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Titrations

Titrations. An acid-base titration is a volumetric analysis in which a solution of one reactant (acid or base) is gradually added to a solution of another reactant (base or acid). One way to describe acid-base titrations is to calculate the [H + ] or pH as a function of the

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Titrations

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  1. Titrations An acid-base titration is a volumetric analysis in which a solution of one reactant (acid or base) is gradually added to a solution of another reactant (base or acid). One way to describe acid-base titrations is to calculate the [H+] or pH as a function of the volume of titrant added. When equivalent amounts of the two reactants have been mixed, na = nb, and this point is known as the equivalence point.

  2. Neutralization Reactions • A neutralization reaction is a double • replacement reaction that occurs between an • acid and a base to form a salt and water. • All ionizable hydrogens from the acid react • with all of the hydroxides from the base. • H+(aq) + OH-(aq) → H2O(l)

  3. Equal chemical equivalent amounts of • sulfuric acid and potassium hydroxide • produce salt and water. • H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l) • 2H+(aq) + SO42-(aq) + 2K+(aq) + 2OH-(aq) → 2K+(aq) + SO42-(aq) + 2H2O(l) • H+(aq) + OH-(aq) → H2O(l)

  4. Neutralization Problem • 25.0 mL of 0.50 M sodium hydroxide is • neutralized by 15.0 mL of sulfuric acid. • Determine the concentration of the sulfuric • acid. • [NaOH] = 0.50 M [H2SO4] = ? • Vb = 25.0 mL Va = 15.0 mL • 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) • H+(aq) + OH-(aq) → H2O(l)

  5. The reason for including the molecular • equation is to illustrate what is involved when • you have a diprotic acid and a monoprotic • base. • The same result is obtained by comparing 1 mol of NaOH to one mole of H2SO4.

  6. 0.50 mol NaOH nb 25.0 mL × × = 1 L 1 mol OH- 1 L • . × 1 mol NaOH 103 mL = 0.0125 mol OH- 1 mol H2SO4 na = 0.0125 mol NaOH × 2 mol NaOH = 6.25 x 10-3 mol H2SO4

  7. 6.25 x 10-3 mol H2SO4 [H2SO4] = • [H2SO4] = n/V 1 L 15.0 mL × 103 mL [H2SO4] = 0.417 M

  8. Strong Acid-Strong Base Titration • (a) Calculate the pH when 24.9 mL • of 0.100 M HI is added to 25.0 mL • of 0.100 M NaOH. • [HI] = 0.100 M [NaOH] = 0.100 M • Va = 24.9 mL Vb = 25.0 mL • HI(aq) H+(aq) + I-(aq) • NaOH(aq) Na+(aq) + OH-(aq) → →

  9. 0.100 mol HI 0.100 mol H+ na × × = 0.100 mol HI L 1 L 1 L 24.9 mL 25.0 mL × × 2.50 x 10-3 mol OH- 2.49 x 10-3 mol H+ = = 103 mL 103 mL • [H+] = n/V 0.100 mol NaOH 0.100 mol OH- nb × × = 0.100 mol NaOH L

  10. nb 0.00249 0.00250 (mol) • H+(aq) + OH-(aq) → H2O(l) • na 0 1.00 x 10-5 (mol) • [OH-] = n/V 1.00 x 10-5 mol OH- [OH-] = = 2.00 x 10-4 M 1 L 49.9 mL × 103 mL

  11. pOH = -log[OH-] = -log(2.00 x 10-4) = 3.700 • pH + pOH = 14.00 • pH = 14.00 – 3.700 = 10.30

  12. 0.100 mol HI 0.100 mol H+ na × × = 0.100 mol HI L 1 L 25.1 mL × 2.51 x 10-3 mol H+ = 103 mL • (b) Calculate the pH when 25.1 mL of 0.100 M • HI is added to 25.0 mL of 0.100 M NaOH. • [H+] = n/V nb 2.50 x 10-3 mol OH- =

  13. nb 0.00251 0.00250 (mol) • H+(aq) + OH-(aq) → H2O(l) • na 1.00 x 10-5 0 (mol) • [H+] = n/V 1.00 x 10-5 mol H+ [H+] = = 2.00 x 10-4 M 1 L 50.1 mL × 103 mL

  14. pH = -log[H+] = -log(2.00 x 10-4) = 3.70 It does not look like a pH of 3.70!

  15. Strong Acid-Strong Base TC • A graph of pH as a function of the volume of • added titrant (a solution of known • concentration which is added to another) is • called a titration curve. • The simplest titration is that of a strong acid • with a strong base. • The following titration curve shows a 50.0 mL sample of 0.10 M HCl being titrated with 0.10 M NaOH.

  16. In the case of a strong acid-strong base titration, the reactant species will be the H+ and OH-. • Remember, the anion of a strong acid and the cation of a strong base are spectator ions. • Spectator ions do not undergo any hydrolysis (reaction with H2O).

  17. pH = 7.00 excess OH- unionized acid

  18. Strong Acid-Strong Base TC • Characteristics of titration curve: • The initial pH is low, ≈ 1.00, which is indicative of a strong acid. • There is a small pH change as the base is initially added. • An equivalent amount of base neutralizes an equivalent amount of acid.

  19. The neutralization of an acid and base produces water which decreases the [H+] and increases the pH. • As you approach the equivalence point the pH changes very rapidly. • The midpoint of the vertical section of the curve is the equivalence point indicating there are equivalents of acid and base.

  20. For a strong acid-base titration, neutralization occurs at a pH = 7 because the [H+] and the [OH-] are equal and form water molecules. H+(aq) + OH-(aq) → H2O(l) Na+ and Cl- do not hydrolyze leaving the solution at a pH = 7. • After the equivalence point, the pH depends on the excess OH- in solution.

  21. Weak Acid-Strong Base Titration • Calculate the pH when 50.0 mL of 1.00 M • LiOH is added to 50.0 mL of 1.00 M • HC2H3O2. • [HC2H3O2] = 1.00 M [LiOH] = 1.00 M • Va = 50.0 mL Vb = 50.0 mL • Kb = 1.8 x 10-5 • LiOH(aq) Li+(aq) + OH-(aq) →

  22. 1.00 mol LiOH 1 mol OH- nb × × = • [OH-] = n/V 1 mol LiOH L 1 L 50.0 mL × 5.00 x 10-2 mol OH- = 103 mL 1.00 mol Ha 1 L 50.0 mL na × × = 103 mL L na 5.00 x 10-2 mol HC2H3O2 =

  23. nb 5.00 x 10-2 5.00 x 10-2 0 (mol) • HC2H3O2(aq) + OH-(aq) → C2H3O2-(aq) + H2O(l) • na 0 0 5.00 x 10-2 (mol) 5.00 x 10-2 mol C2H3O2- [C2H3O2-] = 1 L 100.0 mL × 103 mL [C2H3O2-] = 0.500 M

  24. C2H3O2-(aq) + H2O(l) HC2H3O2(aq) + OH-(aq) • [ ]i 0.5000 0 • [ ]c -x+x+x • [ ]e 0.500-x x x [OH-] [HC2H3O2] Kb = [C2H3O2-] x2 ≈ 1.7 x 10-5 M [OH-] 5.6 x 10-10 = × 0.500 - x

  25. pOH = -log[OH-] = -log(1.75 x 10-5) = 4.76 • pH + pOH = 14.00 • pH = 14.00 – 4.76 = 9.24

  26. Weak Acid-Strong Base TC • A weak acid-strong base titration is more • complicated to analyze due to the buffering • effect. • The following titration curve shows a 50.0 mL sample of 1.00 M HC2H3O2 being titrated with 1.00 M NaOH. • In the case of a weak acid-strong base titration, the reacting species will be HC2H3O2 and OH-.

  27. Remember, the anion of a weak acid is a relatively strong base and the cation of a strong base is a spectator ion. • A spectator ion does not undergo hydrolysis but the anion of a weak base does.

  28. pH of Weak Acid vs Volume of Strong Base pH = 8.10 pH indicative of a weak acid Volume of NaOH (mL)

  29. Weak Acid-Strong Base TC • Characteristics of titration curve: • The initial pH is somewhat higher, > 2.00, which is indicative of a weak acid. • This results from a weak acid not ionizing completely (≈ 5%) which supplies fewer H+ to the solution. • The anion of the weak acid, C2H3O2-, undergoes hydrolysis .

  30. C2H3O2-(aq) + H2O(l) HC2H3O2(aq) + OH-(aq) • This is the cause of the pH rising more rapidly in the early stages when compared to a strong acid-strong base titration. • The combination of the weak acid and the conjugate base of that acid produces a buffering effect.

  31. HC2H3O2(aq) + OH-(aq) → C2H3O2-(aq) + H2O(l) • C2H3O2-(aq) + H2O(l) HC2H3O2(aq) + OH-(aq) • Remember this buffering effect resists changes in pH. • Compared to the strong-acid strong-base titration, there are smaller changes in pH found near the equivalence point.

  32. The acid-base reaction produces both water and salt. • The weaker the acid, the stronger the conjugate base which in turn increases the pH to above 7.00. • After reaching the equivalence point, the titration curve for both the strong acid- strong base and the weak acid-strong base will be identical because the pH depends on the excess OH- in both titrations.

  33. Weak Base-Strong Acid Titration • Calculate the pH when 10.0 mL of 0.100 M • HBr is added to 20.0 mL of 0.100 M NH3. • [HBr] = 0.100 M [NH3] = 0.100 M • Va = 10.0 mL Vb = 20.0 mL • HBr(aq) H+(aq) + Br-(aq) →

  34. 0.100 mol HBr 0.100 mol H+ na × × = • [H+] = n/V 0.100 mol HBr L 1 L 10.0 mL × 1.00 x 10-3 mol H+ = 103 mL 0.100 mol NH3 1 L 20.0 mL nb × × = 103 mL L nb 2.00 x 10-3 mol NH3 =

  35. nb 1.00 x 10-3 2.00 x 10-3 0 (mol) • H+(aq) + NH3(aq) → NH4+(aq) • na 0 mol 1.00 x 10-3 1.00 x 10-3 (mol) 1.00 x 10-3 mol NH3 1.00 x 10-3 mol NH3 1.00 x 10-3 mol NH3 [NH3] = 1 L 30.0 mL × 103 mL [NH3] = 3.33 x 10-2 M

  36. NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) • [ ]i 0.03330.0333 0 • [ ]c -x+x+x • [ ]e 0.0333 - x0.0333 + x x [NH4+] [OH-] Kb = [NH3] 0.0333 - x [H+] 1.8 x 10-5 = × 0.0333 + x ≈ [H+] 1.8 x 10-5 M

  37. Weak Base-Strong Acid TC • A weak base-strong acid titration is more • complicated to analyze due to the buffering • effect. • The following titration curve shows a 50.0 mL sample of 0.10 M NH3 being titrated with 0.10 M HCl. • In the case of a weak base-strong acid titration, the reacting species will be NH3 and H+.

  38. Remember, the conjugate acid of a weak base is a relatively strong acid and the cation of a strong base is a spectator ion. • A spectator ion does not undergo hydrolysis but the conjugate acid of a weak base does.

  39. pH of Weak Base vs Volume of Strong Acid indicative of a weak base pH pH = 5.28 Volume of HCl (mL)

  40. Weak Base-Strong Acid TC • Characteristics of titration curve: • The initial pH is high, ≈ 11.00, which is indicative of a weak base. • This results from a weak base not ionizing completely (≈ 5%) which supplies fewer OH- to the solution. • The added acid neutralizes an equivalent amount of base.

  41. H+(aq) + NH3(aq) → NH4+(aq) • The conjugate acid of the weak base, NH4+, undergoes hydrolysis • NH4+(aq) + H2O(l) H3O+(aq) + NH3(aq) • The combination of the strong acid and the conjugate acid of that base produces a buffering effect. • Remember this buffering effect resists changes in pH.

  42. Compared to the strong acid - strong base titration, there are smaller changes in pH found near the equivalence point. • The midpoint of the vertical section of the curve is the equivalence point indicating there are equivalents of base and acid. • The weaker the base, the stronger the conjugate acid which in turn decreases the pH to below 7.00.

  43. After reaching the equivalence point, the titration curve for both the strong acid- strong base and the weak base-strong acid will be identical because the pH depends on the excess H+ in both titrations.

  44. Acid-Base Color Indicators • An appropriate indicator is a weak acid • whose Ka is close to the concentration of [H+] • or the pH at the equivalence point. • The equivalence point is the point at which • the added solute reacts completely with the • solute present in the solution. • H+(aq) + OH-(aq) → H2O(l) 1.00 mol of a monoprotic acid will completely react with 1.00 mol of a monoprotic base.

  45. Do not confuse the equivalence point with the • endpoint which is the point at which the • indicator changes color. • Knowing the equivalence point helps to • determine the most appropriate indicator. • For the HC2H3O2-NaOH titration the pH ≈ 9.24. • A good choice for indicator would be • phenolphthalein (endpoint ≈ 8-10) or thymol • blue (endpoint ≈ 8-10). • For the HCl-NaOH titration the pH ≈ 7.00. • A good choice for indicator would be • bromthymol blue (endpoint ≈ 6-8).

  46. Diprotic Acid Titration • In the case of a weak diprotic acid (H2SO3) • titrated with a strong base (OH-), the reacting • species are H+, OH-, H2SO3, HSO3-, and SO32-. • For a typical diprotic acid such as • sulfurous acid, H2SO3, you expect to find • two equivalence points. • For 25.0 mL of an equimolar diprotic acid and • strong base, you would expect two • equivalent points at 25.0 and 50.0 mL.

  47. However, the Ka values must be considered. • For H2SO3, the Ka’s are: • Ka1 = 1.50 x 10-2 • Ka2 = 1.0 x 10-7 • If the Ka2 value is extremely small as in the • case of H2S (≈ 10-19), only one equivalence • point will be displayed.

  48. pH of Diprotic Acid (H2SO3) vs Volume of Strong Base E.P. 2 pH E.P. 1 Volume of NaOH (mL)

  49. Triprotic Acid Titration • In the case of a weak triprotic acid (H3PO4) • titrated with a strong base (OH-), the reacting • species are H+, OH-, H3PO4, H2PO4-, HPO42-, • and PO43-. • For a typical triprotic acid such as • phosphoric acid, H3PO4, you expect to find • three equivalence points. • For 25.0 mL of an equimolar triprotic acid and • strong base, you would expect three • equivalent points at 25.0, 50.0, and 75.0 mL.

  50. However, the Ka values must be considered. • For H3PO4, the Ka’s are: • Ka1 = 7.50 x 10-3 • Ka2 = 6.20 x 10-8 • Ka3 = 4.20 x 10-13 • Because of the extremely small Ka3 value, • phosphoric acid displays only two • equivalence points at 25.0 and 50.0 mL. • It is common for polyprotic acids to show only • one or two equivalence points depending on • their Ka values.

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