1 / 55

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.

katen
Download Presentation

Chapter 16 Reactions Between Acids and Bases

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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.

More Related