1 / 77

Acid Base Equilibria , Buffers, Titration

Acid Base Equilibria , Buffers, Titration. AP Chem. Weak Acids & Acid Ionizat i on Constant. Majority of acids are weak. Consider a weak monoprotic acid, HA: The equilibrium constant for the ionization would be:. or. Acid Ionization Constant.

cian
Download Presentation

Acid Base Equilibria , Buffers, Titration

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. Acid Base Equilibria, Buffers, Titration AP Chem

  2. Weak Acids & Acid Ionization Constant • Majority of acids are weak. Consider a weak monoprotic acid, HA: • The equilibrium constant for the ionization would be: or

  3. Acid Ionization Constant • The magnitude of Ka for an acid determines its strength

  4. Determining pH from Ka • Calculate the pH of a 0.50 M HF solution at 25 C. The ionization of HF is given by:

  5. Determining pH from Ka

  6. Determining pH from Ka • This leads to a quadratic equation, so lets simplify: 0.50 – x ≈ 0.50 • The expression becomes:

  7. Determining pH from Ka • At equilibrium, [HF] = (0.50-0.019) = 0.48 M [H+] = 0.019 M [F-] = 0.019 M • This only works if x is less than 5% of 0.50 • Why? Ka are generally only accurate to ±5%

  8. Was the approximation good? • Consider if initial concentration of HF is 0.050 M. Same process above, but x = 6.0 x 10-3 M. This would not be valid. VALID! NOT VALID!

  9. Weak Bases & Base Ionization Constants • Same treatment as for acids. Given ammonia in water: • Reminder: we solve using [OH-], not [H+]

  10. Relationship of Ka to their Conjugate Base Mole Buck Opportunity! Kw = Ka Kb • This leads us to conclude the stronger the acid (larger Ka), the weaker its conjugate base (smaller Kb)

  11. Diprotic & Polyprotic Acids • A more involved process due to stepwise dissociation of hydrogen ion • Each step is like a monoprotic acid • Be sure to think about what is present at each step! • Note the conjugate base is used as the acid in the next step

  12. Diprotic Acid Calculation • Calculate all species present at equilibrium in a 0.10 M solution of oxalic acid (H2C2O4)

  13. Diprotic Acid Calculation

  14. Diprotic Acid Calculation • Making the approximation 0.10 – x  0.10, we get: Quadratic! Approx. good? NOT VALID! STOP

  15. Diprotic Acid Calculation • x = 0.054 M • After the first stage, we have: • [H+] = 0.054 M • [HC2O4-] = 0.054 M • [H2C2O4] = (0.10 – 0.054) M = 0.046 M • Next step: Treat conj. base as acid for 2nd step

  16. Diprotic Acid Calculation • Second dissociation would be:

  17. Diprotic Acid Calculation • Applying the approximation (for both) we obtain: Approx. good? VALID! STOP

  18. Diprotic Acid Calculation! • Finally, at equilibrium: • [H2C2O4] = 0.046 M • [HC2O4-] = (0.054 – 6.1 x 10-5) = 0.054 M • [H+] = (0.054 + 6.1 x 10-5) = 0.054 M • [C2O42-] = 6.1 x 10-5 M

  19. Conclusions on Polyprotic Acids • The past example shows that for diprotic acids, Ka1 >> Ka2 • From this, we can assume the majority of H+ ions are produced in the first stage of ionization • Secondly, concentration of conj. base is numerically equal to Ka2

  20. Molecular Structure and Strength of Acids • Strength of hydrohalic acids (HX) depends on 2 factors: • Bond Strength (called bond enthalpy) • Polarity of Bond • Strength depends on ease of ionization • Stronger the bond, more difficult to ionize • Higher the polarity, better chance of ionizing

  21. Strength of Binary Acids HF << HCl < HBr < HI • Based on polarity, HF might considered strongest, but bond strength opposes this separation of charge • This implies that bond enthalpy is the predominant factor in determining acid strength of binary acids

  22. Strength of Oxoacids • Divide oxoacids into two groups: • Oxoacids having different central atoms from the same group with the same oxidation number • i.e. HClO3and HBrO3 • Oxoacids have same central atom but different number of attached groups • i.e. HClO4and HClO3

  23. Strength of Oxoacids • Oxoacids having different central atoms from the same group with the same oxidation number • In this group, acid strength increases with increasing electronegativity of the central atom • Higher polarity means easier to ionize HClO3 > HBrO3

  24. Strength of Oxoacids • Oxoacids have same central atom but different number of attached groups • In this group, acid strength increases as oxidation number of central atom increases • i.e. the more oxygen, the merrier! HClO4 > HClO3 > HClO2 > HClO

  25. Acid-Base Properties of Salts • Salt hydrolyis is the reaction of an anion or a cation of a salt (or both) with water • Salts can be neutral, basic, or acidic and follow certain trends • Acids mixed with bases forms salt + water! BEWARE!

  26. Salts that Produce Neutral Sol’ns • Alkali metal ion or alkaline earth metal ion (except Be) do not undergo hydrolysis • Conjugate bases of strong acids (or bases) do not undergo hydrolysis • i.e. Cl-, Br-, NO3- • So NaNO3 forms a neutral solution

  27. Salts that Produce Basic Sol’ns • Conjugate bases of a weak acid will react to form OH- ions • For example, NaCH3COO forms Na+ and CH3COO- in solution. Acetate ion is the conjugate base of acetic acid, and undergoes hydrolysis:

  28. Basic Salt Hydrolysis Calculation • Calculate the pH of a 0.15 M solution of sodium acetate (CH3COONa) • Since the dissocation is 1:1 mole ratio, the concentration of the ions is the same

  29. Basic Salt Hydrolysis Calc. • Because acetate ion is the conj. base of a weak acid, it hydrolyzes as:

  30. Basic Salt Hydrolysis Calc. • Applying the approximation, Approx. good? STOP Also known as percent hydrolysis!

  31. Acidic Salt Hydrolysis • Conjugate acid of a weak base will react to form H3O+ ions • Small, highly charged cation will also produce acidic solutions when hydrated • Typically transition metals (Al3+, Cr3+, Fe3+)

  32. Common Ion Effect Revisited • Recall that the addition of a common ion causes an equilibrium to shift • Earlier we related this to the solubility of a salt • The idea is just an application of Le Chatelier’s Principle

  33. pH changes due to Common Ion Effect • Consider adding sodium acetate (NaCH3COO) to a solution of acetic acid: • The addition of a common ion here (CH3COO-) will increase the pH • By consuming H+ions

  34. Henderson-Hasselbalch • Consider: • Rearranging Ka for [H+]we get:

  35. Henderson-Hasselbalch • Take negative log of both sides: or:

  36. Finding pH with common ion present • Calculate the pH of a solution containing both 0.20 M CH3COOH and 0.30 M CH3COONa? The Ka of CH3COOH is 1.8x10-5. • Sodium acetate fully dissociates in solution

  37. Finding pH with common ion present • Can use I.C.E. table, or Henderson-Hasselbach

  38. Effect of Common Ion on pH • Consider calculating the pH for a 0.20 M acetic acid solution • pH = 2.72 • From our last example, its obvious the pH has increased due to the common ion

  39. Buffer Solutions • Buffers are a solution of (1) weak acid or weak base and (2) its salt • Buffers resist changes in pH upon addition of acid or base • Buffer Capacity: refers to the amount of acid or base a buffer can neutralize

  40. Buffer Solutions • Consider a solution of acetic acid and sodium acetate • Upon addition of an acid, H+ is consumed: • Upon addition of a base, OH- is consumed:

  41. Buffer Animation • http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swf

  42. Buffer Problem • (a) Calculate the pH of a buffer system containing 1.0M CH3COOH and 1.0 M CH3COONa. (b) What is the pH of the buffer system after the addition of 0.10 mole of gaseous HCl to 1.0 L of the solution? Assume volume of sol’n does not change.

  43. Buffer Problem • To calculate pH of buffer, I.C.E. or H.H. • With addition of HCl, we are adding 0.10 M H+, which will react completely with acetate ion

  44. Buffer Problem Now acetic acid will still dissociate and amount of H+ formed is the pH of sol’n

  45. Buffer Problem • Using H.H. (or I.C.E.)

  46. Finding Buffers of Specific pH • If concentrations of both species are equal this means: • Using H.H., to find specific pH, search for pKa pH

  47. Finding Buffers of Specific pH • Describe how you would prepare a “phosphate buffer” with a pH of about 7.40

  48. Finding Buffers of Specific pH • Using the HPO42-/H3PO4- buffer: This means a mole ratio of 1.5 moles disodium hydrogen phosphate : 1.0 mole Monosodium dihydrogen phosphate will result in a buffer solution with a 7.4 pH

  49. Finding Buffers of Specific pH • To obtain this solution, disodium hydrogen phosphate (Na2HPO4) and sodium dihydrogen phosphate (NaH2PO4) is in 1.5:1.0 ratio • Meaning it is 1.5 M Na2HPO4 and 1.0 M NaH2PO4 are combined per liter of solution

  50. Acid-Base Titrations • Three situations will be considered: • Strong Acid/ Strong Base • Weak Acid/ Strong Base • Strong Acid/ Weak Base • Titrations of weak acid/base are complicated by hydrolysis

More Related