1 / 129

Chapter 18 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

Chapter 18 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria. More About Chemical Equilibria : Acid–Base & Precipitation Reactions. Stomach Acidity & Acid–Base Reactions. The Common Ion Effect.

makana
Download Presentation

Chapter 18 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

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 18Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria

  2. More About Chemical Equilibria: Acid–Base & Precipitation Reactions

  3. Stomach Acidity & Acid–Base Reactions

  4. The Common Ion Effect • In the previous chapter, you examined the behavior of weak acids and bases in terms of equilibrium involving conjugate pairs. • The pH of a solution was found via Ka or Kb. • What would happen if you started with a solution of acid that was mixed with a solution of its conjugate base? • The change of pH when a significant ammout of conjugate base is present is an example of the “Common Ion Effect”.

  5. The Common Ion Effect What is the effect on the pH of a 0.25M NH3(aq) solution when NH4Cl is added? NH4+ is an ion that is COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the left to reduce the disturbance. This results in a reduciton of the hydroxide ion concentration, which will lower the pH. Hint: NH4+ is an acid!

  6. The Common Ion Effect First let’s find the pH of a 0.25M NH3(aq) Solution:

  7. The Common Ion Effect First let’s find the pH of a 0.25M NH3(aq) Solution:

  8. The Common Ion Effect First let’s find the pH of a 0.25M NH3(aq) Solution: Assuming x is << 0.25, x = [OH] = pOH = 2.67 pH = 14.00  2.67 = 11.33 for 0.25 M NH3

  9. The Common Ion Effect What is the pH of a solution made by adding equal volumes of 0.25M NH3(aq) and 0.10M NH4Cl(aq)? Since the solutions are mixed with one another, the effect of dilution is cancelled out. One can use the initial concentrations of each species in the reaction without calculating new molarities. This also works with mole ratios.

  10. The Common Ion Effect What is the pH of a solution made by adding equal volumes of 0.25 M NH3(aq) and 0.10 M NH4Cl(aq)? Since there is more ammonia than ammonium present, the RXN will proceed to the right.

  11. The Common Ion Effect What is the pH of a solution made by adding equal volumes of 0.25 M NH3(aq) and 0.10 M NH4Cl(aq)?

  12. The Common Ion Effect What is the pH of a solution made by adding equal volumes of 0.25 M NH3(aq) and 0.10 M NH4Cl(aq)? Assuming x is << 0.25 or 0.10 pOH = 4.35 pH = 9.65 The pH drops from11.33 due to the common ion!

  13. Controlling pH: Buffer Solutions HCl is added to pure water. HCl is added to a solution of a weak acid H2PO4- and its conjugate base HPO42-.

  14. Controlling pH: Buffer Solutions • A “Buffer Solution” is an example of the common ion effect. • From an acid/base standpoint, buffers are solutions that resist changes to pH. • A buffer solution requires two components that do not react with one another: • An acid capable of consuming OH • The acid’s conjugate base capable of consuming H3O+

  15. Controlling pH: Buffer Solutions Consider the acetic acid / acetate buffer system. • The ability for the acid to consume OH is seen from the reverse of the base hydrolysis: • Krev is >> 1, indicating that the reaction is product favored. • An hydroxide added will immediately react with the acid so long as it is present.

  16. Controlling pH: Buffer Solutions Consider the acetic acid / acetate buffer system. • Similarly, the conjugate base (acetate) is readily capable of consuming H3O+ • Krev is >> 1, indicating that the reaction is product favored. • An hydronium ion added will immediately react wit the acid so long as it is present.

  17. Buffer Solutions Problem: What is the pH of a buffer that has [CH3CO2H] = 0.700 M and [CH3CO2] = 0.600 M?

  18. Buffer Solutions Problem: What is the pH of a buffer that has [CH3CO2H] = 0.700 M and [CH3CO2] = 0.600 M? Since the concentration of acid is greater than the base, equilibrium will move the reaction to the right.

  19. Buffer Solutions Problem: What is the pH of a buffer that has [CH3CO2H] = 0.700 M and [CH3CO2] = 0.600 M? Assuming that x << 0.700 and 0.600, we find:

  20. Buffer Solutions Problem: What is the pH of a buffer that has [CH3CO2H] = 0.700 M and [CH3CO2] = 0.600 M? [H3O+] = 2.1 105 pH = 4.68

  21. Buffer Solutions The expression for calculating the [H+] of the buffer reduces to: The H3O+ concentration depends only Ka and the ratio of acid to base.

  22. Buffer Solutions Similarly for a basic solution the [OH] of the buffer reduces to: The OH concentration depends only Kb and the ratio of base to acid.

  23. Buffer Solutions: The Henderson-Hasselbalch Equation The result is known as the “Henderson-Hasselbalch” equation. The pH of a buffer can be adjusted by manipulating the ratio of acid to base.

  24. HCOOH (aq) H+(aq) + HCOO-(aq) Initial (M) Change (M) Equilibrium (M) [HCOO-] pH = pKa + log [HCOOH] [0.52] pH = 3.77 + log [0.30] What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Mixture of weak acid and conjugate base! 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x Common ion effect 0.30 – x 0.30 = 4.01 0.52 + x 0.52 HCOOH pKa = 3.77 24

  25. Preparing a Buffer • Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? • A pH of 4.30 corresponds to an [H3O+] = 5.0105 M

  26. Preparing a Buffer • Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? • A pH of 4.30 corresponds to an [H3O+] = 5.0105 M • First choose an acid with a pKa close to the desired pH

  27. Preparing a Buffer • Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? • A pH of 4.30 corresponds to an [H3O+] = 5.0105 M • First choose an acid with a pKa close to the desired pH • Next adjust the ratio of acid to conjugate base to achieve the desired pH.

  28. Preparing a Buffer • Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? • A pH of 4.30 corresponds to an [H3O+] = 5.0105 M • First choose an acid with a pKa close to the desired pH • Acetic acid is the best choice.

  29. Preparing a Buffer • Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? • A pH of 4.30 corresponds to an [H3O+] = 5.0105 M

  30. Preparing a Buffer • Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? • A pH of 4.30 corresponds to an [H3O+] = 5.0105 M • Therefore, if you start with 0.100 mol of acetate ion then add 0.278 mol of acetic acid, will result in a solution with a pH of 4.30.

  31. Preparing a Buffer • Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? • Therefore, if you start with 0.100 mol of acetate ion then add 0.278 mol of acetic acid, will result in a solution with a pH of 4.30. • Since both species are in the same solution (the same volume), the mole ratios are equal to the concentration ratios!

  32. Preparing a Buffer • Suppose you wish to prepare a buffer a solution at pH of 4.30. How would you proceed? • So, by adding 8.20 g of sodium acetate to 2780 ml of a 0.100 M acetic acid solution, one would make a buffer of pH 4.30.

  33. Preparing a Buffer Buffer prepared from 8.4 g NaHCO3 weak acid 16.0 g Na2CO3conjugate base What is the pH?

  34. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and [CH3CO2] = 0.600 M (pH = 4.68)

  35. Adding an Acid to a Buffer • What is the pH when 1.00 mL of 1.00 M HCl is added to: • 1.00 L of pure water • mL L  mols H3O+  M(H3O+)  pH

  36. Adding an Acid to a Buffer • What is the pH when 1.00 mL of 1.00 M HCl is added to: • 1.00 L of buffer that has [CH3CO2H] = 0.700 M and [CH3CO2] = 0.600 M (pH = 4.68) • The acid added will immediately be consumed by the acetate ion. • This in turn increases the acetic acid concentration. • The acetic acid then will react with water to reestablish equilibrium.

  37. Adding an Acid to a Buffer • What is the pH when 1.00 mL of 1.00 M HCl is added to: • 1.00 L of buffer that has [CH3CO2H] = 0.700 M and [CH3CO2] = 0.600 M (pH = 4.68) • The acid added will immediately be consumed by the acetate ion.

  38. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and [CH3CO2] = 0.600 M (pH = 4.68)

  39. Adding an Acid to a Buffer • What is the pH when 1.00 mL of 1.00 M HCl is added to: • 1.00 L of buffer that has [CH3CO2H] = 0.700 M and [CH3CO2] = 0.600 M (pH = 4.68) • This in turn increases the acetic acid concentration. • The acetic acid then will react with water to reestablish equilibrium.

  40. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and [CH3CO2] = 0.600 M (pH = 4.68)

  41. Adding an Acid to a Buffer What is the pH when 1.00 mL of 1.00 M HCl is added to b) 1.00 L of buffer that has [CH3CO2H] = 0.700 M and [CH3CO2] = 0.600 M (pH = 4.68) The pH does not change! The solution absorbs the added acid. It is a buffer!

  42. Commercial Buffers • The solid acid and conjugate base in the packet are mixed with water to give the specified pH. • Note that the quantity of water does not affect the pH of the buffer.

  43. Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3 (a) KF is a weak acid and F- is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO32- is a weak base and HCO3- is its conjugate acid buffer solution 43

  44. 44

  45. 45

  46. NH4+(aq) H+(aq) + NH3(aq) pH = 9.25 + log [0.30] [0.25] 0.028 0.025 [NH3] pH = pKa + log [NH4+] [0.36] [0.28] 0.10 0.10 NH4+(aq) + OH-(aq) H2O (l) + NH3(aq) pH = 9.25 + log [NH3] = [NH4+] = Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? pKa= 9.25 = 9.17 0.024 start (moles) 0.029 0.001 end (moles) 0.028 0.0 0.025 final volume = 80.0 mL + 20.0 mL = 100 mL = 9.20 46

  47. Acid–Base Titrations Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the pH increases very slowly.

  48. Acid–Base Titrations Additional NaOH is added. pH rises as equivalence point is approached.

  49. Acid–Base Titrations Additional NaOH is added. pH increases and then levels off as NaOH is added beyond the equivalence point.

  50. Titration of a Strong Acid with a Strong Base

More Related