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Makeup Midquarter Exams Wed., Mar 9 5:30-7:30 pm 131 Hitchcock Hall You MUST Sign up in 100 CE

Makeup Midquarter Exams Wed., Mar 9 5:30-7:30 pm 131 Hitchcock Hall You MUST Sign up in 100 CE Please do so as soon as possible. Final Exams for Chem 122 - Mathews 2:30 Class: Thursday, Mar. 17 at 1:30 in IH 0100 6:30 Class: Tuesday, Mar. 15 at 5:30pm in MP 1000. Common Ion Effect :

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Makeup Midquarter Exams Wed., Mar 9 5:30-7:30 pm 131 Hitchcock Hall You MUST Sign up in 100 CE

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  1. Makeup Midquarter Exams Wed., Mar 9 5:30-7:30 pm 131 Hitchcock Hall You MUST Sign up in 100 CE Please do so as soon as possible. Final Exams for Chem 122 - Mathews 2:30 Class: Thursday, Mar. 17 at 1:30 in IH 0100 6:30 Class: Tuesday, Mar. 15 at 5:30pm in MP 1000

  2. Common Ion Effect: This is the same as other equilibrium calculations, except there is an initial concentration of two species prior to establishing equilibrium.

  3. Consider a solution which is prepared by mixing 0.10 mole of acetic acid and 0.010 mole of HCl in enough water to make 1.000 L of solution. What are the concentrations of each species? Note that for a solution of 0.10 M acetic acid alone, [OAc- ] = [H3O+] = 1.3 x 10-3 M

  4. Consider a solution made from dissolving 0.30 moles of acetic acid and 0.30 moles of sodium acetate in enough water to make 1.00 L of solution. What is the concentration of Hydronium ions and the pH? HOAc + H2O = H3O+ + OAc- i 0.30 M ~ 0 0.30 M δ -x x x eq (0.30-x) x (0.30+x) Ka = 1.8 x 10-5 = [(x)(0.30+x)] / [(0.30-x)] leads to x = [H3O+ ] = 1.8 x 10-5 M compared to 2.3 x 10-3 M for the acid alone!!!

  5. Consider 1.00L of solution containing 1.00 mol of formic acid (HCOOH) and 0.500 mol of sodium formate (NaHCOO). What is the pH? Ka = 1.77 x 10-4 1.77 x 10-4 = (x)(0.500+x) / (1.00-x) ≈ (x)(0.500) / (1.00) so x = 3.54 x 10-4 = [H3O+] and pH = 3.45

  6. Note, however, that we’re doing the same things again! The expression Ka = [H3O+][A-] / [HA] can be written [H3O+] = Ka [HA] / [A-] ≈ Ka [HA]i / [A-]i or take the negative log of both sides: -log [H3O+] = -logKa - log [HA]i / [A-]i or pH = pKa - log [HA]i / [A-]i or pH = pKa + log[A-] / [HA] Henderson-Hasselbalch Equation(s) Valid ONLY when [H3O+] and [OH-] are very small in comparison to [A-]i and [HA]i

  7. Now let’s add 0.10 mol of strong acid (eg HCl) to the solution (without changing its volume). And pH =pKa+log(0.400)/(1.10) = 3.75+(-0.44) = 3.31=pH or [H3O+] = 4.89x10-4

  8. A similar calculation results when we add enough strong base to the original buffer solution to make the solution 0.10 M in OH- . In this case, [H3O+] = 2.66x10-4 and pH = 3.58

  9. Final Exams for Chem 122 - Mathews 2:30 Class: Thursday, Mar. 17 at 1:30 in IH 0100 6:30 Class: Tuesday, Mar. 15 at 5:30pm in MP 1000

  10. Titration curves: Strong Acid – Strong Base titrate 100.0 mL of 0.1000 M HCl with 0.1000 M NaOH Recognize that if c0 and Vo = concentration and volume of initial acid(basd) and ct and Vt = concentration and volume of base (acid) added Then the ‘equivalence point’ corresponds to the condition coVo = ctVe, where Ve = Vt when nacid = nbase

  11. 1. V = 0 mL of base added n(H30+) = (0.1000M)(0.1000 L) = 1.000 x 10-2 mol H30+ and pH = 1.00 2. V = 30 mL of NaOH added (i.e. 0 < V < Ve ) n(OH-) = (0.1000 M)(0.0300 L) = 3.000 x 10-3 mol OH- which reacts with (neutralizes) the same amount of H30+ i.e. n (H+) = (1.000x10-2 – 3.000x10-3) mol H30+ = 7.000 x 10-3 mol H30+ remaining (+ 3.000 mol Na+ and Cl-) and [H30+] = (7.000x10-3)/(Va + Vb) = (7.000 x 10-3) / (0.1300 L) = 0.05385 M and pH = 1.27

  12. 3. V = 100.0 mL NaOH added (i.e. Vb = Ve ) n(OH- added) = (0.1000 M)(0.1000 L) = 1.000 x 10-2 mol which neutralizes the same amount (all) of H30+ . The significant reaction that has taken place is H+ + OH- H2O therefore [OH-] = [H30+] = 1.000 x 10-7 M and pH = 7.00 Note also that [Na+] = [Cl-] = (1.000 x 10-2 mol) / (0.2000 L) = 0.5000 M

  13. SA – SB, continued: 4. V = 100.05 mL (i.e. Vb > Ve) also, this is about one drop excess easiest to treat by realizing that 100.00 mL gave the result of previous calculation, i.e. neutralized the acid. then n (OH- excess) = [(100.05 - 100.00)mL][0.1000 M] = 5.0 x 10-6 mol OH- ‘extra’ and [OH-] = (5.0 x 10-6 mol) / (100.00 + 100.05) = 2.5 x 10-5 M OH- so that pOH = 4.60 and pH = 9.40 similar calculations can proceed for all excess base

  14. Titration of Strong Acid with Strong Base

  15. Weak Acid + Strong Base 100.0 mL 0.1000 M HOAc titrated with 0.1000 M NaOH Note the definition of the equivalence point remains the same. 1. V = 0.00 mL NaOH usual problem of weak acid, with Ka = 1.8 x 10-5 n(HOAc) = (0.1000 M)(0.1000 L) = 1.000 x 10-2 mol HOAc so [H30+] = 1.3 x 10-3 M and pH = 2.88

  16. 2. 0 < V < Ve acid will be partially neutralized. Since OH- is a stronger base than water, HOAc + OH- = H2O + OAc- K = 1/Kb = Ka / Kw = 2 x 109 >>1 example, 30 mL of NaOH n(OH-) = (0.1000 M)(0.0300 L) = 3.000 x 10-3 mol OH- n(HOAc) = (0.1000 M)(0.1000 L) = 1.000 x 10-2 mol HOAc so after all the OH- has reacted with HOAc, n(OAc-) = n(OH- added) = 3.00 x 10-3 mol n(HOAc remaining) = (1.000x10-2 – 3.000x10-3) mol HOAc = 7.000 x 10-3 mol HOAc

  17. 2. (continued) [OAc-] = (3.000 x 10-3 mol OAC-) / (Va + Vb) = (3.000 x 10-3 mol OAC-) / (0.1300 L) = 2.31 x 10-2 M OAc- and [HOAc] = (7.000 x 10-3) / (0.1300 L) = 5.38 x 10-2 M HOAc Now we’re able to calculate the pH of this buffer soln pH = pKa + log{[OAc-] /[HOAc]} = 4.75 + log{(2.31 x 10-2) / (5.38x10-2)} pH = 4.38 Notice also that when Vb = ½ Va, [HA]=[A-] and pH=pKa

  18. WA-SB titration (cont) 3. V = 100mL NaOH, V = Ve we now have n(NaOAc) = n(OAc-] = (0.1000M)(0.1000L) = 1.000 x 10-2 mol OAc- which is the same as having 1.000 10-2 mol of NaOAc in soln [OAc-] = (1.000x10-2) / (0.2000 L) = 0.05000 M As before, we can calculate the pH based on hydrolysis OAc- + H2O = HOAc + OH- Kb = Kw/Ka = 5.7 x 10-10 and pH = 8.73

  19. WA-SB titration (cont) 4. V >100mL NaOH, i.e, V = Ve just adding more OH- to the solution of NaOAc, results primarily in the [OH-] being dominated by the NaOH being added i.e. this portion of the titration curve looks very similar to the SA-SB titration curve

  20. Titration of Weak Acid with Strong Base, Ve = 50 mL

  21. Titration of Weak Acid with Strong Base, Ve = 50 mL pH=pKa

  22. Anything wrong with this picture???

  23. Discuss weak polyprotic acids, eg H2CO3

  24. Weak polyprotic acids: typical H2CO3 and H3PO4 H2CO3 + H2O = H3O+ + HCO3- Ka1 = 4.3 x 10-7 HCO3- + H2O = H3O+ + CO32- Ka2 = 4.8 x 10-11 Example: Aqueous sol’n sat’d in CO2; [H2CO3] = 0.034 M (1)  [H3O+] = [HCO3-] = 1.2 x 10-4; [H2CO3] = 0.034 M (2)  [H3O+] = [HCO3-] = same ; [CO32-] = 4.8 x 10-11 Note also we can rewrite the eq expressions:

  25. Example: at pH = 10.00, [H3O+] = 1.00 x 10-10

  26. We can calculate the fraction of any species, eg H2CO3

  27. similarly for HCO3- = 0.68 and CO32- = 0.32 And, this could be repeated for all pH values

  28. 100 mL of 0.1000 M H3PO4 titrated with NaOH pKa1 = 2.27, pKa2 = 7.21, pKa3 = 12.25

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