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# Acid Base Equilibria , Buffers, Titration

Acid Base Equilibria , Buffers, Titration. AP Chem. Weak Acids &amp; 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.

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## Acid Base Equilibria , Buffers, Titration

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1. 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

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

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

4. Determining pH from Ka

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

6. 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%

7. 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!

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

9. 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)

10. 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

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

12. Diprotic Acid Calculation

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

14. 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

15. Diprotic Acid Calculation • Second dissociation would be:

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

17. 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

18. 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

19. 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

20. 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

21. 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

22. 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

23. 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

24. 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!

25. 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

26. 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:

27. 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

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

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

30. 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+)

31. 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

32. 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

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

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

35. 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

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

37. 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

38. 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

39. 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:

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

41. 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.

42. 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

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

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

45. 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

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

47. 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

48. 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

49. 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

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