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Acid-Base Equilibrium

Acid-Base Equilibrium

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Acid-Base Equilibrium

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  1. Acid-Base Equilibrium

  2. What You’ll See • Tie together concepts of acids and bases and equilibrium • Weak acids and related calculations • Weak bases and related calculations • Salt solutions (acid-base properties) • Hydrolysis • Common Ion Effect • Buffers & Henderson-Hasselbalch equation • Titrations

  3. Weak Acids • Only slightly ionized • A real equilibrium situation • Examples include • Organic acids (acetic, lactic, formic) • Inorganic where # O exceeds # H by <2 (phosphoric, carbonic, nitrous acids)

  4. Weak acids • Since they are only slightly ionized we can write the process and an equilibrium expression. • For example, for HF it would be • HF + H2O  H3O+ + F- (or HF H+ + F-) • Ka (= Keq for acids) = [H3O+][F-] [HF] We can do several calculations with this(!)

  5. Find Ka from pH • Continue with HF example • A solution of 0.015 M HF has a pH of 2.89. Find the value of Ka for hydrofluoric acid. • A 0.015 M HF solution was made to have 0.015 moles HF/L. A small % of that weak acid ionizes so at equilibrium there is slightly less than 0.015 mole/L. • We can figure this out with an ICE chart

  6. HF + H2O  H3O+ + F- Initial 0.015 0 0 Change -x +x +x Equilibrium 0.015-xx x (we get to leave water out of equilibrium calculations) We need x the [H3O+]. If the pH = 2.89, then [H3O+] = 10-2.89 = 1.29 x 10-3M Plug that back into the chart

  7. HF  H3O+ + F- Initial 0.015 0 0 Change -0.00129 +0.00129 +0.00129 Equilibrium 0.0137 0.00129 0.00129 Plug back into the Ka expression and get Ka = (0.00129)(0.00129)/(0.0137) = 1.2 x 10-4 A small Ka value is what you’d expect for a weak acid (few ions or primarily in un-ionized form) What percent is ionized? (0.00129)/0.015 =0.086 = 8.6 % (far less than the 100% for a strong acid).

  8. Solving Ka, given the pH • The pH of a .85M solution of acetic acid is 2.75. What is Ka for acetic acid • It helps to abbreviate the acid HAc instead of writing out the entire formula • HAc H+ + Ac- • Ka = [H+][Ac-] [HAC]

  9. Start, change, equilibrium HAc  H+ + Ac- Start .85M 0 0 Change -x +x +x Equilibrium .85-x x x But x =[H+], which we know since we know the pH and [H+] = 10-pH = 10-2.75 = 2 x 10-3

  10. Continuing HAc  H+ + Ac- Start .85M 0 0 Change -x +x +x Equilibrium .85-x x x Or .848 .002 .002 Ka = (.002)(.002) .848 = 4.7 x 10-5

  11. Using Ka to get pH • What are the concentrations of H+, F- and HF as well as the pH in a .14M solution of HF? • If this were a strong acid we could know that [H+] was .14M and it would be a simple problem • Alas, that is not the case

  12. Plan of Attack • Write an equilibrium expression • Do an ICE table • Solve for unknowns • Hint- you will be given Ka (6.7 x 10-4)

  13. HF  H+ + F- Initial .14 0 0 Change -x +x +x Equilibrium .14-x x x Ka = 6.7 x 10-4 = [H+][F-] [HF] And it looks like we need the quadratic…

  14. But… 6.7 x 10-4 = (x)(x) (.14-X) If x is small (and it should be for a weak acid) then .14-x is just about .14 This has two benefits • It simplifies the math • It annoys math teachers

  15. So now we get… 6.7 x 10-4 = x2 .14 X2 = 9.4 x 10-5 X = [H+] = [F-] = 9.7 x 10-3 So pH = -log(9.7 x 10-3) = 2.01 And [HF] = .14-.0097 = .13 so it was a pretty good approximation

  16. When can you use that approximation? ( I hear you ask) • In this class almost always, provided you make it clear that you are using a an approximation and that you know it is justified • It is justified when [HA]/Ka >100 • This leads to less than a 5% error

  17. % Ionization of a weak acid • Benzoic acid is a weak, monoprotic acid. A .0135M solution of benzoic acid has a pH of 3.27. What % of the benzoic acid molecules are ionized? What is Ka? • Write the equilibrium equation • Write the Ka expression • Do ICE • solve

  18. % Ionization of Benzoic Acid HBz  H+ + Bz- We often abbreviate formulas like this to let us focus on the ions that are really important to the problem. Ka = [H+][Bz-] [HBz]

  19. Start, change, equilibrium HBz  H+ + Bz- Initial .0135 0 0 Change -x +x +x Equilibrium .0135-x x x X = [H+] = 10-3.27 = 5.37 x 10-4 .0135-x = .0130 (ignoring it would be a 4% error –not necessary b/c you don’t need the quadratic to solve this.

  20. Wrapping it up • Ka = (5.37 x 10-4)2 .0130 • = 2.2 x 10-5 • % ionized = x/[HBz]0 = • 5.37 x 10-4/.0135 = .040 = 4%

  21. Polyprotic Acids • Acids with more than one “donatable”ton • Diprotic • Sulfuric acid • Carbonic acid • Triprotic • Phosphoric acid

  22. Qualitative Approach • In general, the second proton Ka value is much less (factor of 10000 or so) than the first proton’s Ka value • This means you can “usually” ignore the contribution from the ionization of the second hydrogen

  23. Example H2SO3  H+ + HSO3- Ka = 1.3 x 10-2 HSO3-  H+ + SO3-2 Ka = 6.3 x 10-8 Because Ka2<<Ka1 the contribution of the second ionization of hydrogen to the total [H+] is minimal One exception is sulfuric acid Ka for HSO4- = 1.2 x 10-2 Example to follow:

  24. pH of .01 M H2SO4 • We know that the contribution of hydrogen ions from the ionization of H2SO4 is .01M (strong acid: 100% ionized) • For the second ionization of H: • HSO4-  H+ + SO4-2 Ka = 1.2 x 10-2 • Set up an initial, change, equilibrium table

  25. HSO4-  H+ + SO4-2 Iniial .010 .010 0 Change -x +x +x Equilibrium .010-x .010 +x x Note that bisulfate ion and hydrogen ion are not initially zero because of the ionization of H2SO4 Write the Ka (equilibrium) expression

  26. HSO4-  H+ + SO4-2 Ka = [H+][SO4-2] [HSO4-] 1.2 x 10-2 = (.010+x)(x) (.010-x) After a “twist of algebra” you’ll get x =4.5 x 10-3 so that the total [H+] = 1.4 x 10-2 and the pH = 1.85 Since there should be more ions than from just the first ionization (.01M H+), the pH should be slightly less than 2, which it is.

  27. Polyprotic Wrap-Up • Other than that one example, the contribution to the total amount of hydronium ion from the second (or third) hydrogen of a polyprotic acid is so small that we can ignore it. The ffect on the pH is even less since it involes the log of the [hydronium ion]

  28. Weak Base Equilibria • Weak Bases • Few ions • Accept protons • Lead to an increase in hydroxide ions • Involve a Kb expression • Must write as a reaction water (water will act as the Bronsted-Lowry acid)

  29. Sample #1 • Find the pH of a .0035M solution of quinine, a naturally occurring weak base once used to treat malaria. Kb for quinine is 3.3 x 10-6. • Write the ionization equation & Keq (or Kb) expression

  30. Abbreviate quinine Qu Qu + H2O  HQu+ + OH- Initial .0035 0 ~0 Change -x +x +x Equilibrium .0035-x x x (there are a few OH- ions in pure water) Kb = [HQu+][OH-] [Qu] Substituting in give…

  31. 3.3 x 10-6 = (x)(x) (.0035-x) Since [base]/Ka>>100 it is safe to approximate .0035-x = .0035 A bit of algebra and we get x = 1.1 x 10-4 Remember that x = [OH-] Then get [H+] = 1 x 10-14/1.1 x 10-4 Give [H+] = 9.1 x 10-11 and pH = -log (9.1 x 10-11) = 10.0 Which is reasonable for a base

  32. Let’s do one more! • The pH of a .0025M solution of hydrazine (N2H4) is 9.92. What is Kb for hydrazine? • N2H4 + H2O  N2H5+ + OH- Kb = [N2H5+][OH-] [N2H4]

  33. Hydrazine cont’d N2H4 + H2O  N2H5+ + OH- Initial .0025 0 ~0* Change -x +x +x Equilibrium .0025 –x x x (*I should have been doing ~0 for the initial [H+] in acids as well)

  34. Since the pH is 9.92, the [H+] = 10-9.92= 1.20 x 10-10M Then we can get [OH-] =8.33 x 10-5M Substituting we get Kb = (8.33 x 10-5)(8.33 x 10-5) (.00242) = 2.9 x 10-6

  35. Wrap-Up • Follow a similar strategy as acids • Keep in mind that if it is a base, you will be solving for hydroxide ion concentration • pH must be >7 (quick check) • Write ionization equation (water acts as the Bronsted-Lowry acid) • Write Kb, plug in and solve

  36. Salt Solutions- Acid/Base Properties • Salts can be thought of acid being the product of an acid-base neutralization • Salts can be acidic, basic or neutral • We will approach salt solutions first • Qualitatively and then • Quantitatively

  37. First • You need to be able to look at a salt and decide what acid and base its ions could have come from (this is not the same as saying that every salt is made by an acid-base rxn) • Positive ion will be from the base • Negative ion will be from the acid

  38. A couple of examples • NaCN • Sodium hydroxide and hydrogen cyanide • K3PO4 • Potassium hydroxide and phosphoric acid • LiBr • Lithium hydroxide and hydrobromic acid • NH4NO3 • Ammonia and nitric acid (think of ammonia as ammonium hydroxide if it helps)

  39. Qualitative Approach • Great for multiple choice Q’s • A good way to check when you do the quantitative ones later • See table on next slide for a quick summary • Note how recognizing the strong and weak acids is vital

  40. Depends??

  41. Yes It Depends • It has been a while since this was the answer to a question • So the more ionized the weak acid or base the “stronger” a weak acid or base it will be • If Ka>Kb it will be an acidic salt • If Kb>Ka it will be a basic salt

  42. More on “it depends” • Recall that all strong acids and bases are equally ionized (100%) • But weak acids and bases may range from 10% ionized to barely .01% ionized • All strong acids and bases are equally strong • The same is not true for weak acids and bases

  43. Quantitative Approach • Consider a solution of NaF (science problems always begin with “consider…” as in “consider a spherical chicken with radius r…”) • NaF  Na+ + F- (100% ionized) • Na+ ion is nonreactive with water (it is stable when hydrated) • F- ion can react with water

  44. Hydrolysis (that’s the name of it) F- + H2O  HF + OH- This means that the solution should be basic (strong base and weak acid salt) This reaction between the ion of a weak acid or base and water is called hydrolysis

  45. Hydrolysis Kb for F- = Kw/Ka for HF (book does you the “service” of going through the derivation) Kb = 1 x 10-14/6.8 x 10-4 = 1.5 x 10-11 This can then be used to calculate [OH-] and then [H+] and pH As we see in the next few slides

  46. Find the pH of a .4M NaF sol’n NaF  Na+ + F- F- + H2O  HF + OH- Kb = 1.5 x 10-11 (see previous slide) Kb = [HF][OH-] [NaF] Set up an initial, change, equilibrium table (NaF = F-)

  47. F- + H2O  HF + OH- Initial .4M 0 ~0 Change -x +x +x Equilibrium .4-x x x Kb = 1.5 x 10-11 =[HF][OH-] [NaF] 1.5 x 10-11 = (x)(x) (.04-x)

  48. Wrapping it Up… • Since [NaF]/Kb >100 we can approximate .4-x as .4 and get • 1.5 x 10-11 = x2/.4 • X = [OH-] = 2.4 x 10-6 • [H+] = 1 x10-14/2.4 x 10-6 = 4.1 x 10-9 • pH = 8.3 (basic as we would predict) • One more

  49. Morphine is a weak base used for relief of severe pain. Morphine is not soluble in water, but morphine chloride is. What is the pH of a .100M solution of Morphine chloride (MorCl)? Kb for morphine is 1.6 x 10-6 • Plan- first do qualitatively (acidic) • Ionization equation & hydrolysis equation • Start, change, equilibrium • Ka from Kb, then [H+]

  50. MorCl  Mor+ + Cl- • Cl- is the ion of a strong acid (nonreactive) • Mor+ undergoes hydrolysis with water • Mor+ + H2O  H3O+ + Mor • (H3O+ = H+) • Ka = Kw/Kb = 6.3 x 10-9 • Ka = [H3O+][Mor] [Mor+]