1. 1 Acid-Base Equilibria and Solubility Equilibria Chapter 16 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Homogeneous and Heterogeneous Solution Equilibria Homogeneous equilibria: one phase Weak acids and bases in solution do not ionize completely At equilibrium in solution –non-ionized acid HA, H + ions and conjugate base A - –example: CH 3 COOH, H + and CH 3 COO - –non-ionized base B, OH - ions and conjugate acid BH + –example: NH 3, OH - and NH 4 + Heterogeneous equilibria: more than one phase –solution and solid –precipitation of slightly soluble substances –effect of concentration and pH 2

3. The effect of addition of acid or base to … an unbuffered solution or a buffered solution acid addedbase added acid addedbase added Buffer Effect

4. 4 The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). CH 3 COONa (s) Na + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) common ion

5. 5 Consider mixture of salt NaA and weak acid HA. HA (aq) H + (aq) + A - (aq) NaA (s) Na + (aq) + A - (aq) K a = [H + ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a - log [HA] [A - ] -log [H + ] = -log K a + log [A - ] [HA] pH = pK a + log [A - ] [HA] pK a = -log K a Henderson-Hasselbalch equation pH = pK a + log [conjugate base] [acid]

6. 6 What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) 0.300.00 -x-x+x+x 0.30 - x 0.52 +x+x x0.52 + x Common ion effect 0.30 – x  0.30 0.52 + x  0.52 pH = pK a + log [HCOO - ] [HCOOH] HCOOH pK a = 3.77 pH = 3.77 + log [0.52] [0.30] = 4.01 Mixture of weak acid and conjugate base!

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8. 8 A buffer solution is a solution of: 1.A weak acid or a weak base and 2.The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Add strong acid H + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) Add strong base OH - (aq) + CH 3 COOH (aq) CH 3 COO - (aq) + H 2 O (l) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa

9. 9 HCl H + + Cl - HCl + CH 3 COO - CH 3 COOH + Cl -

10. 10 Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) HF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3 - is its conjugate acid buffer solution

11. 11 = 9.20 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH 4 + (aq) H + (aq) + NH 3 (aq) pH = pK a + log [NH 3 ] [NH 4 + ] pK a = 9.25 pH = 9.25 + log [0.30] [0.36] = 9.17 NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (moles) end (moles) 0.0290.001 0.024 0.0280.00.025 pH = 9.25 + log [0.25] [0.28] [NH 4 + ] = 0.028 0.10 final volume = 80.0 mL + 20.0 mL = 100 mL [NH 3 ] = 0.025 0.10

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13. Relationship between buffer capacity and pH change Change in pH after addition of base to an acetic acid buffer system

14. Buffer Capacity and Buffer Range Buffer capacity is the ability to resist pH change. Buffer range is the pH range over which the buffer acts effectively. The more concentrated the components of a buffer, the greater the buffer capacity. A buffer has the highest capacity when the component concentrations are equal. The pH of a buffer is distinct from its buffer capacity. Buffers have a usable range within ± 1 pH unit of the pK a of its acid component.

15. Buffer Capacity and Buffer Range pH of a buffer made of equal volumes of 1.0 M HA and 1.0 M A - pH of a buffer made of equal volumes of 0.1 M HA and 0.1 M A - pH is the same a)at 1.0 M of each pH = pK a + log[base]/[acid] = pK a + log [1.0M]/[1.0M] = pK a b)At 0.1 M each pH = pK a + log [0.1M]/[0.1M] = pK a c)At 0.01 M each pH = pK a + log [0.01M]/[0.01M] = pK a

16. Buffer Capacity and Buffer Range Buffer capacity of a buffer made of equal volumes of 1.0 M HA and 1.0 M A - Buffer capacity of a buffer made of equal volumes of 0.1 M HA and 0. 1 M A - Buffer capacities differ Buffer made of 1.000 M HA and 1.000 M A - [HA] init = [A-] init = 1.000 M  [A-] init /[HA] init = 1.000 add 0.010 mol OH - –[HA] final = 0.990 mol/L –[A-] final = 1.010 mol/L –[A-] final / [HA] final = 1.010/0.990 = 1.02 Apply Henderson-Hasselbalch equation –pH = = pK a + log[base]/[acid] = 4.74 + log (1.02) = 4.75 –Change in ratio of concentrations = (1.02-1.00)/1.00 x 100% = 2% –Change in pH = (4.75 – 4.74)/4.74 x 100% = 0.2% CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq)

17. Buffer Capacity and Buffer Range Buffer made of 0.100 M HA and 0.100 M A - [HA] init = [A-] init = 0.100 M  [A-] init /[HA] init = 1.000 add 0.010 mol OH - –[HA] final = 0.090 mol/L –[A-] final = 0.110 mol/L –[A-] final / [HA] final = 0.110/0.090 = 1.22 Apply Henderson-Hasselbalch equation –pH = 4.74 + log (1.22) = 4.83 –Change in ratio of concentrations = (1.22-1.00)/1.00 x 100% = 22% –Change in pH = (4.83-4.74)/4.74 x 100% = 1.9% CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq)

18. Buffer Capacity and Buffer Range Buffer made of 0.250 M HA and 1.750 M A - [HA] init = 0.250 [A-] init = 1.750 M  [A-] init /[HA] init = 7.00 add 0.010 mol OH - –[HA] final = 0.240 mol/L –[A-] final = 1.760 mol/L –[A-] final / [HA] final = 1.760/0.240 = 7.33 Apply Henderson-Hasselbalch equation –pH = 4.74 + log (7.33) = 5.61 –Change in ratio of concentrations = (7.33-7.00)/7.00 x 100% = 4.7% –Change in pH = (5.61-5.58)/5.58 x 100% = 0.5% CH 3 COOH(aq) + H 2 O(l) CH 3 COO - (aq) + H 3 O + (aq)

19. 19 Titrations (Review) In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink)

20. 20 Alternative Method of Equivalence Point Detection monitor pH

21. 21 Strong Acid-Strong Base Titrations NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) OH - (aq) + H + (aq) H 2 O (l)

22. 22 Weak Acid-Strong Base Titrations CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) At equivalence point (pH > 7):

23. 23 Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq) At equivalence point (pH < 7): H + (aq) + NH 3 (aq) NH 4 + (aq)

24. 24 Calculate the pH of an 0.150 M ammonium/ 0.075 M ammonia buffer solution Ka (NH 4 + ) = 5.6 x 10 -10 What is the pH after adding 25 mL 0.025 M HCl to 250 mL of the ammonium/ammonia buffer solution

25. 25 Exactly 100 mL of 0.10 M HNO 2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O (l) start (moles) end (moles) 0.01 0.0 0.01 NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) 0.050.00 -x-x+x+x 0.05 - x 0.00 +x+x xx [NO 2 - ] = 0.01 0.200 = 0.05 M Final volume = 200 mL K b = [OH - ][HNO 2 ] [NO 2 - ] = x2x2 0.05-x = 2.2 x 10 -11 0.05 – x  0.05x  1.05 x 10 -6 = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02

26. 26 Acid-Base Indicators HIn (aq) H + (aq) + In - (aq)  10 [HIn] [In - ] Color of acid (HIn) predominates  0.10 [HIn] [In - ] Color of conjugate base (In - ) predominates

27. 27 pH Solutions of Red Cabbage Extract

28. Acid-Base Indicators pH HInd H + + Ind -

29. 29 The titration curve of a strong acid with a strong base.

30. 30 Which indicator(s) would you use for a titration of HNO 2 with KOH ? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein

31. 31 Chemistry In Action: Maintaining the pH of Blood Red blood cells in a capillary