1. Polyprotic Acids & Bases A polyprotic acid can donate more than one H + Carbonic acid: H 2 CO 3 (aq); dissolved CO 2 in water Sulfuric acid: H 2 SO 4 (aq) Phosphoric acid: H 3 PO 4 (aq) A polyprotic base: can accept more than one proton Carbonate ion: CO 3 2- (aq) Sulfate ion: SO 4 2- (aq) Phophate ion: PO 4 3- (aq) Treat each step of protonation or deprotonation sequentially

2. H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 - (aq)K a1 = 4.3 x 10 -7 HCO 3 - (aq) + H 2 O(l)  H 3 O + (aq) + CO 3 2- (aq)K a2 = 4.8 x 10 -11 Typically: K a1 >> K a2 >> K a3 >>… Harder to loose a positively charged proton from a negatively charged ion, because of attraction between opposite charges.

4. Calculate the pH of 0.010 M H 2 SO 4 (aq) at 25 o C. Sulfuric acid is the only common polyprotic acid where the first deprotonation step is complete. The second deprotonation step is much weaker and adds slightly to the H 3 O + (aq) concentration. For the first step assume all H 2 SO 4 (aq) deprotonates H 2 SO 4 (aq) + H 2 O(l)  H 3 O + (aq) + HSO 4 - (aq) From the first step [H 3 O + (aq)] = 0.010 M

5. Second deprotonation HSO 4 - (aq) + H 2 O(l)  H 3 O + (aq) + SO 4 2- (aq)K a2 = 0.012 HSO 4 - (aq) SO 4 2- (aq) H 3 O + (aq) Initial0.01000.010 Change-x+ x0.010 + x Equilibrium0.010-xx0.010 + x K a2 = ([H 3 O + (aq)])([SO 4 2- (aq)]) / ([HSO 4 - (aq) ]) 0.012 = (0.010+x)(x) / (0.010-x) Solve the quadratic equation for x. K a2 is large; cannot assume that x << 0.010 [H 3 O + (aq)] = 1.4 x 10 -2 M pH = 1.9

6. Determine the pH of 0.20 M H 2 S(aq) at 25 o C H 2 S (aq) + H 2 O(l)  H 3 O + (aq) + HS - (aq) K a1 = 1.3 x 10 -7 HS - (aq) + H 2 O(l)  H 3 O + (aq) + S 2- (aq) K a2 = 7.1 x 10 -15 For the first deprotonation step determine [H 3 O + (aq)] using equilibrium tables. [H 3 O + (aq)] = 1.6 X 10 -4 M Can assume that x << 0.20 since K a1 is small Second deprotonation constant is very small, so ignore addition of H 3 O + (aq) due to second step. pH determined by first step alone. pH = 3.8

7. Composition and pH For a solution of H 2 CO 3 (aq): at low pH the fully protonated species (H 2 CO 3 ) dominates; at high pH the fully deprotonated form (CO 3 2- ) dominates; and at intermediate pH the intermediate species (HCO 3 - ) dominates. LeChatelier’s principle at work H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 - (aq)K a1 = 4.3 x 10 -7 HCO 3 - (aq) + H 2 O(l)  H 3 O + (aq) + CO 3 2- (aq)K a2 = 4.8 x 10 -11

8. H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 - (aq) K a1 = [H 3 O + (aq)] [HCO 3 - (aq)] / [H 2 CO 3 (aq) ] HCO 3 - (aq) + H 2 O(l)  H 3 O + (aq) + CO 3 2- (aq) K a2 = [H 3 O + (aq)] [CO 3 2- (aq)] / [HCO 3 - (aq) ] Define  (X): fraction of species X  (X) = X [H 2 CO 3 (aq) ] + [HCO 3 - (aq)] + [CO 3 2- (aq)] The fraction of deprotonated species increases as the pH increases

10. Determine the concentration of H 2 CO 3 (aq), HCO 3 - (aq), CO 3 2- (aq), H 3 O + (aq) present and the pH at equilibrium in a solution that is initially 0.010 M in H 2 CO 3. (K a1 = 4.3 x 10 -7, K a2 = 4.8 x 10 -11 ) Step 1 H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 - (aq) H 2 CO 3 (aq) H 3 O + (aq) HCO 3 - (aq) Initial 0.010 0 0 Change -x x x Equilibrium 0.010-x x x 4.3 x 10 -7 = x 2 / (0.010-x) Assume x << 0.010; x = 6.6 x 10 -5 [H 2 CO 3 (aq)] ≈ 0.010 M; [H 3 O + (aq)] = 6.6 x 10 -5 M; [HCO 3 - (aq)] = 6.6 x 10 -5 M

11. HCO 3 - (aq) + H 2 O(l)  H 3 O + (aq) + CO 3 2- (aq) K a2 = [H 3 O + (aq)] [CO 3 2- (aq)] / [HCO 3 - (aq) ] HCO 3 - (aq) H 3 O + (aq) CO 3 2- (aq) Initial 6.6 x 10 -5 6.6 x 10 -5 0 Change -y 6.6 x 10 -5 + y y Equilibrium 6.6 x 10 -5 - y 6.6 x 10 -5 + y y 4.8 x 10 -11 = (6.6 x 10 -5 + y) y / (6.6 x 10 -5 - y) Assume y << 6.6 x 10 -5 4.8 x 10 -11 = (6.6 x 10 -5 y / (6.6 x 10 -5 ); y = 4.8 x 10 -11 At equilibrium: [H 2 CO 3 (aq)] ≈ 0.010 M; [H 3 O + (aq)] = 6.6 x 10 -5 M; [HCO 3 - (aq)] = 6.6 x 10 -5 M; [CO 3 2- (aq)] = 4.8 x 10 -11 M pH = 4.18

13. Buffers Buffer solutions : resists change in pH even with addition of small amounts of acid or base. Buffer solutions are mixed solutions: mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. Human blood has a pH maintained at pH = 7.4 due to a combination of carbonate, phosphate and protein buffers. The ocean is buffered to a pH of ~ 8.4 by buffering that depends the presence of hydrogen carbonates and silicates.

14. Buffer Action An acid buffer is an aqueous solution of a weak acid and its conjugate base. It buffers solutions on the acid side of neutral (pH < 7). Example: solution of CH 3 COOH(aq) + CH 3 COONa(aq) CH 3 COOH / CH 3 COO - A base buffer is an aqueous solution of a weak base and its conjugate acid. It buffers solutions on the basic side of neutral (pH > 7). Example: NH 3 (aq) + NH 4 Cl(aq) NH 3 / NH 4 +

15. Buffer solution of CH 3 COOH(aq) / CH 3 COO - (aq) If a small amount of strong acid is added: H 3 O + (aq) + CH 3 COO - (aq)  CH 3 COOH(aq) + H 2 O(l) K for this reaction = 1/K a (CH 3 COOH(aq)) = 5.5 x 10 4 The CH 3 COO - (aq) acts as a “sink” for the added protons, and the pH remains unchanged. If a small amount of strong base is added OH - (aq) + CH 3 COOH(aq)  CH 3 COO - (aq) + H 2 O(l) K for this reaction = 1/K b (CH3COO - (aq)) = 1.8 x 10 9 The CH 3 COOH(aq) acts as a “sink” for the added OH -, and the pH remains unchanged

16. Designing a buffer Make a solution with particular pH so that it buffers about this pH. Consider a solution of a weak acid and its conjugate base HA(aq) + H 2 O(l)  H 3 O + (aq) + A - (aq) [H 3 O + (aq)] =KaKa [HA(aq)] [A - (aq)] - log [H 3 O + (aq)] =- log K a - log [HA(aq)] [A - (aq)] pH = pK a + log [A - (aq)] [HA(aq)] Henderson-Hasselbach equation Note: for a solution of a weak base/conjugate acid, use K a of the conjugate acid

17. An optimal buffer is one in which the weak acid and its conjugate base have equal concentrations Select a weak acid that has its pK a as close as possible to the desired pH Having chosen the weak acid, use the Henderson Hasselbach equation to determine the ratio of [A - (aq)] and [HA(aq)] that will form solution that buffers around the desired pH

18. Calculate the pH of a buffer solution that is 0.040 M CH 3 COONa (aq) and 0.080 M CH 3 COOH (aq). pK a (CH 3 COOH(aq)) = 4.75 CH 3 COOH(aq) + H 2 O(l)  H 3 O + (aq) + CH 3 COO - (aq) CH 3 COOH(aq) H 3 O + (aq) CH 3 COO - (aq) Initial 0.080 M 00.040M Change-x x0.040 + x Equilibrium 0.080 - x x0.040 + x K a = (0.040 + x) x / (0.080 - x) Assume x << 0.040 x = 3.6 x 10 -5 pH = 4.44

19. Suppose that a solution is made by dissolving 1.2 g NaOH(s) (0.030 moles) in 500 mL of the buffer solution in the previous problem. Calculate the pH of the resulting solution and the change in pH. Assume the volume of the solution to be constant. The OH - (aq) will react with CH 3 COOH(aq) CH 3 COOH(aq) + OH - (aq)  CH 3 COO - (aq) + H 2 O(l) Moles of OH - added = 0.030 moles Moles of CH 3 COOH(aq) = (0.500 L) (0.080 M) = 0.040 mol Amount of unreacted CH 3 COOH(aq) = 0.010 moles Molarity of CH 3 COOH(aq) = 0.020 M

20. Moles of CH 3 COO - (aq) = initial amount + amount formed by reaction of OH - (aq) and CH 3 COOH(aq) = (0.040M x 0.500 L) + (0.030 moles) = 0.050 moles Molarity of CH 3 COO - (aq) = 0.10 M CH 3 COOH(aq) + H 2 O(l)  CH 3 COO - (aq) + H 3 O + (aq) pH = pK a + log ([CH 3 COO - (aq) ] / [CH 3 COOH(aq) ] ) = 5.45 If the solution had contained HCl at pH = 4.4, addition of the NaOH would have raised the pH to 12.8

21. Titrations Strong Acid - Strong Base H 3 O + (aq) + OH - (aq)  2H 2 O(l) pH changes slowly initially, changes rapidly through pH = 7 (equivalence point) and then changes slowly again If the analyte is a strong acid, pH increases as base is added

22. If the analyte is a strong base, pH decreases as acid is added

23. Analyte: 25.00 mL of 0.250 M NaOH(aq) Titrant: 0.340 M HCl(aq) Determine the pH of the solution when 5.00mL of titrant added Answer: pH = 13.18 Determine the amount of titrant that must be added to reach the equivalence point?Answer: 18.4 mL Determine the pH of the solution after the addition of 20.4 mL of titrant. Answer: pH = 1.82

24. Strong Acid-Weak Base and Weak Acid - Strong Base

25. Slow change in pH before equivalence point; solution is a buffer CH 3 COOH(aq)/CH 3 COO - (aq) At halfway point [HA] = [A - ] pH = pK a At equivalence, pH determined by CH 3 COO - (aq) CH 3 COOH(aq) + OH - (aq) -> CH 3 COO - (aq) + H 2 O(l)

26. Changes in pH during a titration of a weak acid/base with a strong base/acid: Halfway to the stoichiometric point, the pH = pK a of the acid The pH is greater than 7 at the equivalence point of the titration of a weak acid and strong base The pH is less that 7 at the equivalence point of the titration of a weak base and strong acid Beyond the equivalence point, the excess strong acid or base will determine the pH of the solution

27. Titration of 100.0 mL of 0.1000 M CH 3 COOH(aq) with 0.1000 M NaOH Before addition of NaOH: pH determined by CH 3 COOH(aq) CH 3 COOH(aq) + H 2 O(l)  H 3 O + (aq) + CH 3 COO - (aq) Answer: pH = 2.88 Before the equivalence point: determine pH for a buffer Addition of 30.00 mL of NaOH(aq) The OH - (aq) reacts with the CH 3 COOH(aq). Determine concentration of CH 3 COOH(aq) and CH 3 COO - (aq) in solution after addition of the base. Answer: pH = 4.38 At half equivalence: [CH 3 COOH(aq)] = [CH 3 COO - (aq)] pH = pKa

28. At equivalence: enough OH - (aq) added to react with all CH 3 COOH(aq). For this problem, equivalence is reached when 100.0mL of OH - is added; i.e. 0.01000 moles of OH - (aq) added Solution contains 0.01000 moles CH 3 COO - (aq) in 200.0 mL solution; [CH 3 COO - (aq)] = 0.05000 M pH determined by CH 3 COO - (aq) + H 2 O(l)  CH 3 COOH(aq) + OH - (aq) pH = 8.72(note greater than 7.0) Beyond equivalence: pH determined by excess OH - (aq)

29. Estimate the pH at the equivalence point of the titration of 25.00 mL of 0.100 M HCOOH(aq) with 0.150 M NaOH(aq) (K a (HCOOH) = 1.8 x 10 -4 ) At the equivalence point, enough NaOH(aq) has been added to react with all the HCOOH(aq) forming HCOO - (aq) The reaction: HCOO - (aq) + H 2 O(l)  HCOOH(aq) + OH - determines the pH at equivalence Answer: 8.26