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Acid-base titration

Acid-base titration. 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

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Acid-base titration

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  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 H3O+ = moles of OH−, the titration has reached its equivalence point Tro: Chemistry: A Molecular Approach, 2/e

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

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

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

  7. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH added 5.0 mL NaOH 5.0 x 10−4 mole NaOH added HCl(aq) + NaOH(aq) NaCl(aq) + H2O(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 Tro: Chemistry: A Molecular Approach, 2/e 7

  8. Continued...

  9. Table

  10. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH • HCl(aq) + NaOH(aq) NaCl(aq) + H2O(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 Tro: Chemistry: A Molecular Approach, 2/e

  11. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH • HCl(aq) + NaOH(aq) NaCl(aq) + H2O(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 2.5 x 10−3 mole NaOH added Tro: Chemistry: A Molecular Approach, 2/e

  12. Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH • HCl(aq) + NaOH(aq) NaCl(aq) + H2O(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

  13. 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 Adding 0.100 M NaOH to 0.100 M HCl added 40.0 mL NaOH 0.00150 mol NaOH pH = 12.36 added 15.0 mL NaOH 0.00100 mol HCl pH = 1.60 added 50.0 mL NaOH 0.00250 mol NaOH pH = 12.52 added 20.0 mL NaOH 0.00050 mol HCl pH = 1.95 Tro: Chemistry: A Molecular Approach, 2/e

  14. 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 HNO3 Tro: Chemistry: A Molecular Approach, 2/e

  15. 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 HNO3 • HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l) • Initial pH = −log(0.250) = 0.60 • Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2 • Before equivalence point added 10.0 mL NaOH Tro: Chemistry: A Molecular Approach, 2/e

  16. Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M HNO3 to reach equivalence Tro: Chemistry: A Molecular Approach, 2/e

  17. Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M HNO3 to reach equivalence • HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l) • Initial pH = −log(0.250) = 0.60 • Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2 • At equivalence point: moles of NaOH = 1.25 x 10−2 Tro: Chemistry: A Molecular Approach, 2/e

  18. 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 HNO3 Tro: Chemistry: A Molecular Approach, 2/e

  19. 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 HNO3 • HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l) • Initial pH = −log(0.250) = 0.60 • Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2 • After equivalence point added 100.0 mL NaOH

  20. 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 HNO3 • HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l) • Initial pH = −log(0.250) = 0.60 • initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2 • After equivalence point added 100.0 mL NaOH Tro: Chemistry: A Molecular Approach, 2/e

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

  22. 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 Ka 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 Kb 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 Tro: Chemistry: A Molecular Approach, 2/e

  23. Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH • HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq) • Initial pH Ka = 1.8 x 10−4 Tro: Chemistry: A Molecular Approach, 2/e

  24. Titration of 25 mL of 0.100 M HCHO2 with 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 • Before equivalence added 5.0 mL NaOH Tro: Chemistry: A Molecular Approach, 2/e

  25. Titration of 25 mL of 0.100 M HCHO2 with 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 [OH−] = 1.7 x 10−6 M

  26. Kb = 5.6 x 10−11

  27. Titration of 25 mL of 0.100 M HCHO2 with 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 • After equivalence added 30.0 mL NaOH 3.0 x 10−3 mole NaOH added 35

  28. 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 added 5.0 mL NaOH 0.00200 mol HCHO2 pH = 3.14 added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96 added 10.0 mL NaOH 0.00150 mol HCHO2 pH = 3.56 initial HCHO2 solution 0.00250 mol HCHO2 pH = 2.37 added 35.0 mL NaOH 0.00100 mol NaOH xs pH = 12.22 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 Tro: Chemistry: A Molecular Approach, 2/e

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

  30. 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 • [H3O+][OH−]=1 x 10−14 Tro: Chemistry: A Molecular Approach, 2/e

  31. Example 16.7a: 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 Tro: Chemistry: A Molecular Approach, 2/e

  32. Example 16.7b: 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 Tro: Chemistry: A Molecular Approach, 2/e

  33. Example 16.7b: 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 Tro: Chemistry: A Molecular Approach, 2/e

  34. Example 16.7b: 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 Tro: Chemistry: A Molecular Approach, 2/e

  35. Example 16.7b: 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 Tro: Chemistry: A Molecular Approach, 2/e

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

  37. Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the initial pH of the NH3 solution. Tro: Chemistry: A Molecular Approach, 2/e

  38. Practice – Titration of 25.0 mL of 0.10 M NH3 with 0.10 M HCl. Calculate the initial pH of the NH3(aq) pKb = 4.75 Kb = 10−4.75 = 1.8 x 10−5 NH3(aq) + HCl(aq)  NH4Cl(aq) Initial: NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq) Tro: Chemistry: A Molecular Approach, 2/e

  39. Practice – Titration of 25.0 mL of 0.10 M NH3 with 0.10 M HCl. Calculate the initial pH of the NH3(aq) pKb = 4.75 Kb = 10−4.75 = 1.8 x 10−5 NH3(aq) + HCl(aq)  NH4Cl(aq) Initial: NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq) Tro: Chemistry: A Molecular Approach, 2/e

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