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ANALYTICAL CHEMISTRY CHEM 3811 CHAPTER 10

ANALYTICAL CHEMISTRY CHEM 3811 CHAPTER 10. DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university. CHAPTER 10 ACID-BASE TITRATIONS. STRONG ACID – STRONG BASE. - Write balanced chemical equation between titrant and analyte

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ANALYTICAL CHEMISTRY CHEM 3811 CHAPTER 10

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  1. ANALYTICAL CHEMISTRY CHEM 3811CHAPTER 10 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university

  2. CHAPTER 10 ACID-BASE TITRATIONS

  3. STRONG ACID – STRONG BASE - Write balanced chemical equation between titrant and analyte - Calculate composition and pH after each addition of titrant - Construct a graph of pH versus titrant added

  4. STRONG ACID – STRONG BASE - Consider titration of base with acid H+ + OH-→ H2O K = 1/Kw = 1/10-14 = 1014 - Equilibrium constant = 1014 - Reaction goes to completion

  5. STRONG ACID – STRONG BASE H+ + OH-→ H2O At equivalence point moles of titrant = moles of analyte (V titrant)(M titrant) = (V analyte)(M analyte)

  6. STRONG ACID – STRONG BASE Consider 50.00 mL of 0.100 M NaOH with 0.100 M HCl - Three regions of titration curve exists - Before the equivalence point where pH is determined by excess OH- in the solution - At the equivalence point where pH is determined by dissociation of water (H+≈ OH-) - After the equivalence point where pH is determined by excess H+ in the solution

  7. STRONG ACID – STRONG BASE First calculate the volume of HCl needed to reach the equivalence point (V HCl)(0.100 M) = (50.00 mL)(0.100 M) Volume HCl = 50.00 mL

  8. STRONG ACID – STRONG BASE Before the equivalence point Initial amount of analyte (NaOH) = 50.00 mL x 0.100 M = 5.00 mmol After adding 1.00 mL of HCl mmol H+ added = mmol OH- consumed mmol H+ = (1.00 mL)(0.100 M) = 0.100 mmol mmol OH- remaining = 5.00 – 0.100 = 4.90 mmol

  9. STRONG ACID – STRONG BASE Before the equivalence point Total volume = 50.00 mL + 1.00 mL = 51.00 mL [OH-] = 4.90 mmol/51.00 mL = 0.0961 M pOH = - log(0.0961) = 1.017 pH = 14.000 - 1.017 = 12.983

  10. STRONG ACID – STRONG BASE - Repeat calculations for all volumes added - Increments can be large initally but must be reduced just before and just after the equivalence point (around 50.00 mL in this case) - Sudden change in pH occurs near the equivalence point - Greatest slope at the equivalence point

  11. STRONG ACID – STRONG BASE At the equivalence point - pH is determined by the dissociation of water H2O ↔ H+ + OH- Kw = x2 = 1.0 x 10-14 x = 1.0 x 10-7 pH = 7.00 (at 25 oC)

  12. STRONG ACID – STRONG BASE After the equivalence point - Excess H+ is present After adding 51.00 mL of HCl Excess HCl present = 51.00 – 50.00 = 1.00 mL Excess H+ = (1.00 mL)(0.100 M) = 0.100 mmol Total volume of solution = 50.00 + 51.00 = 101.00 mL [H+] = 0.100 mmol/101.00 mL = 9.90 x 10-4 M pH = -log(9.90 x 10-4) = 3.004

  13. STRONG ACID – STRONG BASE pH Equivalence point (maximum slope or point of inflection) 7 50.00 Volume of HCl added (mL)

  14. TITRATION CURVE pH 7 Equivalence point (maximum slope or point of inflection) 50.00 Volume of NaOH added (mL)

  15. STOICHIOMETRY AND TITRATION CURVE Compare titration of HCl with NaOH and H2SO4 with NaOH (same volume and same concentration of acid) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) Net ionic equation in both cases: H+(aq) + OH-(aq) → H2O(l) or H3O+(aq) + OH-(aq) → 2H2O(l) 1 mol HCl conctributes 1 mol H3O+ 1 mol H2SO4 contributes 2 mol H3O+

  16. STOICHIOMETRY AND TITRATION CURVE pH HCl H2SO4 Volume of NaOH added (mL)

  17. WEAK ACID – STRONG BASE Consider 50.00 mL 0f 0.0100 M acetic acid with 0.100 M NaOH pKa of acetic acid = 4.76 HC2H3O2 + OH-→ C2H3O2- + H2O For the reverse reaction Kb = Kw/Ka = 5.8 x 10-10 Equilibrium constant = 1/Kb = 1.7 x 109 So large that we can assume the reaction goes to completion

  18. WEAK ACID – STRONG BASE Determine volume of base at equivalence point mmol HC2H3O2≈ mmol OH- (V NaOH)(0.100 M) = (50.00 mL)(0.0100 M) Volume NaOH = 5.00 mL

  19. WEAK ACID – STRONG BASE Four types of calculations to be considered Before OH- is added - pH is determined by equilibrium of weak acid HA ↔ H+ + A- x = 4.1 x 10-4 pH = 3.39

  20. WEAK ACID – STRONG BASE Before equivalence point By adding OH- a buffer solution of HA and A- is formed After adding 0.100 mL OH- HA + OH-→ A- + H2O Initial mmol 0.500 0.0100 0 Final mmol 0.490 0 0.0100

  21. WEAK ACID – STRONG BASE Before equivalence point

  22. WEAK ACID – STRONG BASE Before equivalence point If volume of OH- added is half the volume at equivalence point HA = A- = 0.250 mmol pH = pKa = 4.76

  23. WEAK ACID – STRONG BASE At equivalence point Volume of OH- = 5.00 mL mmol OH- = (5.00 mL)(0.100 M) = 0.500 mmol HA is used up and [HA] = 0 mmol A- = 0.500 mmol

  24. WEAK ACID – STRONG BASE At equivalence point Only A- is present in solution A- + H2O ↔ HA + OH- [A-] = (0.500 mmol)/(50.00 mL + 5.00 mL) = 0.00909 M

  25. WEAK ACID – STRONG BASE At equivalence point Kb = Kw/Ka = 5.8 x 10-10 x = [OH-] = 2.3 x 10-6 pOH = 5.64 pH = 14.00 – 5.64 = 8.36 pH is not 7.00 but greater than 7.00 (pH at equivalence point increases with decreasing strength of acid)

  26. WEAK ACID – STRONG BASE After equivalence point - Strong base (OH-) being added to weak base (A-) - pH is determined by the excess [OH-] (approximation) - After adding 5.10 mL OH- [OH-] = (0.10 mL)(0.100 M)/(50.00 mL + 5.10 mL) = 1.81 x 10-4 pH = 14.00 – pOH = 14.00 – 3.74 = 10.26

  27. WEAK ACID – STRONG BASE Minimum slope pH 8.36 pH = pKa Equivalence point (maximum slope or point of inflection) pKa 2.50 5.00 Volume of NaOH added (mL)

  28. STRONG ACID – WEAK BASE - The reverse of weak acid and strong base B + H+→ BH+ - Similarly assume reaction goes to completion

  29. STRONG ACID – WEAK BASE Consider 50.00 mL of 0.0100 M pyridine with 0.100 M HCl Kb of pyridine = 1.6 x 10-9 Determine volume of acid at equivalence point mmol pyridine ≈ mmol H+ (V HCl)(0.100 M) = (50.00 mL)(0.010 M) Volume HCl = 5.00 mL

  30. STRONG ACID – WEAK BASE Four types of calculations to be considered Before H+ is added - pH is determined by equilibrium of weak base (determined using Kb) B + H2O ↔ BH+ + OH-

  31. STRONG ACID – WEAK BASE Before equivalence point - By adding H+ a buffer solution of B and BH+ is formed - When volume of H+ added = half the volume at equivalent point pH = pKa (for BH+)

  32. STRONG ACID – WEAK BASE At equivalence point - B has been converted into BH+ - B is used up and [B] = 0 pH is calculated by considering BH+ BH+ ↔ B + H+ pH is not 7.00 but less than 7.00

  33. STRONG ACID – WEAK BASE After equivalence point - Strong acid (H+) is being added to weak acid (BH+) pH is determined by the excess [H+] (approximation)

  34. STRONG ACID – WEAK BASE Minimum slope pKa pH = pKa pH Equivalence point (maximum slope or point of inflection) 2.50 5.00 Volume of HCl added (mL)

  35. END POINT Use of Indicators - Indicators are acids or bases so a few drops of dilute solutions are used to minimize indicator errors - Acidic color if pH ≤ pKHIn - 1 - Basic color if pH ≥ pKHIn + 1 - A mixture of both colors if pKHIn - 1 ≤ pH ≤ pKHIn + 1 - Use an indicator whose transition range overlaps the steepest part of the titration curve

  36. END POINT Use of pH Electrodes - The end point is where the slope of the curve is greatest - The end point is the volume at which the first derivative of a titration curve is maximum - The end point is the volume at which the second derivative of a titration curve is zero

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