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Chapter 15 Acids and Bases

Chapter 15 Acids and Bases. Acids and Bases Review. conjugate base. conjugate acid. Bronsted Acid = proton donor. strong acid: HCl + H 2 O → H 3 O + + Cl -. weak acid: HCN + H 2 O ↔ H 3 O + + CN -. Bronsted Base = proton acceptor.

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Chapter 15 Acids and Bases

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  1. Chapter 15 Acids and Bases

  2. Acids and Bases Review conjugate base conjugate acid Bronsted Acid = proton donor strong acid: HCl + H2O → H3O+ + Cl- weak acid: HCN + H2O ↔ H3O+ + CN- Bronsted Base = proton acceptor weak base: NH3 + H2O ↔ NH4+ + OH- strong base: NaOH → Na+ + OH- OH- + H3O+ ↔ 2 H2O

  3. Ionization of water H2O + H2O ↔ H3O+ + OH- K' = [H3O+] [OH-] [H2O]2the [H2O] is constant and added to K' Kw = [H3O+] [OH-] Abbreviated form …… H2O ↔ H+ + OH- Kw = [H+] [OH-] = 1.0 x 10-14 at 25 ˚C In pure water [H+] = [OH-] = 1.0 x 10-7 at 25 ˚C

  4. Kw = [H+] [OH-] = 1.0 x 10-14 at 25 ˚C pH = - log [H+] pH of strong acids: What is the pH, [H+], and [OH-] of … 0.0035 M HCl Initial 0.0035 M 0.0 M 0.0 M HCl → H+ + Cl- Final 0.00 M 0.0035 M 0.0035 M pH = -log(0.0035) = 2.46 How does adding HCl to water change the [OH-]? a) stays same b) increases c) decreases 0.0035 • [OH-] = 1.0 x 10-14 [OH-] = 2.86 x 10-12

  5. What is the [H+] and [OH-] in a) lemon juice b) milk of magnesia lemon juice pH = 2.4 = -log [H+] [H+] = Inv log (-2.4) = 10-2.4. [H+] = 4 x 10-3 M [OH-] [H+] = 1.00 x 10-14. [OH-]= 1.00 x 10-14/4 x 10-3M = 2.5 x 10-12. Or …….. pOH = 14.0 – 2.4 = 11.6 [OH-] = 10-11.6 = 2.5 x 10-12.

  6. Ba(OH)2(s) Ba2+(aq) + 2OH-(aq) Initial 0.0035 M 0.0 M 0.0 M HCl → H+ + Cl- pOH = -log [OH-] = 11.544 pH + pOH = 2.456 + 11.544 = 14.000 Final 0.00 M 0.0035 M 0.0035 M pH = -log(0.0035) = 2.456 [OH-] = 2.86 x 10-12 [H+] [OH-] = 1.0 x 10-14 and pH + pOH = 14.00 What is the pH of a 1.8 x 10-2 MBa(OH)2 solution? Ba(OH)2 is a strong base – 100% dissociation. 0.0 M 0.0 M Initial 0.018 M Equil 0.00 M 0.018 M 0.036 M pOH = - log (0.036) = 1.40 & pH = 14.00 – 1.40 = 12.60

  7. What is the conjugate base of perchloric acid HClO4? a) H2ClO4 b) ClO4- c) ClO4 d) Cl-

  8. Determining the pH of weak acids 1. Write the ionization reaction: HA ↔ H+ + A- 2. Write Ka expression: Ka = [H+] [A-] [HA] • Set up an ICE chart: [H+] [A-] [HA] • I 0 0 0.02 • C +x +x -x • E x x 0.02 - x 4. Solve for x Ka= ___x2__ (0.02-x) 5. Determine [H+] and pH Apply to specific weak acid

  9. What is the [H+] and [OH-] in a) lemon juice b) milk of magnesia lemon juice pH = 2.4 = -log [H+] [H+] = Inv log (-2.4) = 10-2.4. [H+] = 4 x 10-3 M [OH-] [H+] = 1.00 x 10-14. [OH-]= 1.00 x 10-14/4 x 10-3M = 2.5 x 10-12. Or …….. pOH = 14.0 – 2.4 = 11.6 [OH-] = 10-11.6 = 2.5 x 10-12.

  10. Determining the pH of weak acids 1. Write the ionization reaction: HA ↔ H+ + A- 2. Write Ka expression: Ka= [H+] [A-] [HA] • Set up an ICE chart: [H+] [A-] [HA] • I 0 0 0.02 • C +x +x -x • E x x 0.02 - x 4. Solve for x Ka= ___x2__ (0.02-x) 5. Determine [H+] and pH Apply to specific weak acid

  11. Example 0.050M ascorbic acid: C6H8O6 - Ka = 8.3 x 10-5. 1. Write the ionization reaction: C6H8O6↔ H+ + C6H7O6 - 2. Write Ka expression: Ka = [H+] [C6H7O6 -] [C6H8O6] Set up an ICE chart: C6H8O6↔ H+ + C6H7O6 - I 0.050 0 0 C -x +x +x E 0.050 – x x x .00204/.05 • 100 = 4.1 % 4. Solve for x 8.3 x 10-5= x2/(0.050-x) x2 = 4.15 x 10-6. x = [H+] = 2.04 x 10-3. 5. Determine [H+] and pH pH = - log(2.04 x 10-3) = 2.69 Was it ok to ignore the value of x in the denominator? Use 5% rule. a) yes b) no

  12. Determining the pH of weak acids – 0.026 M acetic acid 1. Write the ionization reaction: CH3COOH ↔ H+ + CH3COO- 2. Write Ka expression: 1.85 x 10-5= [H+] [CH3COO-] [CH3COOH] • Set up an ICE chart: [H+] [CH3COO-] [CH3COOH] • I 0 0 0.026 • C +x +x -x • E x x 0.026 - x 4. Solve for x 1.85 x 10-5= ___x2__ (0.026-x) 5. Determine x (= [H+]) and pH Apply the 5% rule assume x << 0.026 x2 = 1.85 x 10-5 • 0.026 = 4.81 x 10-7 x = 6.94 x 10-4 = [H+] 0.000694/.026 x 100 = 2.7% < 5% ok to ignore x pH = - log 0.000694 = 3.16 sig figs & pH approx. Try 0.037 M HF Ka = 7.1 x 10-4

  13. Determining the pH of weak bases – 0.13 M ammonia 1. Write the ionization reaction: NH3 + H2O ↔ NH4+ + OH- 2. Write Kb expression: 1.85 x 10-5= [NH4+] [OH-] [NH3] • Set up an ICE chart: [NH4+] [OH-] [NH3] • I 0 0 0.13 • C +x +x -x • E x x 0.13 - x 4. Solve for x 1.85 x 10-5= ___x2__ (0.13-x) 5. Determine x (= [OH-]) and pOH/pH Apply the 5% rule assume x << 0.13 x2 = 1.85 x 10-5 • 0.13 = 2.41 x 10-6 x = 1.55 x 10-3 = [OH-] 0.00155/.13 x 100 = 1.2% < 5% ok to ignore x pOH = - log 0.00155 = 2.81 & pH = 11.19

  14. Salts of weak acids 0.15 M NaCH3COO (sodium acetate) Is sodium acetate soluble in water? a) yes b) no Na(CH3COO-)  Na+ + CH3COO- An aqueous solution of sodium acetate will be ….. a) acidic b) basic c) neutral CH3COO- + H2O CH3COOH+ OH- Kb = [OH-][CH3COOH]• Ka = [H+] [CH3COO-] [CH3COO-] [CH3COOH] Kb• Ka= [OH-] • [H+] Ka(acetic acid) = 1.8 x 10-5 Kw= 1 x 10-14 = KaKb so Kb (acetate) = Kw/Ka = 5.6 x 10-10

  15. Salts of weak acids are bases referred to as the conjugate base 0.15 M NaCH3COO (sodium acetate) Na(CH3COO-)  Na+ + CH3COO- CH3COO- + H2O CH3COOH+ OH- I 0.15 0 0 C -x +x +x E 0.15 - x xx Kb = [OH-][CH3COOH] = 5.6 x 10-10 [CH3COO-] 5.6 x 10-10 = x2/(0.15 – x) where x = [OH-] = [CH3COOH] x = [OH-] = 9.16 x 10-6 M (0.6% error)pOH = 5.04 pH = 8.96

  16. Try 0.037 M HF Ka = 7.1 x 10-4 HF ↔ H+ + F- Ka = 7.1 x 10-4 = [H+] [F-]/[HF] I 0.037 0 0 C -x +x +x E 0.037 – x x x Ka = 7.1 x 10-4= x2/(0.037 – x) x = 0.0051 M = [H+] however .0051/.037 * 100 = 13.8 % (use quadratic) pH = 2.29 Using quadratic x2 + 0.00071x – 2.63 x 10-5 = 0 x = 0.0048 M = [H+] & pH = 2.32

  17. Salts of weak bases NH3 Kb = 1.8 x 10-5 A solution of 1.5 M ammonium chloride would be …. a) basic b) acidic c) neutral

  18. Polyprotic acids – An acid with 2 or more acidic protons

  19. Polyprotic acids – An acid with 2 or more acidic protons e.g. Phosphoric acid (H3PO4) H3PO4+ H2O H2PO4- + H3O+ Ka1 = 7.5 x 10-3 H2PO4-+ H2O HPO42- + H3O+ Ka2 = 6.2 x 10-8 HPO42-+ H2O PO43- + H3O+ Ka3 = 3.6 x 10-13 What is pH and [ ]s of all species in 0.0011 M H3PO4? This is the [H3PO4] found in a can of cola Use Equilibrium #1 Ka expression to find [H2PO4-] & [H3O+]. Use Equilibrium #2 Ka expression & [H2PO4-] from #1 to find additional [H3O+] and [HPO4-]. Use Equilibrium #3 Ka expression & [HPO4=] to find [PO43-] & additional [H3O+].

  20. 0.0011M Phosphoric acid (H3PO4) H3PO4+ H2O H2PO4- + H3O+ Ka1 = 7.5 x 10-3 I 0.0011 0 0 • C -x1 + x1 +x1 E 0.0011 – x1x1x1 H2PO4-+ H2O HPO42- + H3O+ Ka2 = 6.2 x 10-8 HPO42-+ H2O PO43- + H3O+ Ka3 = 3.6 x 10-13 Use Equilibrium #1 Ka expression to find [H2PO4-] & [H3O+]. Ka1 = 7.5 x 10-3 = [H2PO4-] [H3O+] = ____x12__ [H3PO4] (0.0011 – x1) Need to use quadratic to calculate x1: x12= 0.0011 • 7.5 x 10-3 - 7.5 x 10-3x1 and 0 = x12 + 7.5 x 10-3x – 8.25 x 10-6 x1= -7.5 x 10-3 ± {(7.5 x 10-3)2 – 4 • -8.25 x 10-6}1/2=9.7 x 10-4 M 2 [H2PO4-] = [H3O+] = x1 = 9.7 x 10-4 M [H3PO4] = 0.0011M – 0.00097 M = 0.00013 M H2PO4- is a) an acid b) a base c) both d) neither

  21. 0.0011M Phosphoric acid (H3PO4) H2PO4-+ H2O HPO42- + H3O+ Ka2 = 6.2 x 10-8 I 0.00097 0 0 • C -x2 + x2 +x2 E 0.00097 – x2x2x2 HPO42-+ H2O PO43- + H3O+ Ka3 = 3.6 x 10-13 [H3O+] = [H2PO4-] = 9.7 x 10-4 M from step #1 Use Equilibrium #2 Ka expression & [H2PO4-] from #1 to find additional [H3O+] and [HPO4-]. Ka2 = 6.2 x 10-8 = [H3O+][HPO42-] = (9.7 x 10-4 + x2) • x2 [H2PO4-] (9.7 x 10-4 – x2) Assume x2 << 9.7 x 10-4 ….. x2 = 6.2 x 10-8 = [HPO42-] [H3O+] = x1 + x2 = 0.00097 + 0.000000062 = 0.00097 M

  22. 0.0011M Phosphoric acid (H3PO4) H3PO4+ H2O H2PO4- + H3O+ Ka1 = 7.5 x 10-3 [H3PO4] = 0.00013 M H2PO4-+ H2O HPO42- + H3O+ Ka2 = 6.2 x 10-8 [H2PO4-] = 0.00097 M HPO42-+ H2O PO43- + H3O+ Ka3 = 3.6 x 10-13 [HPO42-] = 6.2 x 10-8 M [PO43-] = 2.3 x 10-17 M [H3O+] = x1 + x2 = 0.00097 M x2 = 6.2 x 10-8 = [HPO42-] Use Equilibrium #3 Ka expression & [HPO4=] to find [PO43-] & additional [H3O+]. Ka3 = 3.6 x 10-13 = [PO43-] [H3O+] = (0.0097 + x3) • x3 [HPO42-] (6.2 x 10-8 – x3) Assume x3 << 6.2 x 10-8 ….. X3 = 3.6 x 10-13• 6.2 x 10-8 = 2.3 x 10-17 M 0.00097 [H3O+] = x1 + x2 + x3 = 0.00097 M pH = - log [H3O+] = - log 0.00097 = 3.01

  23. Polyprotic acids – An acid with 2 or more acidic protons e.g. Phosphoric acid (H3PO4) H3PO4+ H2O H2PO4- + H3O+ Ka1 = 7.5 x 10-3 H2PO4-+ H2O HPO42- + H3O+ Ka2 = 6.2 x 10-8 HPO42-+ H2O PO43- + H3O+ Ka3 = 3.6 x 10-13 A solution of NaH2PO4 is …. a) acidic b) basic c) neutral

  24. What is the pH and [ ]s of all species in 0.044 M sulfurous acid, H2SO3? Ka1 = 1.3 x 10-2 & Ka2 = 6.3 x 10-8

  25. Polyprotic acids – An acid with 2 or more acidic protons

  26. What is the pH and [ ]s of all species in 0.044 M sulfurous acid, H2SO3? H2SO3↔ H+ + HSO3-Ka1 = 1.3 x 10-2. Get as far as you can toward solving this in 5 minutes. What is the pH of the solution? a) 1.74 b) 2.45 c) 5.33 HSO3-↔ H+ + SO32-Ka2 = 6.3 x 10-8. At equilibrium, what species is present in the highest [ ]? a) H2SO3b) HSO3- c) SO32-

  27. What is the pH and [ ]s of all species in 0.044 M sulfurous acid, H2SO3? H2SO3↔ H+ + HSO3-Ka1 = 1.3 x 10-2. I 0.044 0 0 C -x +x +x E 0.044 - x xx x = 0.018M (using quadratic) = 0.024M (assuming x is small) HSO3-↔ H+ + SO32-Ka2 = 6.3 x 10-8. I 0.018 0.018 0 C -x +x +x E 0.018 - x 0.018 + x x x = 6.3 x 10-8M pH = 1.74

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