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Chapter 16 Reactions Between Acids and Bases

Chapter 16 Reactions Between Acids and Bases. Titration of Strong Acids and Bases. Titration : a method used to determine the concentration of a substance known as the analyte by adding another substance, the titrant , which reacts in a known manner with the analyte.

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Chapter 16 Reactions Between Acids and Bases

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  1. Chapter 16Reactions Between Acids and Bases

  2. Titration of Strong Acids and Bases • Titration: a method used to determine the concentration of a substance known as the analyte by adding another substance, the titrant, which reacts in a known manner with the analyte. • analyte + titrant → products

  3. Laboratory Titrations (a) A known volume of acid is measured into a flask. (b) Standard base is added from a buret. (c) The endpoint is indicated by a color change. (d) The volume of base is recorded.

  4. Titration: Strong Acid and Base • Titration Curve: a graph of pH of a solution as titrant is added. • For a titration of a strong acid with a strong base, the pH will start at a very low value and stay low as long as strong acid is still present.

  5. Titration: Strong Acid and Base • The pH will rise sharply to 7 at the equivalence point, where the acid and base are present in stoichiometrically equivalent amounts. • After excess strong base has been added, the pH levels off at a high value.

  6. Titration Curve for a Strong Acid with a Strong Base

  7. The Equivalence Point in a Titration • Calculate the equivalence point in the titration of 20.00 mL of 0.1252 MHCl with 0.1008 MNaOH. HCl + NaOH H2O + NaCl

  8. Test Your Skill • Calculate the equivalence point in the titration of 40.00 mL of 0.2387 M HNO3 with 0.3255 M NaOH.

  9. Units of Millimoles • Millimole: one thousandth of a mole. • If molarity is expressed in moles/liter (M) and volume in milliliters (mL), n will be in millimoles (mmol). . Liters cancel but the milli- multiplier remains.

  10. Calculating a Titration Curve • Calculate the pH in the titration of 20.0 mL of 0.125 M HCl with 0.250 M NaOH after 0, 2.00, 10.00, and 20.00 mL base are added.

  11. Test Your Skill • Calculate the pH in the titration of 20.0 mL of 0.125 M HCl with 0.250 M NaOH after 5.00 mL and 12.00 mL base are added.

  12. Calculating a Titration Curve

  13. Strong Base + Strong Acid Curve • Titration curve of 50.00 mL 0.500 M KOH with 1.00 M HCl.

  14. Stoichiometry and Titration Curves • 10 mL of two different 0.100 M acids titrated with 0.100 M NaOH.

  15. Estimating the pH of Mixtures • Fill in first 3 three lines of sRf table. • Look at the final solution (f-line). • If a strong acid is present, the solution will be strongly acidic. • If a strong base is present the solution will be strongly basic. • If only water is present, the solution will be neutral.

  16. Buffers • Buffer: a solution that resists changes in pH. • A buffer is a mixture of a weak acid or base and its conjugate partner. • HA + OH-→ H2O + A- Weak acid reacts with any added OH-. • A-+ H3O → HA + H2O Weak base reacts with any added H3O+.

  17. The pH of a Buffer System • For the chemical reaction HA + H2O → H3O+ + A-

  18. The pH of a Buffer System • Calculate the pH of a solution of 0.50 M HCN and 0.20 M NaCN, Ka = 4.9 x 10-10.

  19. The pH of a Buffer System • Calculate the pH of a solution of 0.40 M NH3 and 0.10 M NH4Cl. For NH3Kb = 1.8 x 10-5.

  20. Test Your Skill • Calculate the pH of a buffer that is 0.25 M HCN and 0.15 M NaCN, Ka = 4.9 x 10-10 .

  21. The Composition of a pH Buffer • Calculate the amount of sodium acetate that must be added to 250 mL of 0.16 M acetic acid in order to prepare a pH 4.68 buffer. Ka = 1.8 x 10-5

  22. Test Your Skill • How many moles of NaCN should be added to 100 mL of 0.25 M HCN to prepare a buffer with pH = 9.40? Ka = 4.9 x 10-10.

  23. Determining the Response of a Buffer to Added Acid or Base • Calculate the initial and final pH when 10 mL of 0.100 M HCl is added to (a) 100 mL of water, and (b) 100 mL of a buffer which is 1.50 M CH3COOH and 1.20 M CH3COONa.

  24. Test Your Skill • Calculate the final pH when 10 mL of 0.100 M NaOH is added to 100 mL of a buffer which is 1.50 M CH3COOH and 1.25 M CH3COONa.

  25. Qualitative Aspects: Titration: Weak Acid + Strong Base Before any base added. • Part way to equivalence point. • Equivalence point. • Beyond equivalence point.

  26. Titration: Weak Acid + Strong Base • HA + OH- A- + H2O • (a) Before any base is added the solution is a weak acid has a low pH. • Estimated pH = 3 • pH 2-4 is typical • Depends on the concentration of the acid. • Depends on the value of Ka.

  27. Titration: Weak Acid + Strong Base • HA + OH- A- + H2O • (b) After some base is added, but before the equivalence point is reached. • The solution is a mixture of the weak acid HA and its conjugate base A-; therefore, the solution is a buffer. • The estimated pH is equal to pKa.

  28. Titration: Weak Acid + Strong Base • HA + OH- A- + H2O • (c) At the equivalence point the solution is salt of A-, all the HA having been consumed by the stoichiometric amount of OH-. • A- is the weak conjugate base of HA. • The estimated pH is 10.

  29. Titration: Weak Acid + Strong Base • HA + OH- A- + H2O • (d) After excess strong base is added OH- is in excess. • The estimated pH is 13.

  30. pH Estimates

  31. Titration: Weak Acid + Strong Base

  32. Titration Curves for Acids of Different Strengths

  33. Calculating the Titration Curve for a Weak Acid • Calculate the pH in the titration of 20.00 mL of 0.500 M formic acid (HCOOH Ka=1.8 x 10-4) with 0.500 M NaOH after 0, 10.00, 20.00, and 30.00 mL of base have been added. • The titration reaction is HCOOH + OH- HCOO- + H2O

  34. Titration of 25.00 mL of 0.500 M Formic Acid with 0.500 M NaOH

  35. Test Your Skill • Calculate the pH in the titration of 12.00 mL of 0.100 M HOCl with 0.200 M NaOH after 0, 3.00, 6.00, and 9.00 mL of base have been added.

  36. Titration of 20.00 mL of 0.500 M Methylamine with 0.500 M HCl

  37. pH Indicators • Indicator: a substance that changes color at the endpoint of a titration. • pH indicators are weak acids or bases whose conjugate species are a different color.

  38. pH Indicators • HIn + H2O ⇌ H3O+ + In- • pKIn = -log(KIn)

  39. pH Indicators • When pH is lower than pKIn, the indicator will be in the acid form. • When pH is greater than pKIn, the indicator will be in the base form. • An indicator should be chosen which changes at or just beyond the equivalence point.

  40. Properties of Indicators * Thymol blue is polyprotic and has three color forms.

  41. Titration Curves for Strong and Weak Acids

  42. Polyprotic Acids • Polyprotic acids provide more than one proton when they ionize. • Polyprotic acids ionize in a stepwise manner. H2A + H2O ⇌ H3O+ + HA- Step 1 HA- + H2O ⇌ H3O+ + A2- Step 2

  43. Polyprotic Acids • There is a separate acid ionization constant for each step H2A + H2O ⇌ H3O+ + HA- Step 1 HA- + H2O ⇌ H3O+ + A2- Step 2

  44. Polyprotic Acids • HA- is the conjugate base of H2A, so it is a weaker acid than H2A. • Ka1 is always larger than Ka2. • For triprotic acids (such as H3PO4), Ka2 is always larger than Ka3.

  45. Calculating Concentrations of Species in Polyprotic Acid Solutions • When successive Ka values differ by a factor of 1000 or more, each step can be assumed to be essentially unaffected by the occurrence of the subsequent step.

  46. Concentrations of Species in Polyprotic Acid Solutions • Calculate the concentrations of all species in 0.250 M malonic acid, Ka = 1.6 x 10-2. • Consider the first ionization and solve by usual approach. H2C3H2O4 + H2O ⇌ HC3H2O4- + H3O+

  47. Concentrations of Species in Polyprotic Acid Solutions • The second step is needed only to calculate the concentration of C3H2O42- because the concentration of H3O+ is determined by the first step. • You can ignore the effect of the second step on the pH because the Ka1 is so much larger than Ka2.

  48. Test Your Skill • Calculate the pH of a 0.040 M solution of ascorbic acid. (Ka1 = 8.0 x 10-5, Ka2 = 1.6 x 10-12)

  49. Amphoteric Species • Amphoteric: having both acidic and basic properties. • Conjugate bases of weak polyprotic acids are amphoteric. • The hydrogen oxalate ion, HC2O4-, is a weak acid (Ka2 = 1.6 ×10-4). • HC2O4- + H2O ⇌ C2O42- + H3O+

  50. Amphoteric Species • Weak Acid • The hydrogen oxalate ion, HC2O4-, is a weak acid. HC2O4- + H2O ⇌ C2O42- + H3O+ • Ka2 = 1.6 x 10-4 • Weak Base • The hydrogen oxalate ion, HC2O4-, can also act as a weak base. HC2O4- + H2O → H2C2O4 + OH- • Kb = Kw/Ka1 = 1.0 x 10-14 / 5.6 x 10-2 • Kb = 1.9 x 10-13 • Since Ka > Kb, the ion will act as a weak acid in water. • When comparing Ka to Kb note that Ka is Ka2 and Kb is Kw/Ka1.

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