 2009 Brooks/Cole - Cengage Solution 2. K sp = [Pb 2+ ] [I - ] 2 = [Pb 2+ ] {2 [Pb 2+ ]} 2 K sp = 4 [Pb 2+ ] 3 K sp = 4 [Pb 2+ ] 3 Solubility of Lead(II) Iodide = 4 (solubility) 3 K sp = 4 (1.30 x 10 -3 ) 3 = 8.79 x 10 -9 Consider PbI 2 dissolving in water PbI 2 (s) e Pb 2+ (aq) + 2 I - (aq) Calculate K sp if solubility = 0.00130 M

69 © 2009 Brooks/Cole - Cengage Precipitating an Insoluble Salt Hg 2 Cl 2 (s) e Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 If [Hg 2 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 2+ without forming Hg 2 Cl 2 ?

70 © 2009 Brooks/Cole - Cengage Precipitating an Insoluble Salt Hg 2 Cl 2 (s) e Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 Recognize that K sp = product of maximum ion concs. Precip. begins when product of ion concs. EXCEEDS the K sp.

71 © 2009 Brooks/Cole - Cengage Precipitating an Insoluble Salt Hg 2 Cl 2 (s) e Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 = [Hg 2 2+ ] [Cl - ] 2 Solution [Cl - ] that can exist when [Hg 2 2+ ] = 0.010 M, If this conc. of Cl - is just exceeded, Hg 2 Cl 2 begins to precipitate.

72 © 2009 Brooks/Cole - Cengage Precipitating an Insoluble Salt Hg 2 Cl 2 (s) e Hg 2 2+ (aq) + 2 Cl - (aq) K sp = 1.1 x 10 -18 Now raise [Cl - ] to 1.0 M. 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 !

75 © 2009 Brooks/Cole - Cengage Barium Sulfate K sp = 1.1 x 10 -10 (b) BaSO 4 is opaque to x-rays. Drinking a BaSO 4 cocktail enables a physician to exam the intestines. (a) BaSO 4 is a common mineral, appearing a white powder or colorless crystals.

76 © 2009 Brooks/Cole - Cengage 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) e Ba 2+ (aq) + SO 4 2- (aq) Solution Solubility in pure water = [Ba 2+ ] = [SO 4 2- ] = x K sp = [Ba 2+ ] [SO 4 2- ] = x 2 x = (K sp ) 1/2 = 1.1 x 10 -5 M Solubility in pure water = 1.0 x 10 -5 mol/L The Common Ion Effect

77 © 2009 Brooks/Cole - Cengage Solution Solubility in pure water = 1.1 x 10 -5 mol/L. 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?___ The Common Ion Effect 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) e Ba 2+ (aq) + SO 4 2- (aq)

78 © 2009 Brooks/Cole - Cengage Solution [Ba 2+ ][SO 4 2- ] [Ba 2+ ][SO 4 2- ]initialchangeequilib. The Common Ion Effect + y 0.010 0 0.010 + y y 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) e Ba 2+ (aq) + SO 4 2- (aq)

79 © 2009 Brooks/Cole - Cengage Solution K sp = [Ba 2+ ] [SO 4 2- ] = (0.010 + y) (y) Because y < 1.1 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 2+ ion. The Common Ion Effect 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) e Ba 2+ (aq) + SO 4 2- (aq)

80 © 2009 Brooks/Cole - Cengage SUMMARY Solubility in pure water = x = 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! The Common Ion Effect 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) e Ba 2+ (aq) + SO 4 2- (aq)

81 © 2009 Brooks/Cole - Cengage 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 PLAY MOVIE

82 © 2009 Brooks/Cole - Cengage 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? 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 precipitates first. The ion requiring the lesser amount of CrO 4 2- ppts. first.

83 © 2009 Brooks/Cole - Cengage Separating Salts by Differences in K sp [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 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 Calculate [CrO 4 2- ] required by each ion. [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

84 © 2009 Brooks/Cole - Cengage 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 ppts. first. K sp (Ag 2 CrO 4 )= 9.0 x 10 -12 K sp (PbCrO 4 ) = 1.8 x 10 -14 How much Pb 2+ remains in solution when Ag + begins to precipitate? Separating Salts by Differences in K sp

85 © 2009 Brooks/Cole - Cengage 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. How much Pb 2+ remains in solution when Ag + begins to precipitate? Solution Separating Salts by Differences in K sp We know that [CrO 4 2- ] = 2.3 x 10 -8 M to begin to ppt. Ag 2 CrO 4. What is the Pb 2+ conc. at this point? [Pb 2+ ] = K sp / [CrO 4 2- ] = 1.8 x 10 -14 / 2.3 x 10 -8 M = 7.8 x 10 -7 M = 7.8 x 10 -7 M Lead ion has dropped from 0.020 M to < 10 -6 M

86 © 2009 Brooks/Cole - Cengage Separating Salts by Differences in K sp Add CrO 4 2- to solid PbCl 2. The less soluble salt, PbCrO 4, precipitatesAdd CrO 4 2- to solid PbCl 2. The less soluble salt, PbCrO 4, precipitates ePbCl 2 (s) + CrO 4 2- e PbCrO 4 + 2 Cl - SaltK sp PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14

87 © 2009 Brooks/Cole - Cengage Separating Salts by Differences in K sp PbCl 2 (s) + CrO 4 2- e PbCrO 4 + 2 Cl -PbCl 2 (s) + CrO 4 2- e PbCrO 4 + 2 Cl - SaltK sp PbCl 2 1.7 x 10 -5 PbCl 2 1.7 x 10 -5 PbCrO 4 1.8 x 10 -14 PbCl 2 (s) e Pb 2+ + 2 Cl - K 1 = K sp Pb 2+ + CrO 4 2- e PbCrO 4 K 2 = 1/K sp K net = (K 1 )(K 2 ) = 9.4 x 10 8 Net reaction is product-favored

88 © 2009 Brooks/Cole - Cengage Lead Chemistry From Chemistry & Chemical Reactivity, 5th edition Illustrates the transformation of one insoluble compound into an even less soluble compound. PbCl 2 PbI 2 Pb(CO 3 ) 2 PbCrO 4

92 © 2009 Brooks/Cole - Cengage Formation of complex ions explains why you can dissolve a ppt. by forming a complex ion. Dissolving Precipitates by forming Complex Ions AgCl(s) + 2 NH 3 e Ag(NH 3 ) 2 + + Cl - PLAY MOVIE

93 © 2009 Brooks/Cole - Cengage Examine the solubility of AgCl in ammonia. AgCl(s) e Ag + + Cl - K sp = 1.8 x 10 -10 Ag + + 2 NH 3 e Ag(NH 3 ) 2 + K form = 1.6 x 10 7 ------------------------------------- AgCl(s) + 2 NH 3 e Ag(NH 3 ) 2 + + Cl - K net = (K sp )(K form ) = 2.9 x 10 -3 K net = (K sp )(K form ) = 2.9 x 10 -3 By adding excess NH 3, the equilibrium shifts to the right. Dissolving Precipitates by forming Complex Ions

John C. Kotz State University of New York, College at Oneonta John C. Kotz Paul M. Treichel John Townsend Chapter 18 Principles.

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John C. Kotz State University of New York, College at Oneonta John C. Kotz Paul M. Treichel John Townsend Chapter 18 Principles.

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Presentation on theme: "John C. Kotz State University of New York, College at Oneonta John C. Kotz Paul M. Treichel John Townsend Chapter 18 Principles."— Presentation transcript:

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