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Chapter 15: Applications of Aqueous Equilibria - Titrations

GENERAL CHEMISTRY: ATOMS FIRST. Chapter 15: Applications of Aqueous Equilibria - Titrations. John E. McMurray – Robert C. Fay. Prentice Hall. Titration.

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Chapter 15: Applications of Aqueous Equilibria - Titrations

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  1. GENERAL CHEMISTRY: ATOMS FIRST Chapter 15: Applications of Aqueous Equilibria - Titrations John E. McMurray – Robert C. Fay Prentice Hall

  2. Titration • in an acid-base titration, a solution of known concentration (titrant) is slowly added from a burette to a solution of unknown concentration in a flask 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 H3O+ = moles of OH−, the titration has reached its equivalence point

  3. Titration

  4. The Titration Curve • is 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 unknown 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 known solution in the flask (from the burette) is in excess, so the pH approaches its pH

  5. Titration Curve:Known Strong Base Added to a Strong Acid

  6. Strong Acid/Strong BaseTitration of 25 mL of 0.100 M HCl with 0.100 M NaOH HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq) • initial pH = -log(0.100) = 1.00 • initial moles HCl = (0.0250 L)(0.100 mol/L) = 2.50 x 10-3 mol HCl • now add 5.0 mL (0.0050L) NaOH

  7. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH

  8. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH • after 30 mL (0.0300L) NaOH added, after the equivalence point: • when 25 mL NaOH added you reach the equivalence point • no HCl and no NaOH present in flask, pH = 7.00

  9. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH [OH-] = 9.09 x 10-3

  10. added 5.0 mL NaOH 0.00200 mol HCl pH = 1.18 25.0 mL 0.100 M HCl 0.00250 mol HCl pH = 1.00 added 30.0 mL NaOH 0.00050 mol NaOH pH = 11.96 Adding NaOH to HCl added 35.0 mL NaOH 0.00100 mol NaOH pH = 12.22 added 10.0 mL NaOH 0.00150 mol HCl pH = 1.37 added 15.0 mL NaOH 0.00100 mol HCl pH = 1.60 added 40.0 mL NaOH 0.00150 mol NaOH pH = 12.36 added 20.0 mL NaOH 0.00050 mol HCl pH = 1.95 added 50.0 mL NaOH 0.00250 mol NaOH pH = 12.52 added 25.0 mL NaOH equivalence point pH = 7.00

  11. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH • HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(l) • after equivalence point

  12. 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 HAinit and mol A−init using reaction stoichiometry • calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−init • half-neutralization pH = pKa

  13. 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 like Ex 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 • [H3O+][OH−]=1 x 10-14

  14. Titration of a weak acid, 25 mL of 0.100 M HCHO2,with a strong base, 0.100 M NaOH HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(l) Ka = 1.8 x 10-4 = [CHO2-] [H3O+] [HCHO2]

  15. Titration of a weak acid, 25 mL of 0.100 M HCHO2,with a strong base, 0.100 M NaOH HCHO2(aq) + H2O(l) CHO2-(aq) + H3O+(aq) Ka = 1.8 x 10-4

  16. Titration of a weak acid, 25 mL of 0.100 M HCHO2,with a strong base, 0.100 M NaOH HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(l) • initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3 add 5.0 mL NaOH given Ka = 1.8 x 10-4

  17. Titration of a weak acid, 25 mL of 0.100 M HCHO2,with a strong base, 0.100 M NaOH • HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq) • initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3 • at equivalence CHO2−(aq) + H2O(l) HCHO2(aq) + OH−(aq) added 25.0 mL NaOH

  18. Titration of a weak acid, 25 mL of 0.100 M HCHO2,with a strong base, 0.100 M NaOH** • HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq) • initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10-3 • at equivalence CHO2−(aq) + H2O(l) HCHO2(aq) + OH−(aq) Kb = 5.6 x 10-11 [OH-] = 1.7 x 10-6 M

  19. added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96 added 35.0 mL NaOH 0.00100 mol NaOH xs pH = 12.22 initial HCHO2 solution 0.00250 mol HCHO2 pH = 2.37 added 5.0 mL NaOH 0.00200 mol HCHO2 pH = 3.14 added 10.0 mL NaOH 0.00150 mol HCHO2 pH = 3.56 added 25.0 mL NaOH equivalence point 0.00250 mol CHO2− [CHO2−]init = 0.0500 M [OH−]eq = 1.7 x 10-6 pH = 8.23 Adding NaOH to HCHO2 added 12.5 mL NaOH 0.00125 mol HCHO2 pH = 3.74 = pKa half-neutralization added 40.0 mL NaOH 0.00150 mol NaOH xs pH = 12.36 added 15.0 mL NaOH 0.00100 mol HCHO2 pH = 3.92 added 50.0 mL NaOH 0.00250 mol NaOH xs pH = 12.52 added 20.0 mL NaOH 0.00050 mol HCHO2 pH = 4.34

  20. A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the volume of KOH at the equivalence point HNO2 + KOH  NO2 + H2O

  21. A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH HNO2 + KOH  NO2 + H2O 0.00100 0.00300

  22. A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH. HNO2 + H2O  NO2 + H3O+ Table 15.5 Ka = 4.6 x 10-4

  23. A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point HNO2 + KOH  NO2 + H2O at half-equivalence, moles KOH = ½ mole HNO2 0.00200 0.00200

  24. A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point. HNO2 + H2O  NO2 + H3O+ Table 15.5 Ka = 4.6 x 10-4

  25. Titration Curve of a Weak Base with a Strong Acid

  26. Titration of a Polyprotic Acid • if Ka1 >> Ka2, there will be two equivalence points in the titration • the closer the Ka’s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH

  27. 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 [H3O+] • using a probe that specifically measures just H3O+ • 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 anindicator

  28. Monitoring pH During a Titration

  29. Monitoring a Titration with an Indicator • for most titrations, the titration curve shows a very large change in pH for very small additions of base 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 • pKa of H-Indicator ≈ pH at equivalence point

  30. Indicators • many dyes change color depending on the pH of the solution • these dyes are weak acids, establishing an equilibrium with the H2O and H3O+ in the solution H-Ind(aq) + H2O(l) Ind(aq) + H3O+(aq) • the color of the solution depends on the relative concentrations of Ind:HInd • when Ind:H-Ind ≈ 1, the color will be mix of the colors of Ind and HInd • when Ind:H-Ind > 10, the color will be mix of the colors of Ind • when Ind:H-Ind < 0.1, the color will be mix of the colors of H-Ind

  31. Phenolphthalein

  32. Methyl Red

  33. Acid-Base Indicators

  34. GENERAL CHEMISTRY: ATOMS FIRST Chapter 15: Applications of Aqueous Equilibria-Buffers John E. McMurray – Robert C. Fay Prentice Hall

  35. Buffers • buffers are solutions that resist changes in pH when an acid or base is added • they act by neutralizing the added acid or base • there is a limit to their neutralizing ability, eventually the pH changes • Solution made by mixing a weak acid with a soluble salt containing its conjugate base anion

  36. Formation of an Acid Buffer

  37. How Acid Buffers WorkHA(aq) + H2O(l) A−(aq) + H3O+(aq) • buffers follow Le Châtelier’s Principle • buffers contain significant amount of weak acid, HA • The HA molecules react with added base to neutralize it • the H3O+ combines with OH− to make H2O • H3O+ is then replaced by the shifting equilibrium • buffer solutions also contain significant amount of conjugate base anion, A− • The A− molecules react with added acid to make more HA and keep H3O+ constant

  38. H2O How Buffers Work new HA HA HA A− A−  H3O+ + Added H3O+

  39. H2O How Buffers Work new A− A− HA A− HA  H3O+ + Added HO−

  40. Common Ion Effect HA(aq) + H2O(l) A−(aq) + H3O+(aq) • adding a salt, NaA, containing the acid anion, shifts the position of equilibrium to the left (A− is the conjugate base of the acid) • this lowers the H3O+ ion concentration and causes the pH to be higher Le Châtelier’s Principle

  41. Common Ion Effect

  42. What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10-5 H2O + HC2H3O2 C2H3O2 + H3O+

  43. What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10-5 H2O + HC2H3O2 C2H3O2 + H3O+ x +x +x x 0.100 x 0.100 + x

  44. What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10-5 0.100 x 0.100 +x

  45. What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10-5 x = 1.8 x 10-5 the approximation is valid

  46. What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10-5 0.100 x 0.100 + x x x = 1.8 x 10-5

  47. What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10-5

  48. What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10-5 the values match

  49. What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10-4 H2O + HF  F + H3O+

  50. What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF? Ka for HF = 7.0 x 10-4 H2O + HF  F + H3O+ x +x +x x 0.14 x 0.071 + x

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