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Chapter 17: Additional Aqueous Equilibria

Chemistry: The Molecular Science Moore, Stanitski and Jurs. Chapter 17: Additional Aqueous Equilibria. Buffer Solutions. # #. * *. # 5 pH units~100,000X change in [H 3 O + ] *added 10 mL of 1.0 M HCl (to 1.00L buffer (0.50mol HOAc/NaOAc). Buffer Solutions.

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Chapter 17: Additional Aqueous Equilibria

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  1. Chemistry: The Molecular Science Moore, Stanitski and Jurs Chapter 17:Additional Aqueous Equilibria

  2. Buffer Solutions # # * * # 5 pH units~100,000X change in [H3O+] *added 10 mL of 1.0 M HCl (to 1.00L buffer (0.50mol HOAc/NaOAc)

  3. Buffer Solutions A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. Buffers contain either a weak acid and its conjugate base (or a weak base and its conjugate acid) in approximately equal proportion and in equilibrium (they do not react with each other).

  4. Buffer Solutions • Consider a buffer with equal molar amounts of HA and its conjugate base A-. • When H3O+ is added to the buffer it reacts with the base A-. • When OH- is added to the buffer it reacts with the acid HA.

  5. Human blood has a normal pH of 7.40  0.05. Below 7.35 - acidosis, Above 7.45 - alkalosis – both life threatening conditions. The most important blood buffer is provided by CO2 CO2(aq) + H2O(l) H2CO3(aq) H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)

  6. The pH of Buffer Solutions Calculate the pH of a solution that contains 0.0025 mol carbonic acid and 0.025 mol of hydrogen carbonate ion in 1.00L of water. The Ka of carbonic acid is 4.2 × 10-7 H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq) To calculate pH we need the concentration of hydronium ion in the solution

  7. H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq) 0.025 0.0025 0 -x +x +x Initial Change Equilibrium 0.0025 - x x 0.025 + x pH = - log [H3O+] = - log (4.2 x 10-8) = 7.38

  8. Consider a buffer of a weak acid HA and its conjugate base A-. The acid ionization equilibrium is: The pH of Buffer Solutions The Henderson-Hasselbalch Equation • The acid ionization constant is:

  9. The pH of Buffer Solutions The Henderson-Hasselbalch Equation

  10. The Henderson-Hasselbalch Equation OR pH = pKa when [HA] = [A-], since log(1)=0 Useful buffer range: pH = pKa ±1 (10:1 or 1:10 ratio of [A-]/[HA]).

  11. Example Calculate the pH of a solution that contains 0.0025 mol carbonic acid and 0.025 mol of hydrogen carbonate ion in 1.00L of water. The Ka of carbonic acid is 4.2 × 10-7 H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)

  12. Buffer Solutions A buffer that is needed for maintaining a desired pH can be chosen by examining pKa values

  13. Example – Prepare a pH=4 buffer From table 17.1 lactic acid, CH3CHOHCOOH has the closest value pKa = 3.85. What proportion of CH3CHOHCOO-/CH3CHOHCOOH should you use? CH3CHOHCOO-/CH3CHOHCOOH = 1, pH=pKa CH3CHOHCOO-/CH3CHOHCOOH <1, more acidic – pH<3.85 CH3CHOHCOO-/CH3CHOHCOOH >1, more basic – pH>3.85

  14. Example – Prepare a pH=4 buffer Using Henderson-Hasselbach equation…

  15. Example The concentration of lactate ion has to be 1.4 times that of lactic acid. For example if we use 0.10 molar lactic acid we will need 0.14 molar lactate ion from a soluble salt such as sodium lactate, CH3CHOHCOONa.

  16. Preparing Buffers There are two ways of preparing a buffer: • Combine the conjugate acid/base pairs directly. 0.25 M CH3CHOHCOOH combined with about 0.25 M NaCH3CHOHCOO 2. Start with one of the conjugate pairs and add strong acid or strong base to create the other. • Begin with 0.50 M CH3CHOHCOOHsolution. • Add 0.25 M NaOH. • This produces a solution containing: 0.25 M CH3CHOHCOOH + 0.25 M CH3CHOHCOO - CH3CHOHCOOH + OH - CH3CHOHCOO - + H2 O

  17. Practice • 1. (17-21) To buffer a solution at a pH of 4.57, what mass of sodium acetate (NaCH3COO). would you add to 500. mL of a 0.150 M solution of acetic acid (CH3COOH). • A. Choose an acid/base combination to prepare a pH=7 buffer. • B. Describe how you would prepare the buffer by starting with 1.0L of a 0.10 M solution of the weak acid and adding 1.0M NaOH.

  18. Buffer Capacity Buffer Capacity- the ability of the buffer to prevent changes in pH when small amounts of acid or base are added. The amount of an acid or a base that can be added to a buffer without causing a significant pH change ( more than 1 unit). The ratio of [A-]/[HA] determines the pH of the buffer whereas the magnitude of [A-] and [HA] determine the buffer capacity.

  19. Buffer Capacity 1 mol of acetate ion + 1 mol of acetic acid

  20. Buffer Capacity • A 1 liter solution contains: • 0.390 M hydrocyanic acid + 0.350 M sodium cyanide. • What is the pH of the buffer? Ka HCN = 4.0x10-10 • How will the pH change if we add 0.200 moles of NaOH? • (neglect the change in volume) HCN(aq) + H2O H3O+ + CN-(aq) BEFORE: 0.390 M 0.350 M

  21. Buffer Capacity 2. How will the pH change if we add 0.200 moles of NaOH? (neglect the change in volume) HCN(aq) + H2O H3O+ + CN-(aq) BEFORE: 0.390 M 0.350 M AFTER: -0.200 M +0.200 M I C E 0.190 0 0.550 -x +x + x 0.190-x + x 0.550 + x

  22. Buffer Capacity HCN(aq) + H2O H3O+ + CN-(aq) E 0.190-x + x 0.550 + x [H3O+] = 1.4 x 10-10 The pH value changes slightly. The buffer capacity has not been exceeded pH = -log (1.4 x 10-10) = 9.86

  23. Buffer Capacity A 1 liter solution contains 0.390 M hydrocyanic acid and 0.350 M sodium cyanide. Ka = 4.0 x 10-10 3. What will be pH if we add 0.450 moles of hydrochloric acid? (neglect the change in volume) 4. Will the buffer capacity be exceeded? HCN(aq) + H2O H3O+ + CN-((aq) BEFORE: 0.390 M 0.350 M AFTER: +0.450 -0.450 0 0.740 I C E 0.100 -x +x + x 0.740-x 0.100 + x + x

  24. Buffer Capacity HCN(aq) + H2O H3O+ + CN-(aq) E 0.740-x 0.100 + x + x [H3O+] = 0.100+3.0x10-9 The pH value changes significantly. The buffer capacity has been exceeded. pH = -log 0.100 = 1.000

  25. Example - Buffer Capacity A buffer is prepared using 0.25 mol H2PO4- and 0.15 mol HPO42- in 1000.0 mL solution. pKa of H2PO4- = 7.21 a.) What is the pH of the buffer? b.) What will be the pH of a solution if 6.2 g KOH are added to this buffer. Will the buffer capacity be exceeded? c.) What will be the pH of a solution if 20.0 mL of 6.00 M HCl are added to this buffer. Will the buffer capacity be exceeded? H2PO4-(aq) + H2O(l) H3O+(aq)+ HPO42-(aq) We can calculate the pH of the buffer using Henderson-Hasselbach equation:

  26. Moles of KOH: a.) What will be the pH of a solution if 6.2 g KOH are added to this buffer. Will the buffer capacity be exceeded? H2PO4-(aq) + H2O(l) H3O+ + HPO42-(aq) Before adding base: Reaction with base: After adding base: 0.25 0.15 -0.11 +0.11 0.14 0.26 The buffer capacity was not exceeded

  27. What will be the pH of a solution if 20.0 mL of 6.00 M HCl are added to this buffer. Will the buffer capacity be exceeded? Moles of HCl: b.) mol = M × V = 6.0M × 0.0200L = 0.12 mol HCl H2PO4-(aq) + H2O(l) H3O+ + HPO42-(aq) Before adding acid: Reaction with acid: After adding acid: 0.25mol 0.15mol + 0.12mol - 0.12mol 0.37mol 0.03mol The buffer capacity was not exceeded

  28. Buffer Capacity A pH=4.0 buffer was prepared with 0.10 mol lactic acid and 0.14 mole sodium lactate in 1.00L solution. The Ka of lactic acid is 1.4x10-4. • Calculate the pH after addition of 4.2mL of 12M HCl. • Calculate the pH of the buffer after 6.25mL of 8M NaOH to the original buffer. (In both cases, neglect volume changes.)

  29. Acid-Base Titration Equivalence point – is reached when the stoichimetric amount of titrant (solution in the buret) has been added. End point – occurs when the indicator changes color. We want to use indicators that give an end point as close as possible to the equivalence point.

  30. Acid-Base Titration Curves An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid). • Such curves are used to gain insight into the titration process. • We can use titration curves to choose an appropriate indicator that will show when the titration is complete.

  31. Titration Curve for a strong acid by a strong base.

  32. Example – strong acid and strong base A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M HCl. Calculate the pH: a.) before the titrant is added b.) After 40.0mL of titrant has been added c.) After 50.0mL of titrant has been added d.) After 50.2mL of titrant has been added HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) H+(aq) + OH-(aq)  H2O(l)

  33. Example – strong acid and strong base A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M HCl. Calculate the pH: a.) before the titrant is added HCl is a strong acid, the H3O+ concentration = HCl concentration pH = -log[H3O+] = -log(0.100) = 1.00

  34. Calculate moles of acid: nacid = 0.0500L × 0.100 M = 5.00 x 10-3 mol Calculate moles of base: nbase = 0.0400L × 0.100 M = 4.00 x 10-3 mol b.) After 40.0mL of titrant (NaOH) has been added 0.00400 moles of base reacts with 0.00400 mol of acid 0.00500 molacid- 0.00400 molbase = 1.00 x 10-3mol acid left pH = -log[H3O+] = -log(0.0111) = 1.95

  35. A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M HCl. Calculate the pH: c.) After 50.0mL of titrant has been added Calculate moles of acid: nacid = 0.0500L × 0.100 M = 5.00 x 10-3 mol Calculate moles of base: nbase = 0.0500L × 0.100 M = 5.00 x 10-3 mol 5.00 x 10-3 mol of base react with 5.00 x 10-3 mol mol of acid HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) This is the equivalence point. The salt produced is a neutral salt so the pH is 7.00

  36. A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M HCl. Calculate the pH: d.) After 50.2mL of titrant has been added. Calculate moles of acid: nacid = 0.0500L × 0.100 M = 5.00 x 10-3 mol Calculate moles of base: nbase = 0.0502L × 0.100 M = 5.02 x 10-3 mol 5.02 x 10-3 mol of base react with 5.00 x 10-3 mol of acid 5.02 x 10-3 molbase- 5.00 x 10-3 molacid = 0.02 x 10-3mol base pOH = -log(2.00 x 10-4)=3.70 pH = 14.00 - 3.70 = 10.3

  37. Titration Curve for a strong acid by a strong base.

  38. Titration Curve for a weak acid by a strong base.

  39. Example – weak acid and strong base A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M HCH3COO. Calculate the pH: a.) before the titrant is added b.) After 40.0mL of titrant has been added c.) After 50.0mL of titrant has been added d.) After 50.2mL of titrant has been added

  40. Example – weak acid and strong base A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M HCH3COO. Calculate the pH: a.) before the titrant is added CH3COOH(aq) + H2O(l) CH3COO-(aq) +H3O+(aq) b.) After 40.0mL of titrant has been added CH3COOH(aq) + H2O(l) CH3COO-(aq) +H3O+(aq) Buffer!

  41. c.) After 50.0mL of titrant has been added CH3COOH(aq) + OH-(aq) CH3COO-(aq) +H2O(l) d.) After 50.2mL of titrant has been added solution has xs OH-

  42. a.) CH3COOH is a weak acid: -x +x +x 0.100-x x x 1.8 × 10 -5 = x = [H3O+] = 1.34 × 10-3 pH = -log(1.34 × 10-3) = 2.87

  43. b.) After 40.0 mL of titrant has been added - moles of acid: nacid = 0.0500L × 0.100 M = 5.00 x 10-3 mol moles of base (NaOH) added = moles of base (acetate ion) formed: nbase = 0.0400L × 0.100 M = 4.00 x 10-3 mol CH3COO- moles of acid left: nacid = 5.00 x 10-3 – 4.00 x 10-3 = 1.00 x 10-3 mol CH3COOH A BUFFER!

  44. 1.00 x 10-3mol 4.00 x 10-3 Ka for acetic acid = 1.8 × 10-5, pKa = 4.74

  45. A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M CH3COOH. Calculate the pH: c.) After 50.0 mL of titrant has been added moles of acid: nacid = 0.0500L × 0.100 M = 0.00500 mol moles of base (acetate ion) formed: nbase = 0.0500L × 0.100 M = 0.00500 mol We are at the equivalence point now. There is no original acetic acid left. The acetate ion will hydrolize: We need to calculate the concentration of the hydroxide ion to calculate the pH.

  46. -x +x +x 0.0500-x x x Remember to use the total volume. M = 0.00500mol/0.100L = 0.0500M x = [OH-] = 5.3 × 10-6 pH = 14.00 – pOH = 14.00 – [-log(5.3 × 10-6)] = 14.00 – 5.28 = 8.72 Notice that the pH is not 7 since CH3COONa is a basic salt.

  47. A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M CH3COOH. Calculate the pH: d.) After 50.2mL of titrant has been added Calculate moles of acid: nacid = 0.0500L × 0.100 M = 5.00 x 10-3 mol Calculate moles of base: nbase = 0.05020L × 0.100 M = 5.02 x 10-3 mol 5.00 x 10-3 mol of base react with 5.00 x 10-3 mol of acid 5.02 x 10-3 molbase- 5.00 x 10-3 molacid = 2 x 10-5 mol base pH = 14.00 – pOH = 14.00 -log[OH-] = -log(2.00 x 10-4) = 10.3

  48. Curve for the titration of a weak acid by a strong base.

  49. Titration of a weak acid by a strong base. • Before the titration the pH is higher for a weak acid; the acid and its conjugate base are at equilibrium • At the equivalence point pH is higher because of the hydrolysis of a basic salt formed. pH=pKa A slighter change in pH requires careful choice of indicator.

  50. Curve for the titration of a weak base by a strong acid.

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