1. Acid-base titration

2. Titration In an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete – when the reaction is complete we have reached the endpoint of the titration An indicator may be added to determine the endpoint – an indicator is a chemical that changes color when the pH changes When the moles of H 3 O + = moles of OH −, the titration has reached its equivalence point 2 Tro: Chemistry: A Molecular Approach, 2/e

3. Titration 3 Tro: Chemistry: A Molecular Approach, 2/e

4. Titration Curve A plot of pH vs. amount of added titrant The inflection point of the curve is the equivalence point of the titration Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH The pH of the equivalence point depends on the pH of the salt solution – equivalence point of neutral salt, pH = 7 – equivalence point of acidic salt, pH < 7 – equivalence point of basic salt, pH > 7 Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH 4 Tro: Chemistry: A Molecular Approach, 2/e

5. Titration Curve: Unknown Strong Base Added to Strong Acid 5 Tro: Chemistry: A Molecular Approach, 2/e

6. Before Equivalence (excess acid) After Equivalence (excess base) Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH Equivalence Point equal moles of HCl and NaOH pH = 7.00  Because the solutions are equal concentration, and 1:1 stoichiometry, the equivalence point is at equal volumes 6 Tro: Chemistry: A Molecular Approach, 2/e

7. HClNaClNaOH mols before2.50E-305.0E-4 mols change −5.0E-4+5.0E-4−5.0E-4 mols end 2.00E-35.0E-40 molarity, new 0.06670.0170 HClNaClNaOH mols before2.50E-305.0E-4 mols change mols end molarity, new HClNaClNaOH mols before2.50E-305.0E-4 mols change −5.0E-4+5.0E-4−5.0E-4 mols end 2.00E-35.0E-40 molarity, new 5.0 x 10 −4 mole NaOH added Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 Before equivalence point added 5.0 mL NaOH 7Tro: Chemistry: A Molecular Approach, 2/e

8. Continued...

9. Table HClNaOHNaCl mols before2.50E-35.0E-40 mols change −5.0E-4 +5.0E-4 mols end 2.00E-305.0E-4 molarity, new 0.0667M00.017M

10. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (aq) To reach equivalence, the added moles NaOH = initial moles of HCl = 2.50 x 10 −3 moles At equivalence, we have 0.00 mol HCl and 0.00 mol NaOH left over Because the NaCl is a neutral salt, the pH at equivalence = 7.00 10 Tro: Chemistry: A Molecular Approach, 2/e

11. HClNaClNaOH mols before2.50E-302.5E-3 mols change mols end molarity, new HClNaClNaOH mols before2.50E-302.5E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-30 molarity, new HClNaClNaOH mols before2.50E-302.5E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-30 molarity, new 00.0500 2.5 x 10 −3 mole NaOH added Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 At equivalence point added 25.0 mL NaOH 11 Tro: Chemistry: A Molecular Approach, 2/e

13. HClNaClNaOH mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new 00.0450.0091 HClNaClNaOH mols before2.50E-303.0E-3 mols change mols end molarity, new HClNaClNaOH mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 After equivalence point added 30.0 mL NaOH 3.0 x 10 −3 mole NaOH added

15. added 5.0 mL NaOH 0.00200 mol HCl pH = 1.18 added 10.0 mL NaOH 0.00150 mol HCl pH = 1.37 added 25.0 mL NaOH equivalence point pH = 7.00 added 40.0 mL NaOH 0.00150 mol NaOH pH = 12.36 added 50.0 mL NaOH 0.00250 mol NaOH pH = 12.52 added 15.0 mL NaOH 0.00100 mol HCl pH = 1.60 added 20.0 mL NaOH 0.00050 mol HCl pH = 1.95 Adding 0.100 M NaOH to 0.100 M HCl 15 Tro: Chemistry: A Molecular Approach, 2/e

17. Practice – Calculate the pH of the solution that results when 10.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO 3 17 Tro: Chemistry: A Molecular Approach, 2/e

18. HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-3 mols change −1.5E-3+1.5E-3−1.5E-3 mols end 1.1E-31.5E-30 molarity, new HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-3 mols change −1.5E-3+1.5E-3−1.5E-3 mols end 1.1E-31.5E-30 molarity, new 0.0180.0250 Practice – Calculate the pH of the solution that results when 10.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO 3 HNO 3 (aq) + NaOH(aq)  NaNO 3 (aq) + H 2 O(l) Initial pH = −log(0.250) = 0.60 Initial mol of HNO 3 = 0.0500 L x 0.25 mol/L=1.25 x 10 −2 Before equivalence point added 10.0 mL NaOH 18 Tro: Chemistry: A Molecular Approach, 2/e

20. Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M HNO 3 to reach equivalence 20 Tro: Chemistry: A Molecular Approach, 2/e

21. Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M HNO 3 to reach equivalence HNO 3 (aq) + NaOH(aq)  NaNO 3 (aq) + H 2 O(l) Initial pH = −log(0.250) = 0.60 Initial mol of HNO 3 = 0.0500 L x 0.25 mol/L=1.25 x 10 −2 At equivalence point: moles of NaOH = 1.25 x 10 −2 21 Tro: Chemistry: A Molecular Approach, 2/e

22. Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO 3 22 Tro: Chemistry: A Molecular Approach, 2/e

23. HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-2 mols change −1.25E-2+1.25E-2−1.25E-2 mols end 01.25E-20.0025 molarity, new 00.08330.017 HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-2 mols change −1.25E-2+1.25E-2−1.25E-2 mols end 01.25E-20.0025 molarity, new Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO 3 HNO 3 (aq) + NaOH(aq)  NaNO 3 (aq) + H 2 O(l) Initial pH = −log(0.250) = 0.60 Initial mol of HNO 3 = 0.0500 L x 0.25 mol/L=1.25 x 10 −2 After equivalence point added 100.0 mL NaOH

25. HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-2 mols change −1.25E-2+1.25E-2−1.25E-2 mols end 01.25E-20.0025 molarity, new 00.08330.017 HNO 3 NaNO 3 NaOH mols before1.25E-201.5E-2 mols change −1.25E-2+1.25E-2−1.25E-2 mols end 01.25E-20.0025 molarity, new Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO 3 HNO 3 (aq) + NaOH(aq)  NaNO 3 (aq) + H 2 O(l) Initial pH = −log(0.250) = 0.60 initial mol of HNO 3 = 0.0500 L x 0.25 mol/L=1.25 x 10 −2 After equivalence point added 100.0 mL NaOH 25 Tro: Chemistry: A Molecular Approach, 2/e

27. Titration of a Strong Base with a Strong Acid If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one just shown 27 Tro: Chemistry: A Molecular Approach, 2/e

28. Titration of a Weak Acid with a Strong Base Titrating a weak acid with a strong base results in differences in the titration curve at the equivalence point and excess acid region The initial pH is determined using the K a of the weak acid The pH in the excess acid region is determined as you would determine the pH of a buffer The pH at the equivalence point is determined using the K b of the conjugate base of the weak acid The pH after equivalence is dominated by the excess strong base – the basicity from the conjugate base anion is negligible 28 Tro: Chemistry: A Molecular Approach, 2/e

29. Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH HCHO 2(aq) + NaOH (aq)  NaCHO 2(aq) + H 2 O (aq) Initial pH [HCHO 2 ][CHO 2 − ][H 3 O + ] initial 0.1000.000≈ 0 change −x−x+x+x+x+x equilibrium 0.100 − xxx K a = 1.8 x 10 −4 29 Tro: Chemistry: A Molecular Approach, 2/e

30. Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq) Initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 Before equivalence added 5.0 mL NaOH HAA−A− OH − mols before2.50E-300 mols added ––5.0E-4 mols after 2.00E-35.0E-4≈ 0 30 Tro: Chemistry: A Molecular Approach, 2/e

31. HAA−A− OH − mols before2.50E-300 mols added ––2.50E-3 mols after 02.50E-3≈ 0 Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH added 25.0 mL NaOH CHO 2 − (aq) + H 2 O (l)  HCHO 2(aq) + OH − (aq) [OH − ] = 1.7 x 10 −6 M HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq) Initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 At equivalence

33. K b = 5.6 x 10 −11

35. HAA−A− NaOH mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new HAA−A− NaOH mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new 00.0450.0091 HAA−A− NaOH mols before2.50E-303.0E-3 mols change mols end molarity, new Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH 35 added 30.0 mL NaOH 3.0 x 10 −3 mole NaOH added HCHO 2(aq) + NaOH (aq)  NaCHO 2(aq) + H 2 O (aq) Initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 After equivalence

37. added 35.0 mL NaOH 0.00100 mol NaOH xs pH = 12.22 initial HCHO 2 solution 0.00250 mol HCHO 2 pH = 2.37 added 5.0 mL NaOH 0.00200 mol HCHO 2 pH = 3.14 added 10.0 mL NaOH 0.00150 mol HCHO 2 pH = 3.56 added 25.0 mL NaOH equivalence point 0.00250 mol CHO 2 − [CHO 2 − ] init = 0.0500 M [OH − ] eq = 1.7 x 10 − 6 pH = 8.23 added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96 added 20.0 mL NaOH 0.00050 mol HCHO 2 pH = 4.34 added 15.0 mL NaOH 0.00100 mol HCHO 2 pH = 3.92 added 12.5 mL NaOH 0.00125 mol HCHO 2 pH = 3.74 = pK a half-neutralization Adding NaOH to HCHO 2 added 40.0 mL NaOH 0.00150 mol NaOH xs pH = 12.36 added 50.0 mL NaOH 0.00250 mol NaOH xs pH = 12.52 37 Tro: Chemistry: A Molecular Approach, 2/e

38. Titrating Weak Acid with a Strong Base The initial pH is that of the weak acid solution – calculate like a weak acid equilibrium problem e.g., 15.5 and 15.6 Before the equivalence point, the solution becomes a buffer – calculate mol HA init and mol A − init using reaction stoichiometry – calculate pH with Henderson-Hasselbalch using mol HA init and mol A − init Half-neutralization pH = pK a 38 Tro: Chemistry: A Molecular Approach, 2/e

39. Titrating Weak Acid with a Strong Base At the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established – mol A − = original mole HA calculate the volume of added base as you did in Example 4.8 – [A − ] init = mol A − /total liters – calculate like a weak base equilibrium problem e.g., 15.14 Beyond equivalence point, the OH is in excess – [OH − ] = mol MOH xs/total liters – [H 3 O + ][OH − ]=1 x 10 −14 39 Tro: Chemistry: A Molecular Approach, 2/e

40. Example 16.7a: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the volume of KOH at the equivalence point. 40 write an equation for the reaction for B with HA use stoichiometry to determine the volume of added B HNO 2 + KOH  NO 2  + H 2 O Tro: Chemistry: A Molecular Approach, 2/e

41. Example 16.7b: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH. 41 write an equation for the reaction for B with HA determine the moles of HA before & moles of added B make a stoichiometry table and determine the moles of HA in excess and moles A  made HNO 2 + KOH  NO 2  + H 2 O HNO 2 NO 2 − OH − mols before 0.004000≈ 0 mols added –– 0.00100 mols after ≈ 0≈ 0 0.00300 0.00100 Tro: Chemistry: A Molecular Approach, 2/e

42. Example 16.7b: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH. 42 write an equation for the reaction of HA with H 2 O determine K a and pK a for HA use the Henderson- Hasselbalch equation to determine the pH HNO 2 + H 2 O  NO 2  + H 3 O + HNO 2 NO 2 − OH − mols before 0.004000≈ 0 mols added –– 0.00100 mols after 0.003000.00100 ≈ 0≈ 0 Table 15.5 K a = 4.6 x 10 −4 Tro: Chemistry: A Molecular Approach, 2/e

44. Example 16.7b: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point. 44 write an equation for the reaction for B with HA determine the moles of HA before & moles of added B make a stoichiometry table and determine the moles of HA in excess and moles A  made HNO 2 + KOH  NO 2  + H 2 O HNO 2 NO 2 − OH − mols before 0.004000≈ 0 mols added –– 0.00200 mols after ≈ 0≈ 0 0.00200 at half-equivalence, moles KOH = ½ mole HNO 2 Tro: Chemistry: A Molecular Approach, 2/e

45. Example 16.7b: A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point. 45 write an equation for the reaction of HA with H 2 O determine K a and pK a for HA use the Henderson- Hasselbalch equation to determine the pH HNO 2 + H 2 O  NO 2  + H 3 O + HNO 2 NO 2 − OH − mols before 0.004000≈ 0 mols added –– 0.00200 mols after 0.00200 ≈ 0≈ 0 Table 15.5 K a = 4.6 x 10 -4 Tro: Chemistry: A Molecular Approach, 2/e

47. Titration Curve of a Weak Base with a Strong Acid 47 Tro: Chemistry: A Molecular Approach, 2/e

48. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the initial pH of the NH 3 solution. 48 Tro: Chemistry: A Molecular Approach, 2/e

49. Practice – Titration of 25.0 mL of 0.10 M NH 3 with 0.10 M HCl. Calculate the initial pH of the NH 3 (aq) 49 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial: NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH − (aq) [HCl][NH 4 + ][NH 3 ] initial 000.10 change +x+x+x+x−x−x equilibrium xx0.10−x pK b = 4.75 K b = 10 −4.75 = 1.8 x 10 −5 Tro: Chemistry: A Molecular Approach, 2/e

50. Practice – Titration of 25.0 mL of 0.10 M NH 3 with 0.10 M HCl. Calculate the initial pH of the NH 3 (aq) 50 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial: NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH − (aq) [HCl][NH 4 + ][NH 3 ] initial 000.10 change +x+x+x+x−x−x equilibrium xx0.10−x pK b = 4.75 K b = 10 −4.75 = 1.8 x 10 −5 Tro: Chemistry: A Molecular Approach, 2/e

51. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl. 51 Tro: Chemistry: A Molecular Approach, 2/e

52. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl. 52 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 Before equivalence: after adding 5.0 mL of HCl NH 3 NH 4 ClHCl mols before2.50E-305.0E-4 mols change −5.0E-4 mols end 2.00E-35.0E-40 molarity, new 0.06670.0170 NH 4 + (aq) + H 2 O (l)  NH 4 + (aq) + H 2 O (l) pK b = 4.75 pK a = 14.00 − 4.75 = 9.25 Tro: Chemistry: A Molecular Approach, 2/e

53. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl. NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 Before equivalence: after adding 5.0 mL of HCl NH 3 NH 4 ClHCl mols before2.50E-305.0E-4 mols change −5.0E-4 mols end 2.00E-35.0E-40 molarity, new 0.06670.0170 53 Tro: Chemistry: A Molecular Approach, 2/e

54. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. 54 Tro: Chemistry: A Molecular Approach, 2/e

55. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. 55 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 At equivalence mol NH 3 = mol HCl = 2.50 x 10 −3 added 25.0 mL HCl NH 3 NH 4 ClHCl mols before2.50E-302.5E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-30 molarity, new 00.0500 Tro: Chemistry: A Molecular Approach, 2/e

56. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. 56 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) at equivalence [NH 4 Cl] = 0.050 M [NH 3 ][NH 4 + ][H 3 O + ] initial 00.050≈ 0 change +x+x−x−x+x+x equilibrium x0.050−xx NH 4 + (aq) + H 2 O (l)  NH 3(aq) + H 3 O + (aq) Tro: Chemistry: A Molecular Approach, 2/e

57. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 30.0 mL of HCl. 57 Tro: Chemistry: A Molecular Approach, 2/e

58. Practice – Titration of 25.0 mL of 0.10 M NH 3 (pK b = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 30.0 mL of HCl. 58 NH 3(aq) + HCl (aq)  NH 4 Cl (aq) Initial mol of NH 3 = 0.0250 L x 0.100 mol/L = 2.50 x 10 −3 After equivalence: after adding 30.0 mL HCl NH 3 NH 4 ClHCl mols before2.50E-303.0E-3 mols change −2.5E-3+2.5E-3−2.5E-3 mols end 02.5E-35.0E-4 molarity, new 00.0450.0091 when you mix a strong acid, HCl, with a weak acid, NH 4 +, you only need to consider the strong acid Tro: Chemistry: A Molecular Approach, 2/e

59. Titration of a Polyprotic Acid If K a1 >> K a2, there will be two equivalence points in the titration – the closer the K a ’s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of 0.100 M H 2 SO 3 with 0.100 M NaOH 59 Tro: Chemistry: A Molecular Approach, 2/e

60. Monitoring pH During a Titration The general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H 3 O + ] – using a probe that specifically measures just H 3 O + The endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve If you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator 60 Tro: Chemistry: A Molecular Approach, 2/e

61. Monitoring pH During a Titration 61 Tro: Chemistry: A Molecular Approach, 2/e

62. Indicators Many dyes change color depending on the pH of the solution These dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution HInd (aq) + H 2 O (l)  Ind  (aq) + H 3 O + (aq) The color of the solution depends on the relative concentrations of Ind  :HInd – when Ind  :HInd ≈ 1, the color will be mix of the colors of Ind  and HInd – when Ind  :HInd > 10, the color will be mix of the colors of Ind  – when Ind  :HInd < 0.1, the color will be mix of the colors of HInd 62 Tro: Chemistry: A Molecular Approach, 2/e

63. Phenolphthalein 63 Tro: Chemistry: A Molecular Approach, 2/e

64. Methyl Red 64 Tro: Chemistry: A Molecular Approach, 2/e

65. Monitoring a Titration with an Indicator For most titrations, the titration curve shows a very large change in pH for very small additions of titrant near the equivalence point An indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH – pK a of HInd ≈ pH at equivalence point 65 Tro: Chemistry: A Molecular Approach, 2/e

66. Acid-Base Indicators 66 Tro: Chemistry: A Molecular Approach, 2/e

67. Solubility Equilibria All ionic compounds dissolve in water to some degree – however, many compounds have such low solubility in water that we classify them as insoluble We can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water 67 Tro: Chemistry: A Molecular Approach, 2/e

68. Solubility Product The equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, K sp For an ionic solid M n X m, the dissociation reaction is: M n X m (s)  nM m+ (aq) + mX n− (aq) The solubility product would be K sp = [M m+ ] n [X n− ] m For example, the dissociation reaction for PbCl 2 is PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) And its equilibrium constant is K sp = [Pb 2+ ][Cl − ] 2 68 Tro: Chemistry: A Molecular Approach, 2/e

69. 69 Tro: Chemistry: A Molecular Approach, 2/e

70. Molar Solubility Solubility is the amount of solute that will dissolve in a given amount of solution – at a particular temperature The molar solubility is the number of moles of solute that will dissolve in a liter of solution – the molarity of the dissolved solute in a saturated solution for the general reaction M n X m (s)  nM m+ (aq) + mX n− (aq) 70 Tro: Chemistry: A Molecular Approach, 2/e

71. Example 16.8: Calculate the molar solubility of PbCl 2 in pure water at 25  C 71 write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Pb 2+ ][Cl − ] Initial00 Change+S+2S EquilibriumS2S PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) K sp = [Pb 2+ ][Cl − ] 2 Tro: Chemistry: A Molecular Approach, 2/e

72. Example 16.8: Calculate the molar solubility of PbCl 2 in pure water at 25  C 72 substitute into the K sp expression find the value of K sp from Table 16.2, plug into the equation, and solve for S [Pb 2+ ][Cl − ] Initial00 Change+S+2S EquilibriumS2S K sp = [Pb 2+ ][Cl − ] 2 K sp = (S)(2S) 2 Tro: Chemistry: A Molecular Approach, 2/e

73. Practice – Determine the K sp of PbBr 2 if its molar solubility in water at 25  C is 1.05 x 10 −2 M 73 Tro: Chemistry: A Molecular Approach, 2/e

74. Practice – Determine the K sp of PbBr 2 if its molar solubility in water at 25  C is 1.05 x 10 −2 M 74 write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Pb 2+ ][Br − ] initial00 change+(1.05 x 10 −2 )+2(1.05 x 10 −2 ) equilibrium(1.05 x 10 −2 )(2.10 x 10 −2 ) PbBr 2 (s)  Pb 2+ (aq) + 2 Br − (aq) K sp = [Pb 2+ ][Br − ] 2 Tro: Chemistry: A Molecular Approach, 2/e

75. Practice – Determine the K sp of PbBr 2 if its molar solubility in water at 25  C is 1.05 x 10 −2 M 75 substitute into the K sp expression plug into the equation and solve K sp = [Pb 2+ ][Br − ] 2 K sp = (1.05 x 10 − 2 )(2.10 x 10 − 2 ) 2 [Pb 2+ ][Br − ] initial00 change+(1.05 x 10 −2 )+2(1.05 x 10 −2 ) equilibrium(1.05 x 10 −2 )(2.10 x 10 −2 ) Tro: Chemistry: A Molecular Approach, 2/e

76. K sp and Relative Solubility Molar solubility is related to K sp But you cannot always compare solubilities of compounds by comparing their K sp s To compare K sp s, the compounds must have the same dissociation stoichiometry 76 Tro: Chemistry: A Molecular Approach, 2/e

77. The Effect of Common Ion on Solubility Addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt For example, addition of NaCl to the solubility equilibrium of solid PbCl 2 decreases the solubility of PbCl 2 PbCl 2 (s)  Pb 2+ (aq) + 2 Cl − (aq) addition of Cl − shifts the equilibrium to the left 77 Tro: Chemistry: A Molecular Approach, 2/e

78. Example 16.10: Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25  C 78 write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Ca 2+ ][F − ] initial00.100 change+S+2S equilibriumS0.100 + 2S CaF 2 (s)  Ca 2+ (aq) + 2 F − (aq) K sp = [Ca 2+ ][F − ] 2 Tro: Chemistry: A Molecular Approach, 2/e

79. Example 16.10: Calculate the molar solubility of CaF 2 in 0.100 M NaF at 25  C 79 substitute into the K sp expression, assume S is small find the value of K sp from Table 16.2, plug into the equation, and solve for S [Ca 2+ ][F − ] initial00.100 change+S+2S equilibriumS0.100 + 2S K sp = [Ca 2+ ][F − ] 2 K sp = (S)(0.100 + 2S) 2 K sp = (S)(0.100) 2 Tro: Chemistry: A Molecular Approach, 2/e

81. Practice – Determine the concentration of Ag + ions in seawater that has a [Cl − ] of 0.55 M K sp of AgCl = 1.77 x 10 −10 81 Tro: Chemistry: A Molecular Approach, 2/e

82. Practice – Determine the concentration of Ag + ions in seawater that has a [Cl − ] of 0.55 M write the dissociation reaction and K sp expression create an ICE table defining the change in terms of the solubility of the solid [Ag + ][Cl − ] initial00.55 change+S+S+S+S equilibriumS0.55 + S AgCl(s)  Ag + (aq) + Cl − (aq) K sp = [Ag + ][Cl − ] 82 Tro: Chemistry: A Molecular Approach, 2/e

83. Practice – Determine the concentration of Ag + ions in seawater that has a [Cl − ] of 0.55 M 83 substitute into the K sp expression, assume S is small find the value of K sp from Table 16.2, plug into the equation, and solve for S [Ag + ][Cl − ] Initial00.55 Change+S+S+S+S EquilibriumS0.55 + S K sp = [Ag + ][Cl − ] K sp = (S)(0.55 + S) K sp = (S)(0.55) Tro: Chemistry: A Molecular Approach, 2/e

84. [Ag + ][Cl − ] Initial00.55 Change+S+S+S+S EquilibriumS0.55 + S

85. The Effect of pH on Solubility For insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide – and the lower the pH, the higher the solubility – higher pH = increased [OH − ] M(OH) n (s)  M n+ (aq) + nOH − (aq) For insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M 2 (CO 3 ) n (s)  2 M n+ (aq) + nCO 3 2− (aq) H 3 O + (aq) + CO 3 2− (aq)  HCO 3 − (aq) + H 2 O(l) 85 Tro: Chemistry: A Molecular Approach, 2/e

86. Precipitation Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound If we compare the reaction quotient, Q, for the current solution concentrations to the value of K sp, we can determine if precipitation will occur – Q = K sp, the solution is saturated, no precipitation – Q < K sp, the solution is unsaturated, no precipitation – Q > K sp, the solution would be above saturation, the salt above saturation will precipitate Some solutions with Q > K sp will not precipitate unless disturbed – these are called supersaturated solutions 86 Tro: Chemistry: A Molecular Approach, 2/e

87. precipitation occurs if Q > K sp a supersaturated solution will precipitate if a seed crystal is added 87 Tro: Chemistry: A Molecular Approach, 2/e

88. Selective Precipitation A solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others A successful reagent can precipitate with more than one of the cations, as long as their K sp values are significantly different 88 Tro: Chemistry: A Molecular Approach, 2/e

89. Example 16.12: Will a precipitate form when we mix Pb(NO 3 ) 2(aq) with NaBr (aq) if the concentrations after mixing are 0.0150 M and 0.0350 M respectively? write the equation for the reaction determine the ion concentrations of the original salts determine the K sp for any “insoluble” product write the dissociation reaction for the insoluble product calculate Q, using the ion concentrations compare Q to K sp. If Q > K sp, precipitation Pb(NO 3 ) 2(aq) + 2 NaBr (aq) → PbBr 2(s) + 2 NaNO 3(aq) K sp of PbBr 2 = 4.67 x 10 –6 PbBr 2(s)  Pb 2+ (aq) + 2 Br − (aq) Pb(NO 3 ) 2 = 0.0150 M  Pb 2+ = 0.0150 M, NO 3 − = 2(0.0150 M) NaBr = 0.0350 M  Na + = 0.0350 M, Br − = 0.0350 M Q < K sp, so no precipitation 89 Tro: Chemistry: A Molecular Approach, 2/e

90. Practice – Will a precipitate form when we mix Ca(NO 3 ) 2(aq) with NaOH (aq) if the concentrations after mixing are both 0.0175 M? K sp of Ca(OH) 2 = 4.68 x 10 −6 90 Tro: Chemistry: A Molecular Approach, 2/e

91. Practice – Will a precipitate form when we mix Ca(NO 3 ) 2(aq) with NaOH (aq) if the concentrations after mixing are both 0.0175 M? write the equation for the reaction determine the ion concentrations of the original salts determine the K sp for any “insoluble” product write the dissociation reaction for the insoluble product calculate Q, using the ion concentrations compare Q to K sp. If Q > K sp, precipitation Ca(NO 3 ) 2(aq) + 2 NaOH (aq) → Ca(OH) 2(s) + 2 NaNO 3(aq) K sp of Ca(OH) 2 = 4.68 x 10 –6 Ca(OH) 2(s)  Ca 2+ (aq) + 2 OH − (aq) Ca(NO 3 ) 2 = 0.0175 M  Ca 2+ = 0.0175 M, NO 3 − = 2(0.0175 M) NaOH = 0.0175 M  Na + = 0.0175 M, OH − = 0.0175 M Q > K sp, so precipitation 91 Tro: Chemistry: A Molecular Approach, 2/e

92. 92 Example 16.13: What is the minimum [OH − ] necessary to just begin to precipitate Mg 2+ (with [0.059]) from seawater? precipitating may just occur when Q = K sp Mg(OH) 2(s)  Mg 2+ (aq) + 2 OH − (aq) Tro: Chemistry: A Molecular Approach, 2/e

93. Practice – What is the minimum concentration of Ca(NO 3 ) 2(aq) that will precipitate Ca(OH) 2 from 0.0175 M NaOH (aq) ? K sp of Ca(OH) 2 = 4.68 x 10 −6 93 Tro: Chemistry: A Molecular Approach, 2/e

94. 94 Practice – What is the minimum concentration of Ca(NO 3 ) 2(aq) that will precipitate Ca(OH) 2 from 0.0175 M NaOH (aq) ? precipitating may just occur when Q = K sp [Ca(NO 3 ) 2 ] = [Ca 2+ ] = 0.0153 M Ca(NO 3 ) 2(aq) + 2 NaOH (aq) → Ca(OH) 2(s) + 2 NaNO 3(aq) Ca(OH) 2(s)  Ca 2+ (aq) + 2 OH − (aq) Tro: Chemistry: A Molecular Approach, 2/e

95. 95 Example 16.14: What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? precipitating may just occur when Q = K sp Ca(OH) 2(s)  Ca 2+ (aq) + 2 OH − (aq) Tro: Chemistry: A Molecular Approach, 2/e

96. 96 Example 16.14: What is the [Mg 2+ ] when Ca 2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg 2+ begins when [OH − ] = 1.9 x 10 − 6 M precipitating Ca 2+ begins when [OH − ] = 2.06 x 10 − 2 M when Ca 2+ just begins to precipitate out, the [Mg 2+ ] has dropped from 0.059 M to 4.8 x 10 − 10 M Mg(OH) 2(s)  Mg 2+ (aq) + 2 OH − (aq) Tro: Chemistry: A Molecular Approach, 2/e

97. Practice – A solution is made by mixing Pb(NO 3 ) 2(aq) with AgNO 3(aq) so both compounds have a concentration of 0.0010 M. NaCl (s) is added to precipitate out both AgCl (s) and PbCl 2(aq). What is the [Ag + ] concentration when the Pb 2+ just begins to precipitate? 97 Tro: Chemistry: A Molecular Approach, 2/e

98. 98 Practice – What is the [Ag + ] concentration when the Pb 2+ (0.0010 M) just begins to precipitate? precipitating may just occur when Q = K sp AgNO 3(aq) + NaCl (aq) → AgCl (s) + NaNO 3(aq) AgCl (s)  Ag + (aq) + Cl − (aq) Tro: Chemistry: A Molecular Approach, 2/e

99. 99 Practice – What is the [Ag + ] concentration when the Pb 2+ (0.0010 M) just begins to precipitate? precipitating may just occur when Q = K sp Pb(NO 3 ) 2(aq) + 2 NaCl (aq) → PbCl 2(s) + 2 NaNO 3(aq) PbCl 2(s)  Pb 2+ (aq) + 2 Cl − (aq) Tro: Chemistry: A Molecular Approach, 2/e

100. 100 Practice – What is the [Ag + ] concentration when the Pb 2+ (0.0010 M) just begins to precipitate precipitating Ag + begins when [Cl − ] = 1.77 x 10 − 7 M precipitating Pb 2+ begins when [Cl − ] = 1.08 x 10 − 1 M when Pb 2+ just begins to precipitate out, the [Ag + ] has dropped from 0.0010 M to 1.6 x 10 − 9 M AgCl (s)  Ag + (aq) + Cl − (aq) Tro: Chemistry: A Molecular Approach, 2/e

101. 101 Qualitative Analysis An analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme – wet chemistry A sample containing several ions is subjected to the addition of several precipitating agents Addition of each reagent causes one of the ions present to precipitate out Tro: Chemistry: A Molecular Approach, 2/e

102. 102 Qualitative Analysis Tro: Chemistry: A Molecular Approach, 2/e

103. 103 Tro: Chemistry: A Molecular Approach, 2/e

104. 104 Group 1 Group one cations are Ag +, Pb 2+ and Hg 2 2+ All these cations form compounds with Cl − that are insoluble in water – as long as the concentration is large enough – PbCl 2 may be borderline molar solubility of PbCl 2 = 1.43 x 10 −2 M Precipitated by the addition of HCl Tro: Chemistry: A Molecular Approach, 2/e

105. 105 Group 2 Group two cations are Cd 2+, Cu 2+, Bi 3+, Sn 4+, As 3+, Pb 2+, Sb 3+, and Hg 2+ All these cations form compounds with HS − and S 2− that are insoluble in water at low pH Precipitated by the addition of H 2 S in HCl Tro: Chemistry: A Molecular Approach, 2/e

106. 106 Group 3 Group three cations are Fe 2+, Co 2+, Zn 2+, Mn 2+, Ni 2+ precipitated as sulfides; as well as Cr 3+, Fe 3+, and Al 3+ precipitated as hydroxides All these cations form compounds with S 2− that are insoluble in water at high pH Precipitated by the addition of H 2 S in NaOH Tro: Chemistry: A Molecular Approach, 2/e

107. 107 Group 4 Group four cations are Mg 2+, Ca 2+, Ba 2+ All these cations form compounds with PO 4 3− that are insoluble in water at high pH Precipitated by the addition of (NH 4 ) 2 HPO 4 Tro: Chemistry: A Molecular Approach, 2/e

108. 108 Group 5 Group five cations are Na +, K +, NH 4 + All these cations form compounds that are soluble in water – they do not precipitate Identified by the color of their flame Tro: Chemistry: A Molecular Approach, 2/e

109. Complex Ion Formation Transition metals tend to be good Lewis acids They often bond to one or more H 2 O molecules to form a hydrated ion – H 2 O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag + (aq) + 2 H 2 O(l)  Ag(H 2 O) 2 + (aq) Ions that form by combining a cation with several anions or neutral molecules are called complex ions – e.g., Ag(H 2 O) 2 + The attached ions or molecules are called ligands – e.g., H 2 O 109 Tro: Chemistry: A Molecular Approach, 2/e

110. Complex Ion Equilibria If a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand Ag(H 2 O) 2 + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) + 2 H 2 O (l) – generally H 2 O is not included, because its complex ion is always present in aqueous solution Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) 110 Tro: Chemistry: A Molecular Approach, 2/e

111. Formation Constant The reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) The equilibrium constant for the formation reaction is called the formation constant, K f 111 Tro: Chemistry: A Molecular Approach, 2/e

112. Formation Constants 112 Tro: Chemistry: A Molecular Approach, 2/e

113. Example 16.15: 200.0 mL of 1.5 x 10 −3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Write the formation reaction and K f expression. Look up K f value determine the concentration of ions in the diluted solutions Cu 2+ (aq) + 4 NH 3 (aq)  Cu(NH 3 ) 2 2+ (aq) 113 Tro: Chemistry: A Molecular Approach, 2/e

114. Example 16.15: 200.0 mL of 1.5 x 10 −3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Create an ICE table. Because K f is large, assume all the Cu 2+ is converted into complex ion, then the system returns to equilibrium. [Cu 2+ ][NH 3 ][Cu(NH 3 ) 2 2+ ] initial6.7E-40.110 change≈−6.7E-4≈−4(6.7E-4)≈+6.7E-4 equilibriumx0.116.7E-4 Cu 2+ (aq) + 4 NH 3 (aq)  Cu(NH 3 ) 2 2+ (aq) 114 Tro: Chemistry: A Molecular Approach, 2/e

115. Example 16.15: 200.0 mL of 1.5 x 10 -3 M Cu(NO 3 ) 2 is mixed with 250.0 mL of 0.20 M NH 3. What is the [Cu 2+ ] at equilibrium? Cu 2+ (aq) + 4 NH 3 (aq)  Cu(NH 3 ) 2 2+ (aq) substitute in and solve for x confirm the “x is small” approxima- tion [Cu 2+ ][NH 3 ][Cu(NH 3 ) 2 2+ ] initial6.7E-40.110 change≈−6.7E-4≈−4(6.7E-4)≈+6.7E-4 equilibriumx0.116.7E-4 2.7 x 10 −13 << 6.7 x 10 −4, so the approximation is valid 115 Tro: Chemistry: A Molecular Approach, 2/e

117. Practice – What is [HgI 4 2− ] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) 117 Tro: Chemistry: A Molecular Approach, 2/e

118. Practice – What is [HgI 4 2− ] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) Write the formation reaction and K f expression. Look up K f value determine the concentration of ions in the diluted solutions Hg 2+ (aq) + 4 I − (aq)  HgI 4 2− (aq) 118 Tro: Chemistry: A Molecular Approach, 2/e

119. Practice – What is [HgI 4 2− ] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) Create an ICE table. Because K f is large, assume all the lim. rgt. is converted into complex ion, then the system returns to equilibrium. [Hg 2+ ][I − ][HgI 4 2− ] initial3.75E-46.25E-40 change ≈¼ ( −6.25E-4)≈−(6.25E-4)≈¼ ( +6.25E-4) equilibrium2.19E-4x1.56E-4 Hg 2+ (aq) + 4 I − (aq)  HgI 4 2− (aq) I − is the limiting reagent 119 Tro: Chemistry: A Molecular Approach, 2/e

120. Practice – What is [HgI 4 2− ] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl 2 ? 4 KI(aq) + HgCl 2 (aq)  2 KCl(aq) + K 2 HgI 4 (aq) substitute in and solve for x confirm the “x is small” approximation 2 x 10 −8 << 1.6 x 10 −4, so the approximation is valid Hg 2+ (aq) + 4 I − (aq)  HgI 4 2− (aq) [HgI 4 2− ] = 1.6 x 10 −4 [Hg 2+ ][I − ][HgI 4 2− ] initial3.75E-46.25E-40 change ≈¼ ( −6.25E-4)≈−(6.25E-4)≈¼ ( +6.25E-4) equilibrium2.19E-4x1.56E-4 120 Tro: Chemistry: A Molecular Approach, 2/e

122. 122 The Effect of Complex Ion Formation on Solubility The solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands AgCl (s)  Ag + (aq) + Cl − (aq) K sp = 1.77 x 10 −10 Ag + (aq) + 2 NH 3(aq)  Ag(NH 3 ) 2 + (aq) K f = 1.7 x 10 7 Adding NH 3 to a solution in equilibrium with AgCl (s) increases the solubility of Ag + Tro: Chemistry: A Molecular Approach, 2/e

123. 123 Tro: Chemistry: A Molecular Approach, 2/e

124. 124 Solubility of Amphoteric Metal Hydroxides Many metal hydroxides are insoluble All metal hydroxides become more soluble in acidic solution – shifting the equilibrium to the right by removing OH − Some metal hydroxides also become more soluble in basic solution – acting as a Lewis base forming a complex ion Substances that behave as both an acid and base are said to be amphoteric Some cations that form amphoteric hydroxides include Al 3+, Cr 3+, Zn 2+, Pb 2+, and Sb 2+ Tro: Chemistry: A Molecular Approach, 2/e

125. 125 Al 3+ Al 3+ is hydrated in water to form an acidic solution Al(H 2 O) 6 3+ (aq) + H 2 O (l)  Al(H 2 O) 5 (OH) 2+ (aq) + H 3 O + (aq) Addition of OH − drives the equilibrium to the right and continues to remove H from the molecules Al(H 2 O) 5 (OH) 2+ (aq) + OH − (aq)  Al(H 2 O) 4 (OH) 2 + (aq) + H 2 O (l) Al(H 2 O) 4 (OH) 2 + (aq) + OH − (aq)  Al(H 2 O) 3 (OH) 3(s) + H 2 O (l) Tro: Chemistry: A Molecular Approach, 2/e

126. 126 Tro: Chemistry: A Molecular Approach, 2/e