1. 19-1 Chapter 19 Ionic Equilibria in Aqueous Systems

2. 19-2 Ionic Equilibria in Aqueous Systems 19.1 Equilibria of acid-base buffer systems 19.2 Acid-base titration curves 19.3 Equilibria of slightly soluble ionic compounds 19.4 Equilibria involving complex ions 19.5 Application of ionic equilibria to chemical analysis

3. 19-3 Figure 19.1 The effect of addition of acid or base to … an unbuffered solution. a buffered solution. acid addedbase added acid addedbase added Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

4. 19-4 Acid-Base Buffer Systems Buffers function by reducing changes in [H 3 O + ] that result from additions of acid or base to the solution. Buffers are composed of the conjugate acid-base pair of a weak acid. Buffers function via the common ion effect. CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq) The common ion effect occurs when a reactant containing a given ion is added to an equilibrium mixture that already contains that ion and the position of the equilibrium shifts away from forming more of it.

5. 19-5 Table 19.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH 3 COOH] initial [CH 3 COO - ] added % dissociation*pH * % dissociation = [CH 3 COOH] dissoc [CH 3 COOH] initial x 100 0.100.00 0.10 0.050 0.10 0.15 1.3 0.036 0.018 0.012 2.89 4.44 4.74 4.92 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

6. 19-6 Figure 19.2 How an acetic acid/acetate buffer works buffer with equal concentrations of weak acid and its conjugate base OH - H3O+H3O+ buffer after addition of H 3 O + H 2 O + CH 3 COOH H 3 O + + CH 3 COO - buffer after addition of OH - CH 3 COOH + OH - H 2 O + CH 3 COO - Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. HA scavenges OH -, A - scavenges H +

7. 19-7 Some Details K a = [CH 3 COO - ][H 3 O + ]/[CH 3 COOH] [H 3 O + ] = K a x [CH 3 COOH]/[CH 3 COO - ] Since K a is constant, [H 3 O + ] depends directly on the ratio of the concentrations of HA and its conjugate base, A -.

8. 19-8 Sample Problem 19.1 Calculating the effect of added H 3 O + and OH - on buffer pH PROBLEM:Calculate the pH of the following solutions. (a) A buffer solution consisting of 0.50 M CH 3 COOH and 0.50 M CH 3 COONa (b) After adding 0.020 mol of solid NaOH to 1.0 L of the buffer solution in part (a) (c) After adding 0.020 mol of HCl to 1.0 L of the buffer solution in part (a) K a of CH 3 COOH = 1.8 x 10 - 5 (assume the additions cause negligible volume changes) PLAN:PLAN: We know K a and can find initial concentrations of conjugate acid and base. Make assumptions about the amount of acid dissociating relative to its initial concentration. Proceed stepwise through changes in the system. initial change equilibrium 0.50 + x 0.50 - x - - - 0.500 + x 0.50 + xx - x SOLUTION: CH 3 COOH( aq ) + H 2 O( l ) CH 3 COO - ( aq ) + H 3 O + ( aq )concentration (M) (a) ___________________________________________________________

9. 19-9 Sample Problem 19.1(continued) [CH 3 COOH] eq ≈ 0.50 M[CH 3 COO - ] eq ≈ 0.50 M[H 3 O + ] = x K a = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] [H 3 O + ] = x = K a [CH 3 COO - ] [CH 3 COOH] = 1.8 x 10 - 5 M Check the assumption:1.8 x 10 - 5 /0.50 X 100 = 3.6 x 10 - 3 % CH 3 COOH( aq ) + OH - ( aq ) CH 3 COO - ( aq ) + H 2 O ( l )concentration (M) before addition addition after addition (b) [OH - ] added = 0.020 mol 1.0 L soln = 0.020 M NaOH 0.50- - - - -0.020- 0.4800.52 ___________________________________________________________ pH = 4.74

10. 19-10 Sample Problem 19.1(continued) Set up a reaction table with the new values. CH 3 COOH( aq ) + H 2 O( l ) CH 3 COO - ( aq ) + H 3 O + ( aq )concentration (M) initial change equilibrium 0.48- - x- x 0.48 - x - - 0.520 x + x 0.52 + x [H 3 O + ] = 1.8 x 10 - 5 x 0.48 0.52 = 1.7 x 10 - 5 pH = 4.77 CH 3 COO - ( aq ) + H 3 O + ( aq ) CH 3 COOH( aq ) + H 2 O ( l )concentration (M) before addition addition after addition (c)[H 3 O + ] added = 0.020 mol 1.0 L soln = 0.020 M H 3 O + 0.50- - - - -0.020- 0.4800.52 ___________________________________________________________

11. 19-11 Sample Problem 19.1(continued) Set up a reaction table with the new values. CH 3 COOH( aq ) + H 2 O( l ) CH 3 COO - ( aq ) + H 3 O + ( aq )concentration (M) initial change equilibrium 0.52- - x- x 0.52 - x - - 0.480 x + x 0.48 + x [H 3 O + ] = 1.8 x 10 - 5 x 0.48 0.52 = 2.0 x 10 - 5 pH = 4.70 ___________________________________________________________ Note that this buffer resists changes in pH from additions of either strong acid or strong base! (What is the solution pH of 0.020 M HCl or 0.020 M NaOH?)

12. 19-12 The Henderson-Hasselbalch Equation An equation that relates pH, pK a and the ratio of weak acid to its conjugate base for a buffer solution. HA + H 2 O H 3 O + + A - K a = [H 3 O + ][A - ]/[HA] [H 3 O + ] = K a x [HA]/[A - ] -log [H 3 O + ] = -log K a - log ([HA]/[A - ]) pH = pK a + log ([A - ]/[HA]) pH = pK a + log ([base]/[acid]) Special case: when [base] = [acid], the pH of the buffer solution equals the pK a of the weak acid. or

13. 19-13 Figure 19.3 The relationship between buffer capacity and pH change Buffer capacity refers to the ability of a buffer to resist pH change; buffer capacity increases as the concentrations of its components (i.e., the weak acid and its conjugate base) increase. Buffer pH and buffer capacity are different concepts. For an acetic acid/ acetate buffer

14. 19-14 A buffer has the highest capacity when the concentrations of HA and A - are equal. For a given addition of acid or base, the concentration ratio ([A - ]/[HA]) changes less for similar buffer component concentrations than it does for different concentrations. Key Concepts Buffer range: the pH range over which the buffer acts effectively; defined as pK a +/- 1 pH unit A buffer whose pH is equal to or near the pK a of its acid component has the highest buffer capacity.

15. 19-15 Sample Problem 19.2Preparing a buffer SOLUTION: PROBLEM:An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of acid rain on limestone-rich soils. How many grams of Na 2 CO 3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO 3 to prepare the buffer? K a of HCO 3 - is 4.7 x 10 - 11. PLAN:We know the K a and the conjugate acid-base pair. Convert pH to [H 3 O + ], find the number of moles of carbonate and convert to mass. HCO 3 - ( aq ) + H 2 O( l ) CO 3 2 - ( aq ) + H 3 O + ( aq ) K a = [CO 3 2 - ][H 3 O + ] [HCO 3 - ] pH = 10.00; [H 3 O + ] = 1.0 x 10 - 10 4.7 x 10 - 11 = [CO 3 2 - ][10 - 10 ] [0.20] [CO 3 2 - ] = 0.094 M moles of Na 2 CO 3 = (1.5 L)(0.094 mol/L) = 0.14 moles 0.14 moles 105.99 g mol = 15 g Na 2 CO 3 x

16. 19-16 Acid-Base Titration Curves Acid-Base Indicator: a weak organic acid (HIn) that has a different color than its conjugate base (In - ); small amounts are used in acid-base titrations; change color over different pH ranges

17. 19-17 Figure 19.4 Colors and approximate pH ranges of some common acid-base indicators

18. 19-18 Figure 19.5 The color change of the indicator bromthymol blue acidic basic change occurs over ~2 pH units Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. pH 6 pH 7.5

19. 19-19 Figure 19.6 Curve for a strong acid-strong base titration Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 40 mL of 0.1000 M HCl

20. 19-20 Figure 19.7 Curve for a weak acid- strong base titration Titration of 40.00 mL of 0.1000 M HPr with 0.1000 M NaOH [HPr] = [Pr - ] pH = 8.80 at equivalence point pK a of HPr = 4.89 methyl red HPr = CH 3 CH 2 COOH

21. 19-21 Sample Problem 19.3 Calculating the pH during a weak acid-strong base titration PROBLEM:Calculate the pH during the titration of 40.00 mL of 0.1000 M propanoic acid (HPr; K a = 1.3 x 10 - 5 ) after adding the following volumes of 0.1000 M NaOH: (a) 0.00 mL(b) 30.00 mL(c) 40.00 mL(d) 50.00 mL PLAN:The amounts of HPr and Pr - will be changing during the titration. Remember to adjust the total volume of solution after each addition. SOLUTION:(a) Find the starting pH using the methods of Chapter 18. K a = [Pr - ][H 3 O + ]/[HPr][Pr - ] = x = [H 3 O + ] x = 1.1 x 10 - 3 ; pH = 2.96 (b) before addition addition after addition 0.004000 0.003000 0.001000 0- - -0 - -- HPr( aq ) + OH - ( aq ) Pr - ( aq ) + H 2 O ( l )amount (mol) _________________________________________________

22. 19-22 Sample Problem 19.3(continued) [H 3 O + ] = 1.3 x 10 - 5 x 0.001000 mol 0.003000 mol = 4.3 x 10 - 6 M pH = 5.37 (c) When 40.00 mL of NaOH are added, all of the HPr will be reacted and the [Pr - ] will be: 0.004000 mol 0.04000 L + 0.04000 L = 0.05000 M K a x K b = K w K b = K w /K a = 1.0 x 10 - 14 /1.3 x 10 - 5 = 7.7 x 10 - 10 [H 3 O + ] = K w / = 1.6 x 10 - 9 M pH = 8.80 (d) 50.00 mL of NaOH will produce an excess of OH -. mol excess OH - = (0.1000 M)(0.05000 L - 0.04000 L) = 0.00100 mol M = 0.00100 mol 0.0900 L M = 0.01111 [H 3 O + ] = 1.0 x 10 - 14 /0.01111 = 9.0 x 10 - 13 M pH = 12.05

23. 19-23 Figure 19.8 Curve for a weak base- strong acid titration Titration of 40.00 mL of 0.1000 M NH 3 with 0.1000 M HCl pH = 5.27 at equivalence point pK a of NH 4 + = 9.25

24. 19-24 pK a2 = 7.19 pK a1 = 1.85 Figure 19.9 Curve for the titration of a weak polyprotic acid. Titration of 40.00 mL of 0.1000 M H 2 SO 3 with 0.1000 M NaOH sulfurous acid: a diprotic weak acid

25. 19-25 Figure 19.10 Sickle shape of red blood cells in sickle cell anemia Single-site mutations in the hemoglobin molecule can change the net charge on the protein, which causes protein aggregation and a consequent change in cell morphology

26. 19-26 Dealing with Solubility Equilibria Slightly soluble ionic compounds Ion-Product Expressions (Q sp ); Solubility-Product Constants (K sp ) PbSO 4 (s) Pb 2 + (aq) + SO 4 2 - (aq) Q c = [Pb 2 + ][SO 4 2 - ]/[PbSO 4 ] Q sp = [Pb 2 + ][SO 4 2 - ] At saturation, Q sp attains a constant value (equilibrium has been established); thus, Q sp = K sp Generally, for a saturated solution of a slightly soluble ionic compound with formula M p X q : Q sp = [M n + ] p [X z - ] q = K sp

27. 19-27 A slightly soluble ionic compound PbCl 2

28. 19-28 Sample Problem 19.4 Writing ion-product expressions for slightly soluble ionic compounds SOLUTION: PROBLEM:Write the ion-product expression for each of the following: (a) magnesium carbonate(b) iron(II) hydroxide (c) calcium phosphate(d) silver sulfide PLAN:Write an equation which describes a saturated solution. Take note of the unusual behavior of the sulfide ion produced in (d). K sp = [Mg 2 + ][CO 3 2 - ](a) MgCO 3 ( s ) Mg 2 + ( aq ) + CO 3 2 - ( aq ) K sp = [Fe 2 + ][OH - ] 2 (b) Fe(OH) 2 ( s ) Fe 2 + ( aq ) + 2OH - ( aq ) K sp = [Ca 2 + ] 3 [PO 4 3 - ] 2 (c) Ca 3 (PO 4 ) 2 ( s ) 3Ca 2 + ( aq ) + 2PO 4 3 - ( aq ) (d) Ag 2 S( s ) 2Ag + ( aq ) + S 2 - ( aq ) S 2 - ( aq ) + H 2 O( l ) HS - ( aq ) + OH - ( aq ) Ag 2 S( s ) + H 2 O( l ) 2Ag + ( aq ) + HS - ( aq ) + OH - ( aq )K sp = [Ag + ] 2 [HS - ][OH - ]

29. 19-29 Table 19.2 Solubility-Product Constants (K sp ) of Selected Ionic Compounds at 25 o C Name and FormulaK sp aluminum hydroxide, Al(OH) 3 cobalt(II) carbonate, CoCO 3 iron(II) hydroxide, Fe(OH) 2 lead(II) fluoride, PbF 2 lead(II) sulfate, PbSO 4 silver sulfide, Ag 2 S zinc iodate, Zn(IO 3 ) 2 3 x 10 - 34 1.0 x 10 - 10 4.1 x 10 - 15 3.6 x 10 - 8 1.6 x 10 - 8 4.7 x 10 - 29 8 x 10 - 48 mercury(I) iodide, Hg 2 I 2 3.9 x 10 - 6 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The magnitude of K sp is a measure of how far to the right the dissolution proceeds at equilibrium (i.e., at saturation).

30. 19-30 Sample Problem 19.5Determining K sp from solubility data PROBLEM:(a) Lead(II) sulfate is a key component in car batteries. Its solubility in water at 25 o C is 4.25 x 10 - 3 g/100 mL solution. What is the K sp of PbSO 4 ? (b) When lead(II) fluoride (PbF 2 ) is shaken with pure water at 25 o C, the solubility is found to be 0.64 g/L. Calculate the K sp of PbF 2. PLAN:Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into the solubility product constant expression. K sp = [Pb 2 + ][SO 4 2 - ] 4.25 x 10 - 3 g 100 mL soln 1000 mL L 303.3 g PbSO 4 mol PbSO 4 = 1.40 x 10 - 4 M PbSO 4 K sp = [Pb 2 + ][SO 4 2 - ] = (1.40 x 10 - 4 ) 2 = 1.96 x 10 - 8 SOLUTION:PbSO 4 ( s ) Pb 2 + ( aq ) + SO 4 2 - ( aq )(a) xx

31. 19-31 Sample Problem 19.5(continued) (b) PbF 2 (s) Pb 2 + (aq) + 2F - (aq) K sp = [Pb 2 + ][F - ] 2 0.64 g L soln 245.2 g PbF 2 mol PbF 2 = 2.6 x 10 - 3 M PbF 2 K sp = (2.6 x 10 - 3 )(5.2 x 10 - 3 ) 2 = 7.0 x 10 - 8 x [Pb 2 + ][F - ]

32. 19-32 Sample Problem 19.6Determining solubility from K sp PROBLEM:Calcium hydroxide is a major component of mortar, plaster and cement, and solutions of Ca(OH) 2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH) 2 in water at 25 o C if its K sp is 6.5 x 10 - 6. PLAN:Write a dissociation equation and K sp expression; find the molar solubility (S) using a table. SOLUTION: Ca(OH) 2 ( s ) Ca 2 + ( aq ) + 2OH - ( aq ) K sp = [Ca 2 + ][OH - ] 2 -initial change equilibrium - - 00 +S+S+ 2S S2S2S K sp = (S)(2S) 2 = 4S 3 S = = 1.2 x 10 - 2 M Ca(OH) 2 ( s ) Ca 2 + ( aq ) + 2OH - ( aq )concentration (M) ____________________________________________________

33. 19-33 K sp and Solubilities K sp values are used to determine relative solubilities provided that comparisons are made between compounds whose formulas contain the same total number of ions. The compound having the higher K sp is more soluble.

34. 19-34 Table 19.3 Relationship Between K sp and Solubility at 25 o C no. of ionsformulacation:anionK sp solubility (M) 2 MgCO 3 1:1 3.5 x 10 - 8 1.9 x 10 - 4 2 PbSO 4 1:1 1.6 x 10 - 8 1.3 x 10 - 4 2 BaCrO 4 1:1 2.1 x 10 - 10 1.4 x 10 - 5 3 Ca(OH) 2 1:2 6.5 x 10 - 6 1.2 x 10 - 2 3 BaF 2 1:2 1.5 x 10 - 6 7.2 x 10 - 3 3 CaF 2 1:2 3.2 x 10 - 11 2.0 x 10 - 4 3Ag 2 CrO 4 2:1 2.6 x 10 - 12 8.7 x 10 - 5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

35. 19-35 Common Ion Effects on Solubility The presence of a common ion decreases the solubility of a slightly soluble ionic compound. PbCrO 4 (s) Pb 2 + (aq) + CrO 4 2 - (aq) Add Na 2 CrO 4 (a very soluble salt; a strong electrolyte) Result: equilibrium shifts to the left (LeChâtelier’s principle)

36. 19-36 Figure 19.11 The effect of a common ion on solubility PbCrO 4 ( s ) Pb 2 + ( aq ) + CrO 4 2 - ( aq ) CrO 4 2 - added Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

37. 19-37 Sample Problem 19.7 Calculating the effect of a common ion on solubility PROBLEM:In Sample Problem 19.6, we calculated the solubility of Ca(OH) 2 in water. What is its solubility in 0.10 M Ca(NO 3 ) 2 ? K sp of Ca(OH) 2 is 6.5 x 10 - 6. PLAN:Set up a reaction equation and table for the dissolution of Ca(OH) 2. The Ca(NO 3 ) 2 will supply extra [Ca 2 + ], which will influence the solubility of the Ca(OH) 2 through the common ion effect. SOLUTION:Ca(OH) 2 ( s ) Ca 2 + ( aq ) + 2OH - ( aq )concentration (M) initial change equilibrium - - - 0.100 +S+S+2S 0.10 + S2S2S K sp = 6.5 x 10 - 6 = (0.10 + S)(2S) 2 = (0.10)(2S) 2 (S << 0.10) S ≈ = 4.0 x 10 - 3 M Check the assumption: 4.0 x 10 - 3 M 0.10 M x 100 = 4.0 % _____________________________________________ (6.5 x 10 -5 /4) 0.5 = 1.2 x 10 - 2 M

38. 19-38 The Effect of pH on Solubility If the compound contains the anion of a weak acid, addition of H 3 O + (from a strong acid) increases its solubility (LeChâtelier’s principle) CaCO 3 (s) Ca 2 + (aq) + CO 3 2 - (aq) CO 3 2 - (aq) + H 3 O + (aq) HCO 3 - (aq) + H 2 O(l) HCO 3 - (aq) + H 3 O + (aq) H 2 CO 3 (aq) + H 2 O(l) H 2 CO 3 (aq) 2H 2 O(l) + CO 2 (g) Adding H 3 O + shifts the equilibrium to the right.

39. 19-39 Figure 19.12 Test for the presence of a carbonate Effect of the addition of a strong acid (release of carbon dioxide)

40. 19-40 Sample Problem 19.8 Predicting the effect of adding a strong acid on solubility PROBLEM:Write balanced equations to explain whether addition of H 3 O + from a strong acid will affect the solubility of the following ionic compounds: (a) lead(II) bromide(b) copper(II) hydroxide(c) iron(II) sulfide PLAN:Write dissolution equations and consider how strong acid would affect the anion component. Br - is the anion of a strong acid. No effect. SOLUTION:(a) PbBr 2 ( s ) Pb 2 + ( aq ) + 2Br - ( aq ) (b) Cu(OH) 2 ( s ) Cu 2 + ( aq ) + 2OH - ( aq ) The OH - reacts with the added hydronium ion to form water. The equilibrium will shift to the right and thus solubility will increase. (c) FeS( s ) Fe 2 + ( aq ) + S 2 - ( aq ) S 2 - is the anion of a weak acid and will react with water to produce OH -. Both weak acids serve to increase the solubility of FeS. FeS( s ) + H 2 O( l ) Fe 2 + ( aq ) + HS - ( aq ) + OH - ( aq )

41. 19-41 Predicting Whether a Precipitate Will Form Compare Q sp with K sp. When Q sp = K sp, the solution is saturated and no change occurs. When Q sp > K sp, a precipitate forms until the solution is saturated. When Q sp < K sp, the solution is unsaturated and no precipitate forms.

42. 19-42 Sample Problem 19.9Predicting whether a precipitate will form PROBLEM:A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100 L of 0.30 M Ca(NO 3 ) 2 is mixed with 0.200 L of 0.060 M NaF? PLAN:Write out a reaction equation to see which salt could form. Look up the K sp values in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < K sp. Remember to consider the final diluted solution when calculating concentrations. SOLUTION: CaF 2 (s) Ca 2 + (aq) + 2F - (aq) K sp = 3.2 x 10 - 11 mol Ca 2 + = 0.100 L(0.30 mol/L) = 0.030 mol[Ca 2 + ] = 0.030 mol/0.300 L = 0.10 M mol F - = 0.200 L(0.060 mol/L) = 0.012 mol[F - ] = 0.012 mol/0.300 L = 0.040 M Q = [Ca 2 + ][F - ] 2 =(0.10)(0.040) 2 = 1.6 x 10 - 4 Q is >> K sp, thus CaF 2 will precipitate.

43. 19-43 Equilibria Involving Complex Ions A complex ion consists of a central metal ion covalently bonded to two or more anions (or molecules) called ligands. Ionic ligands include hydroxide (OH - ), chloride (Cl - ) and cyanide (CN - ) anions. Water, CO and NH 3 are examples of molecular ligands. All complex ions are Lewis adducts. The metal acts as a Lewis acid and the ligand acts as a Lewis base. We will consider equilibria of hydrated ions with ligands other than water.

44. 19-44 Figure 19.13 Cr(NH 3 ) 6 3+ is a typical complex ion. Cr 3 + is the central metal, and is surrounded by six NH 3 ligands

45. 19-45 Formation Constants, K f, of Complex Ions M(H 2 O) 4 2 + (aq) + 4NH 3 (aq) M(NH 3 ) 4 2 + (aq) + 4H 2 O(l) K c = [M(NH 3 ) 4 2 + ][H 2 O] 4 /[M(H 2 O) 4 2 + ][NH 3 ] 4 K f = K c /[H 2 O] 4 = [M(NH 3 ) 4 2 + ]/[M(H 2 O) 4 2 + ][NH 3 ] 4 In fact, four sequential reactions occur, defined by four K f values, for the systematic substitution of H 2 O ligands by NH 3 ligands. Thus, K f = K f1 x K f2 x K f3 x K f4

46. 19-46 Figure 19.14 The stepwise exchange of NH 3 for H 2 O in M(H 2 O) 4 2 + M(H 2 O) 4 2 + M(H 2 O) 3 (NH 3 ) 2 + M(NH 3 ) 4 2 + NH 3 3NH 3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

47. 19-47

48. 19-48 Sample Problem 19.10Calculating the concentration of a complex ion SOLUTION: PROBLEM:An industrial chemist converts Zn(H 2 O) 4 2 + to the more stable Zn(NH 3 ) 4 2 + by mixing 50.0 L of 0.0020 M Zn(H 2 O) 4 2 + and 25.0 L of 0.15 M NH 3. What is the final [Zn(H 2 O) 4 2 + ]? K f of Zn(NH 3 ) 4 2 + is 7.8 x 10 8. PLAN:Write the reaction equation and K f expression. Use a reaction table to list various concentrations. Remember that components will be diluted when mixed as you calculate final concentrations. The large excess of NH 3 will drive the reaction to completion. Zn(H 2 O) 4 2 + ( aq ) + 4NH 3 ( aq ) Zn(NH 3 ) 4 2 + ( aq ) + 4H 2 O( l ) K f = [Zn(NH 3 ) 4 2 + ] [Zn(H 2 O) 4 2 + ][NH 3 ] 4 [Zn(H 2 O) 4 2 + ] initial = (50.0 L)(0.0020 M) 75.0 L = 1.3 x 10 - 3 M [NH 3 ] initial =(25.0 L)(0.15 M) 75.0 L = 5.0 x 10 - 2 M

49. 19-49 Sample Problem 19.10(continued) Zn(H 2 O) 4 2 + ( aq ) + 4NH 3 ( aq ) Zn(NH 3 ) 4 2 + ( aq ) + 4H 2 O( l ) concentration (M) initial change equilibrium 1.3 x 10 - 3 5.0 x 10 - 2 0- ~(-1.3 x 10 - 3 )~(-5.2 x 10 - 3 ) Since we assume that all of the Zn(H 2 O) 4 2 + has reacted, it would consume four times its amount in NH 3.. [NH 3 ] used = 4 (1.3 x 10 - 3 M) = 5.2 x 10 - 3 M ~(+1.3 x 10 - 3 )- x 4.5 x 10 - 2 1.3 x 10 - 3 [Zn(H 2 O) 4 2 + ] remaining = x (a very small amount) - K f = [Zn(NH 3 ) 4 2 + ] [Zn(H 2 O) 4 2 + ][NH 3 ] 4 7.8 x 10 8 = (1.3 x 10 - 3 ) x (4.5 x 10 - 2 ) 4 x = 4.1 x 10 - 7 M _____________________________________________________________ A result of the very large K f !

50. 19-50 Solubility Issues A ligand increases the solubility of a slightly soluble ionic compound if it forms a complex ion with the cation. ZnS(s) + H 2 O(l) Zn 2 + (aq) + HS - (aq) + OH - (aq) K sp = 2.0 x 10 - 22 Upon addition of some 1.0 M NaCN: Zn 2 + (aq) + 4CN - (aq) Zn(CN) 4 2 + (aq) K f = 4.2 x 10 19 The overall equation is: ZnS(s) + 4CN - (aq) + H 2 O(l) Zn(CN) 4 2 + (aq) + HS - (aq) + OH - (aq) K overall = K sp x K f = 8.4 x 10 - 3

51. 19-51 Sample Problem 19.11Calculating the effect of complex ion formation on solubility SOLUTION: PROBLEM:In black-and-white film developing, excess AgBr is removed from the film negative by “hypo”, an aqueous solution of sodium thiosulfate (Na 2 S 2 O 3 ), through formation of the complex ion Ag(S 2 O 3 ) 2 3 -. Calculate the solubility of AgBr in (a) H 2 O; (b) 1.0 M hypo. K f of Ag(S 2 O 3 ) 2 3 - is 4.7 x 10 13 and K sp AgBr is 5.0 x 10 - 13. PLAN:Write equations for the reactions involved. Use K sp to find S, the molar solubility. Consider the shift in the equilibrium upon the addition of the complexing agent. AgBr( s ) Ag + ( aq ) + Br - ( aq ) K sp = [Ag + ][Br - ] = 5.0 x 10 - 13 S = [AgBr] dissolved = [Ag + ] = [Br - ] K sp = S 2 = 5.0 x 10 - 13 ; S = 7.1 x 10 - 7 M (a) (b)AgBr( s ) Ag + ( aq ) + Br - ( aq ) Ag + ( aq ) + 2S 2 O 3 2 - ( aq ) Ag(S 2 O 3 ) 2 3 - ( aq ) AgBr( s ) + 2S 2 O 3 2 - ( aq ) Br - ( aq ) + Ag(S 2 O 3 ) 2 3 - ( aq )

52. 19-52 Sample Problem 19.11(continued) K overall = K sp x K f = [Br - ][Ag(S 2 O 3 ) 2 3 - ] [S 2 O 3 2 - ] 2 = (5.0 x 10 - 13 )(4.7 x 10 13 )= 24 AgBr( s ) + 2S 2 O 3 2 - ( aq ) Br - ( aq ) + Ag(S 2 O 3 ) 2 3 - ( aq )concentration (M) initial change equilibrium - - - 1.0 -2S 1.0 - 2S 00 +S SS K overall = S2S2 (1.0 - 2S) 2 = 24 S 1.0 - 2S = (24) 1/2 S = [Ag(S 2 O 3 ) 2 3 - ] = 0.45 M __________________________________________________________

53. 19-53 Complex Ions of Amphoteric Hydroxides Amphoteric hydroxides: compounds that dissolve very little in water but to a much greater extent in acidic and basic solutions Al(OH) 3 (s) Al 3 + (aq) + 3OH - (aq) K sp = 3 x 10 - 34 Al(OH) 3 dissolves in acid: 3H 3 O + (aq) + 3OH - (aq) 6H 2 O(l) Al(OH) 3 dissolves in base through the formation of a complex ion: Al(OH) 3 (s) + OH - (aq) Al(OH) 4 - (aq) Al(OH) 3 (s) + 3H 3 O + (aq) Al 3 + (aq) + 6H 2 O(l)

54. 19-54 Figure 19.15 The amphoteric behavior of aluminum hydroxide

55. 19-55 Selective Precipitation Selection of an ion in solution by precipitation; achieved by adding a precipitating agent (ion) to the solution until the Q sp of the more soluble compound is almost equal to its K sp (but Q sp K sp precipitate.

56. 19-56 Sample Problem 19.12Separating ions by selective precipitation SOLUTION: PROBLEM:A solution consists of 0.20 M MgCl 2 and 0.10 M CuCl 2. Calculate the [OH - ] that would separate the metal ions as their hydroxides. K sp of Mg(OH) 2 = is 6.3 x 10 - 10 ; K sp of Cu(OH) 2 = 2.2 x 10 - 20. PLAN:Both precipitates have the same ion ratio, 1:2, so we can compare their K sp values to determine which has the greater solubility. Cu(OH) 2 will precipitate first (it has the smaller K sp ) so we calculate the [OH - ] needed for a saturated solution of Mg(OH) 2. This will ensure that we do not precipitate Mg(OH) 2. We can then check how much Cu 2 + remains in solution. Mg(OH) 2 ( s ) Mg 2 + ( aq ) + 2OH - ( aq ) K sp = 6.3 x 10 - 10 Cu(OH) 2 ( s ) Cu 2 + ( aq ) + 2OH - ( aq ) K sp = 2.2 x 10 - 20 [OH - ] needed for a saturated Mg(OH) 2 solution = = 5.6 x 10 - 5 M

57. 19-57 Sample Problem 19.12(continued) Use the K sp for Cu(OH) 2 to find the amount of Cu 2 + remaining in solution. [Cu 2 + ] = K sp /[OH - ] 2 = 2.2 x 10 - 20 /(5.6 x 10 - 5 ) 2 =7.0 x 10 - 12 M Since the solution was 0.10 M CuCl 2, virtually none of the Cu 2 + remains in solution.

58. 19-58 Qualitative Analysis Self-Study: pp. 841-845 in textbook Only an overview of this material will be provided in class. You will be responsible, however, for this material for the Final Exam.

59. 19-59 Figure 19.16 General procedure for separating ions in qualitative analysis add precipitating ion centrifuge add precipitating ion centrifuge Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

60. 19-60 Figure 19.17 A qualitative analysis scheme for separating cations into five ion groups add 6 M HCl centrifuge acidify to pH 0.5; add H 2 S centrifuge add NH 3 /NH 4 + buffer(pH 8) centrifuge add (NH 4 ) 2 HPO 4 centrifuge Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

61. 19-61

62. 19-62 Tests to determine the presence of cations in ion group 5 (a) flame test for Na + : yellow-orange (b) flame test for K + : violet (c) NH 4 + + OH - : pH test for NH 3 gas Figure 19.18

63. 19-63 Figure 19.18 Step 1: Add NH 3 ( aq ) centrifuge Step 2: Add HCl Step 3: Add NaOH centrifuge Step 4: Add HCl, Na 2 HPO 4 Step 5: Dissolve in HCl and add KSCN A qualitative analysis scheme for Ag +,Al 3 +,Cu 2 +, and Fe 3 +

64. 19-64 End of Assigned Material

65. 19-65 Figure B19.1 A view inside Carlsbad Caverns, New Mexico

66. 19-66 Figure B19.2 Formation of acidic precipitation

67. 19-67 Figure B19.4 The effect of acid rain on statuary Location: New York City

68. 19-68 A forest damaged by acid rain Figure B19.3 A forest damaged by acid rain