Acid-Base Equilibrium (Mono- and Polyprotic) [Chapter 10,11]

1. Acid-Base Equilibrium (Mono- and Polyprotic) [Chapter 10,11] THE TRUTH, THE WHOLE TRUTH ANDNOTHING BUT THE TRUTH.

2. Strong Acids and Bases pH + pOH = - log Kw = pKw = 14.00 for Kw = 1.0x10-14 pH + pOH = - log Kw = pKw = 13.996 for Kw = 1.01x10-14 at 25oC

3. EXAMPLE: What is the pH of a 1.0x10-8M solution of HCl? MHCl = 1.0 x 10-8 M [H+]HCl = 1.0 x 10-8 M pH = 8.00 WRONG!!!!!!!! HCl is an acid! Acids have pH less than 7! OH NO! WHAT SHOULD WE DO !

4. What's wrong? We ignored the fact that water is also an ACID! [H+]total = [H+]water + [H+]HCl near pH 7, the contribution of [H+] from water becomes the dominate source for the [H+] in the solution

5. [H+]total = [H+]water + [H+]HCl [H+]total = [H+]water + 1.0 x 10-8M let [H+]water = x = [OH-] Kw = [H+]total[OH-] = 1.01 X 10-14 at 25oC [H+]total[OH-] = 1.01 X 10-14 ([H+]water + 1.0 x 10-8 M) * [OH-] = 1.01 X 10-14 ( x + 1.0 x 10-8M) * x = 1.01 X 10-14

6. ( x + 1.0 x 10-8M) * x = 1.01 X 10-14M2 by quadratic x = 9.51 x 10-8M [H+]total = 9.51 x 10-8M + 1.0 x 10-8M [H+]total = 10.5 x 10-8M = 1.05 x 10-7M pH = 6.98

7. Fractional Ionization of a Monoprotic Weak Acid Weak acids are those that are not leveled, or completely ionized in the solvent. Acetic acid is a relatively weak acid. The degree of ionization in aqueous solution depends on the formal concentration of HOAc, as well as the existence of other acid or base species that may be in solution.

8. If we divide both sides by volume, and use the definition of formal Which is a mathematical statement of mass balance, i.e., the sum of the molar concentrations of all acetic species equals the formula weights we put into solution.

9. A Little Algebra…. and substituted into the formal mass balance equation to yield The fraction of acid in acetate form is [OAc-]/FHOAc. Solving for this fraction results in the amount of HOAc in the form of the acetate a a & The ratios are often called the "alpha“ (a) of acetate and acetic acid respectively.

10. Fractional Composition (a) plot for acetic acid What is the pH when [HA] = [A-]??? 4.75 Which is the pKa….

11. Weak-Acid Equilibria Typical Weak-Acid Problem [HA] = CHA - [A-] = CHA - [H3O+] substituting into the Ka expression [H3O+]2 Ka = ----------------- CHA - [H3O+] produces the quadratic equation [H3O+]2 + Ka[H3O+] - KaCHA = 0

12. Weak-Acid Equilibria Typical Weak-Acid Problem [HA] = CHA - [A-] = CHA - [H3O+] produces the quadratic equation [H3O+]2 + Ka[H3O+] - KaCHA = 0

13. Weak-Acid Equilibria Typical Weak-Acid Problem simplifying assumption if CHA >> [H3O+] and CHA - [H3O+] ~ CHA [H3O+]2 Ka = ----------- thus CHA

14. Weak-Acid Equilibria Typical Weak-Acid Problem simplifying assumption Assume valid if CHA >>> Ka where each > signifies a power of 10

15. EXAMPLE: Calculate the [H3O+] in an aqueous 0.140 M acetic acid solution. HC2H3O2 + H2O H3O+ + C2H3O2- CHA = 0.140 M let [H3O+] ~ [C2H3O2-] [HC2H3O2] = 0.140M - [H3O+]

16. EXAMPLE: Calculate the [H3O+] in an aqueous 0.140 M acetic acid solution. HC2H3O2 + H2O H3O+ + C2H3O2-CHA = 0.140 M let [H3O+] ~ [C2H3O2-]; [HC2H3O2] = 0.140M - [H3O+] using the quadratic equation

17. EXAMPLE: Calculate the [H3O+] in an aqueous 0.140 M acetic acid solution. CHA = 0.140 M Ka = 1.75 X 10-5 M using the quadratic equation

18. CHA = 0.140 M Ka = 1.75 X 10-5M using the quadratic equation

19. CHA = 0.140 M Ka = 1.75 X 10-5M using the quadratic equation

20. EXAMPLE: Calculate the [H3O+] in an aqueous 0.140 M acetic acid solution. CHA = 0.140 M Ka = 1.75 X 10-5M assume [HC2H3O2] = 0.140M - [H3O+] ~ 0.140M

21. EXAMPLE: Calculate the [H3O+] in an aqueous 0.140 M acetic acid solution. CHA = 0.140 M Ka = 1.75 X 10-5M assume [HC2H3O2] = 0.140M - [H3O+] ~ 0.140M [H3O+] = 1.57 x 10-3M check assumption [HC2H3O2] = 0.140M - [H3O+] = 0.140M - 1.57 x 10-3M = 0.138 M assumption valid

22. Fraction of Dissociation EXAMPLE: Calculate the fraction of dissociation in an aqueous 0.140 M acetic acid solution. 1.11% dissociated

23. Use of Acid/Base Distributions in pH Problems

26. Interesting Features of a plots • First, notice that the pH where two species concentrations are the same is around the pKa for that equilibrium. In fact, for polyprotic acids with pKa's that differ by over 3 to 4 units, the pH is equal to the pKa.

27. Take for example the point where [H3PO4]=[H2PO4-]. The equilibrium equation relating these two species is If we take the -log10, or "p", of this equation Since [H3PO4]=[H2PO4-], and log10(1) = 0, pH=pKa1 pKa1 pKa3 pKa2

28. Second, you might notice that the concentrations of the conjugate bases are maximum half-way between the pKa points. For example, the point where [H2PO4-] is a maximum lies half-way between between pKa1 and pKa2. Since H2PO4- is the major species present in solution, the major equilibrium is the disproportionation reaction. This equilibrium cannot be used to solve for pH because [H3O+] doesn't occur in the equilibrium equation. We solve the pH problem adding the first two equilibria equations +

29. Note that when we add chemical equilibria, we take the product of the equilibrium equations. Taking the -log10 of the last equation Since the disproportionation reaction predicts [H3PO4]=[HPO42-]

30. What about the “Z” word?? • Zwitterions – (German for “Double Ion”) – a molecule that both accepts and losses protons at the same time. • EXAMPLES??? • How about – AMINO ACIDS zwitterion Both groups protonated neutral amino-protonated carboxylic-deprotonated

31. Let’s look at the simplest of the amino acids, glycine K1 K2 Gly- glycinate H2Gly+ glycinium HGly In water the charge balance would be, Combining the autoprotolysis of water and the K1 and K2 expressions into the charge balance yields:

32. Example • Calculate the pH and the concentrations of the various species in a 0.100 M glycine solution. What is the most prevalent form of glycine in This solution?

33. HGly Gly- H2Gly+

34. Buffers Abuffered solution resists changes in pH when acids and bases are added or when dilution occurs. definition - a buffer solution is composed of: a weak-acid and its salt (conjugate base) or a weak-base and its salt (conjugate acid)

35. Buffers Henderson-Hasselbalch Equation [H3O+] = Ka ([HA]/[A-]) pH = pKa + log([A-]/[HA]) when the [A-] = [HA] pH = pKa AH HA! REMEMBER THE RESULT FROM THE a plots

36. Biological Buffers • Biochemical reactions are especially sensitive to pH. Most biological molecules contain groups of atoms that may be charged or neutral depending on pH, and whether these groups are charged or neutral has a significant effect on the biological activity of the molecule. •      In all multicellular organisms, the fluid within the cell and the fluids surrounding the cells have a characteristic and nearly constant pH. This pH is maintained in a number of ways, and one of the most important is through buffer systems. Two important biological buffer systems are the dihydrogen phosphate system and the carbonic acid system.

37. The phosphate buffer system •      The phosphate buffer system operates in the internal fluid of all cells. This buffer system consists of dihydrogen phosphate ions (H2PO4-) as hydrogen-ion donor (acid) and hydrogen phosphate ions (HPO42-) as hydrogen-ion acceptor (base). • H2PO4-(aq) H+(aq) + HPO42-(aq) Ka = [H +] [HPO42-] =6.23 × 10-8 at 25oC [H2PO4-] • If additional hydrogen ions enter the cellular fluid, they are consumed in the reaction with HPO42-, and the equilibrium shifts to the left. If additional hydroxide ions enter the cellular fluid, they react with H2PO4-, producing HPO42-, and shifting the equilibrium to the right.

38. when the concentrations of H2PO4- and HPO42- are the same, what will the pH equal? 7.21 Buffer solutions are most effective at maintaining a pH near the value of the pKa. In mammals, cellular fluid has a pH in the range 6.9 to 7.4, and the phosphate buffer is effective in maintaining this pH range.

39. Carbonate Buffer      Another biological fluid in which a buffer plays an important role in maintaining pH is blood plasma. In blood plasma, the carbonic acid and hydrogen carbonate ion equilibrium buffers the pH. In this buffer, carbonic acid (H2CO3) is the hydrogen-ion donor (acid) and hydrogen carbonate ion (HCO3-) is the hydrogen-ion acceptor (base). H2CO3(aq) H+(aq) + HCO3-(aq) Additional H+ is consumed by HCO3- and additional OH- is consumed by H2CO3. Ka for this equilibrium is 7.9 × 10-7, and the pKa is 6.1 at body temperature. In blood plasma, the concentration of hydrogen carbonate ion is about twenty times the concentration of carbonic acid. The pH of arterial blood plasma is 7.40. If the pH falls below this normal value, a condition called acidosis is produced. If the pH rises above the normal value, the condition is called alkalosis.

40.      The concentrations of hydrogen carbonate ions and of carbonic acid are controlled by two independent physiological systems. Carbonic acid concentration is controlled by respiration, that is through the lungs. Carbonic acid is in equilibrium with dissolved carbon dioxide gas. H2CO3(aq) CO2(aq) + H2O(l) An enzyme called carbonic anhydrase catalyzes the conversion of carbonic acid to dissolved carbon dioxide. In the lungs, excess dissolved carbon dioxide is exhaled as carbon dioxide gas. CO2(aq) CO2(g)

41. The concentration of hydrogen carbonate ions is controlled through the kidneys. Excess hydrogen carbonate ions are excreted in the urine.      The much higher concentration of hydrogen carbonate ion over that of carbonic acid in blood plasma allows the buffer to respond effectively to the most common materials that are released into the blood. Normal metabolism releases mainly acidic materials: carboxylic acids such as lactic acid (HLac). These acids react with hydrogen carbonate ion and form carbonic acid. HLac(aq) + HCO3-(aq) Lac-(aq) + H2CO3(aq) The carbonic acid is converted through the action of the enzyme carbonic anhydrase into aqueous carbon dioxide. H2CO3(aq) CO2(aq) + H2O(l)

42. An increase in CO2(aq) concentration stimulates increased breathing, and the excess carbon dioxide is released into the air in the lungs.      The carbonic acid-hydrogen carbonate ion buffer works throughout the body to maintain the pH of blood plasma close to 7.40. The body maintains the buffer by eliminating either the acid (carbonic acid) or the base (hydrogen carbonate ions). Changes in carbonic acid concentration can be effected within seconds through increased or decreased respiration. Changes in hydrogen carbonate ion concentration, however, require hours through the relatively slow elimination through the kidneys

43. EXAMPLE: What is the ratio of [H2CO3]/[HCO3-] in the blood buffered to a pH of 7.40? Ka = 4.4 x 10-7M [H3O+] = 10-pH = 10-7.4 = 4.0 x 10-8M

44. Buffer Capacity • refers to the ability of the buffer to retard changes in pH when small amounts of acid or base are added • the ratio of [A-]/[HA] determines the pH of the buffer whereas the magnitude of [A-] and [HA] determine the buffer capacity