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Chapter 13 Applications of Aqueous Equilibria

Chapter 13 Applications of Aqueous Equilibria

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Chapter 13 Applications of Aqueous Equilibria

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  1. Chapter 13Applications of Aqueous Equilibria 13.1 Solutions of Acids or Bases Containing a Common Ion 13.2 Buffered Solutions 13.3 Exact Treatment of Buffered Solutions (skip) 13.4 Buffer Capacity 13.5 Titrations and pH Curves 13.6 Acid-Base Indicators 13.7 Titration of Polyprotic Acids (skip) 13.8 Solubility Equilibria and the Solubility Product 13.9 Precipitation and Qualitative Analysis (skip) 13.10 Complex Ion Equilibria (skip) Zumdahl Chapter 8

  2. The Common Ion Effect (1) This applies to weak acids, weak bases and solubility of salts when a common ion is added to the equilibrium reaction. You can change the pH, but adding salt. HF (aq) H+ (aq) + F- (aq) Ka = 7.2 x 10-4 NaF (s) → Na+ (aq) + F– (aq) When F- is added (from NaF), then [H+] must decrease (Le Chatelier’s principle). The pH increases. See example 13.1 in the text to see how to quantitatively determine this effect, with an ICE box)

  3. The Common Ion Effect (2) If a solution and a salt to be dissolved in it have an ion in common, then the solubility of the salt is depressed relative to pure. General equation AB (s) A+ (aq) + B- (aq) If you add BH(aq), which dissociates into B- and H+, then the [A+]decreases, and AB is driven out of solution.

  4. Applications of Aqueous Equilibrium Human blood is a buffered solution Buffered Solutions The Common Ion Effect on Buffering Buffers: resist change in pH when acid or base is added. Buffer Solutions: contain a common ion and are important in biochemical and physiological processes Organisms (and humans) have built-in buffers to protect them against changes in pH. Blood: (pH 7.4)Death = 7.0 <pH > 7.8 = Death Human blood is maintained by a combination of CO3-2, PO4-3 and protein buffers.

  5. Unbuffered solution A solution that is buffered by acetic acid/acetate

  6. How Do Buffers Work? HA H+ + A – HA = generic acid • Ka = [H+][A-]/[HA] => [H+] = Ka[HA]/[A-] • If Ka is small (weak acid) then [H+] does not change much when [HA] and [A-] change. • If [HA] and [A-] are large, and [HA]/[A-] ≈ 1,then small additions of acid ([H+]) or base ([OH-]) don’t change the ratio much.

  7. Example: A solution of 0.5 M acetic acid plus 0.5 M acetate Ka = 1.8x10-5 pKa = 4.7 HA H+ + A- Ka = [H+][A-]/[HA] pH = pKa + log[A-]/[HA] Use an ICE box to calculate the pH [AH]ini = 0.5, [A-]ini = 0.5, [H+]ini = 0 => => pH = 4.74 i.e., pH=pKa

  8. Example: A solution of 0.5 M acetic acid plus 0.5 M acetate Ka = 1.8x10-5 pKa = 4.74 HA H+ + A- Ka = [H+][A-]/[HA] pH = pKa + log[A-]/[HA] Use an ICE box to calculate the pH [AH]ini = 0.5, [A-]ini = 0.5, [H+]ini = 0 => => pH = 4.74 i.e., pH=pKa Now add NaOH to 0.01 M (in an unbuffered solution this would give a pOH of 2 and pH of 12) Use HA + OH-→H2O + A- Redo the ICE Box [AH]ini = 0.49, [A-]ini = 0.51 => => pH = 4.76

  9. Buffer Calculation: Add Acid (#1) to a Buffered Solution Acetic Acid: Ka = 1.8x10-5 pKa = 4.74 HA H+ + A- Ka = [H+][A-]/[HA] => -pKa = -pH + log[A-]/[HA] => pH = pKa + log[A-]/[HA] Case 1) [CH3COOH]tot = [CH3COOH] + [CH3COO-] = 1.0M pH = pKa => [CH3COOH] = [CH3COO-] = 0.5 Now add 0.01 M HCl (strong acid) [CH3COOH] = 0.51 [CH3COO-] = 0.49 pH = pKa + log[A-]/[HA] = 4.74 + log(0.49/0.51) = 4.74 – 0.02 = 4.72 ΔpH = -0.02

  10. Buffer Calculation: Add Acid (#2) to a dilute Buffered Solution Acetic Acid: Ka = 1.8x10-5 pKa = 4.74 HA H+ + A- Ka = [H+][A-]/[HA] => -pKa = -pH + log[A-]/[HA] => pH = pKa + log[A-]/[HA] Case 2) [CH3COOH]tot = [CH3COOH] + [CH3COO-] = 0.10 M pH = pKa => [CH3COOH] = [CH3COO-] = 0.05 Now add 0.01 M HCl (strong acid) [CH3COOH] = 0.06 [CH3COO-] = 0.04 pH = pKa + log[A-]/[HA] = 4.74 + log(0.04/0.06) = 4.74 – 1.5 = 3.54 ΔpH = -1.5

  11. Adding Base to a Buffered Solution:OH- ions do not accumulate but are replaced by A- ions. OH– + HA ⇌ A– + H2O

  12. When the OH- is added, the concentrations of HA and A- change, but only by small amounts. Under these conditions the [HA]/[A-] ratio and thus the [H+] stay virtually constant. OH– + HA ⇌ A– + H2O When protons are added to a buffered solution, the conjugate base (A–) reacts H+ + A–⇌ HA

  13. Characteristics of Buffered solutions • Contain relatively large concentrations of a weak acid and its conjugate base. • When acid is added, it reacts with the conjugate base • When base is added, it reacts with the acid. • pH is determined by the ratio of the base and acid.

  14. Buffer Capacity • Buffer capacity is the amount of protons or hydroxide ions that can be absorbed without a significant change in pH. • pH is determined by the ratio of [A–]/[HA] and pKa • Capacity is determined by the magnitudes of [HA] and [A–].

  15. Equivalence Point

  16. Buffer Summary Buffer Design: Add known amount of HA (weak acid) and salt of HA (its conjugate base, A─) [H+] or pH depends on Kaand the ratio of acid to salt or [A─]. Thus if both conc. HA and A- are large then small additions of acid or base don’t change the ratio much Zumdahl Chapter 8

  17. How to Actually Make a Buffer (a buffered solution in a lab) HA H+ + A – • Assume you want to make a 1L of of 1M buffer at pH 4.74. Use acetic acid, because the pKa of acetic acid is 4.74. • To make the Buffer: • The easy way (this is how we actually do it in the lab): Weigh out the appropriate amount of Na+Acetate- (70 g). Add it to a volumetric flask. Add water to around 0.9 Liters. Add a stir bar and stir. Slowly add HCl, while monitoring the pH with a pH meter. When the pH reaches the pKa (4.74), stop. Take out the stir bar and add enough water to give 1L. • The hard way. Start with two solutions. Solution A is 1M acetic acid. Solution B is 1M Na+Acetate-. Add solution A to solution B. Then check the pH with a pH meter. Adjust the pH with HCl or NaOH as required, to get the pH to 4.74 But this will change the concentration of A- and HA.

  18. Acid-Base Titrations A controlled addition of measured volumes of a solution of known concentration (the Titrant) from a buret to a second solution of unknown concentration under conditions in which the solutes react cleanly (without side reactions), completely, and rapidly. Titration: A titration is complete when the second solute is fully consumed Completion is signaled by a change in some physical property, such as the color of the reacting mixture or the color of an indicator that has been added to it. [NaOH] Indicator phenolphthalein “X” Zumdahl Chapter 8

  19. Titrations and pH Curves Zumdahl Chapter 8

  20. Titration of a weak acid with a strong base. • The base reacts with the acid. • . OH– + HA ⇌ A– + H2O Strong acid Zumdahl Chapter 8

  21. Equivalence point determined defined by the Stoichiometry, not by the pKa. nbase = nacid number of moles of acid = number of moles of base Zumdahl Chapter 8

  22. For a variety of weak acids: The Equivalence point occurs at the same stoichiometric amount of base added The weaker the acid, the greater the pH value for the equivalent point.

  23. Similar for titration of weak bases with strong acids Zumdahl Chapter 8

  24. The indicator phenolphthalein is pink in basic solution and colorless in acidic solution. Acid-Base Indicators

  25. Indicators (Weak Acid Equilibria) pH = -log10[H3O+] Zumdahl Chapter 8

  26. Indicators A soluble compound, generally an organic dye, that changes its color noticeably over a fairly short range of pH. Typically, Indicators are a weak organic acid that has a different color than its conjugate base. HIn(aq) + H2O(l)↔ H3O+(aq) + In–(aq) Acid: HIn (aq) Conjugate base: In– (aq) Phenolphtalein Indicator denoted by In Zumdahl Chapter 8

  27. Conjugate base: In- (aq) Acid: HIn (aq) HIn(aq) + H2O(l)↔ H3O+(aq) + In-(aq) [H3O+][OH-] = Kw pH = -log10[H3O+]

  28. Methyl Red Bromothymol blue Phenolphtalein Zumdahl Chapter 8 HIn(aq) + H2O(l)↔H3O+(aq) + In-(aq)

  29. The useful pH ranges for several common indicators HIn(aq) + H2O(l)↔H3O+(aq) + In–(aq) Zumdahl Chapter 8

  30. Indicator Selection • Want indicator color change and titration equivalence point to be as close as possible • Easier with a large pH change at the equivalence point Strong acid Weak acid Could use either indicator Methyl red changes color to early Zumdahl Chapter 8

  31. Solubility Product Ksp Describes a chemical equilibrium in which an excess solid salt is in equilibrium with a saturated aqueous solution of its separated ions. General equation AB (s) ↔ A+ (aq) + B- (aq) The solubility expression controls the amount of solid that will dissolve Ksp = Ksp = Zumdahl Chapter 8

  32. Ksp Values at 25°C for Common Ionic Solids Zumdahl Chapter 8

  33. The Solubility of Ionic Solids The Solubility Product AgCl(s) ↔Ag+(aq) + Cl-(aq) excess Ksp = The solid AgCl, which is in excess, is understood to have a concentration of 1 mole per liter. Ksp = 1.6  10-10at 25oC Zumdahl Chapter 8

  34. The Solubility of Ionic Solids The Solubility Product Ag2SO4(s) ↔2Ag+(aq) + SO42-(aq) excess Ksp = Fe(OH)3(s) ↔Fe+3(aq) + 3OH-1(aq) excess Ksp = Zumdahl Chapter 8

  35. Solubility and Ksp Determine the mass of lead(II) iodate dissolved in 2.50 L of a saturated aqueous solution of Pb(IO3)2 at 25oC. The Ksp of Pb(IO3)2 is 2.6  10-13. Pb(IO3)2(s) ↔ Pb2+(aq) + 2 IO3-(aq) Let “y” = molar solubility in mol/L Zumdahl Chapter 8

  36. Determine the mass of lead(II) iodate dissolved in 2.50 L of a saturated aqueous solution of Pb(IO3)2 at 25oC. The Ksp of Pb(IO3)2 is 2.6  10-13. Pb(IO3)2(s) ↔ Pb2+(aq) + 2 IO3-(aq) [Pb2+][IO3-]2 = Ksp “y” = molar solubility [Pb2+][IO3-]2 = y = 4.0  10-5  [Pb(IO3)2] = [Pb2+] = y = 4.0  10-5 mol L-1  [IO3-] = 2y = 8.0  10-5 mol L-1 Gram solubility of Lead (II) iodate = (4.0  10-5 mol L-1)  (557 g mol-1) = 0.0223 g L-1  2.50 L Molar Mass of lead (II) iodate Pb(IO3)2 = 557g per mole Zumdahl Chapter 8

  37. The Solubility of Salts Solubility and Ksp Exercise 9-3 Compute the Ksp of silver sulfate (Ag2SO4) at 25oC if its mass solubility is 8.3 g L-1. 1 Ag2SO4 (s) ↔ 2 Ag+(aq) + 1 SO42-(aq) Zumdahl Chapter 8

  38. Compute the Ksp of silver sulfate (Ag2SO4) at 25oC if its mass solubility is 8.3 g L-1. 1 Ag2SO4 (s) ↔ 2 Ag+(aq) + 1 SO42-(aq) [y] = (8.3 g Ag2SO4 L-1)  (1 mol Ag2SO4/311.8 g) [y] = 2.66  10-2 mol Ag2SO4 L-1 [Ag+]2[SO42-] = Ksp Ksp = Zumdahl Chapter 8

  39. The Nature of Solubility Equilibria Dissolution and precipitation are reverse of each other. Dissolution (Solubility) General reaction X3Y2 (s) ↔ 3X+2 (aq) + 2Y-3 (aq) [s] Ksp = s = molar solubility expressed in moles per liter Zumdahl Chapter 8

  40. Relative Solubilities A salt’s Ksp value gives us information about its solubility. Salt ↔ Cation + Anion If the salts being compared produce the same number of ions, eg., AgI, CuI, CaSO4 s = [cation] s = [anion] Ksp = [cation] [anion] = s2 Salt molar solubility = s = (Ksp)1/2 Salt Ksp Solubility AgI 1.5 x 10-16 1.2 x 10-8 CuI 5.0 x 10-12 2.2 x 10-6 CaSO4 6.1 x 10-5 7.8 x 10-3 Solubility CaSO4 > CuI > AgI Zumdahl Chapter 8

  41. The Effects of pH on Solubility Solubility of Hydroxides Many solids dissolve more readily in more acidic solutions Zn(OH)2(s) ↔Zn2+(aq) + 2 OH-(aq) [Zn2+][OH-]2 = Ksp = 4.5  10-17 If pH decreases (or made more acidic), the [OH-] decreases. In order to maintain Ksp the [Zn2+] must increase and consequently more solid Zn(OH)2 dissolves. Make more acidic: Zinc hydroxide is more soluble in acidic solution than in pure water. [Zn2+][OH-]2 = Ksp [Zn2+][OH-]2 = Ksp [Zn2+][OH-]2 = Ksp Zumdahl Chapter 8

  42. The Effects of pH on Solubility Estimate the molar solubility of Fe(OH)3 in a solution that is buffered to a pH of 2.9. Lookup Ksp = 1.1x10-36 Fe(OH)3(s) ↔Fe3+(aq) + 3 OH-(aq) [OH-] = 7.9  10-12 mol L-1 (pH = 2.9 and) pOH = 11.1 [Fe3+][OH-]3 = Ksp [Fe3+] = Ksp/[OH-]3 = 1.1  10-36 / (7.9  10-12)3 [Fe3+] = [Fe(OH)3] =2.2  10-3 mol L-1 answer In pure water: [Fe3+] = y [OH-] = 3y y(3y)3 = 27y4 = Ksp = 1.1  10-36 y = 4.5  10-10 mol L-1 = [Fe3+] = [Fe(OH)3]= [OH-] = 3y = 1.3  10-9 mol L-1 pOH = 8.87 (and pH = 5.13) In pure water, Fe(OH)3 is 5 x 10 6 less soluble than at pH = 2.9 Zumdahl Chapter 8

  43. [Tl+] (mol L-1)[IO3-] (mol L-1) Initial concentration Change in concentration Equilibrium concentration The Common Ion Effect The Ksp of thallium(I) iodate (TlO3) is 3.1  10-6 at 25oC. Determine the molar solubility of TlIO3 in 0.050 mol L-1 KIO3 at 25oC. TlIO3(s) ↔ Tl+(aq) + IO3-(aq) [Tl+][IO3-] = Ksp s = [TlIO3] = 6.2 × 10-5 mol L-1 = molar solubility Zumdahl Chapter 8

  44. The Common Ion Effect [s] [s] [s] The Ksp of thallium(I) iodate (TlO3) is 3.1  10-6 at 25oC. Determine the molar solubility of TlIO3 in 0.050 mol L-1 KIO3 at 25oC. TlIO3(s) ↔ Tl+(aq) + IO3-(aq) With common ion (from previous calculation) s = [TlIO3] = 6.2 × 10-5 mol L-1 =molar solubility What if no common ion is added?i.e., dissolve thallium iodate in pure water s= 1.76x10-3 mol L-1 = molar solubility [Tl+][IO3-] = Ksp With common ion, s = [TlIO3]= 6.2 × 10-5 mol L-1 Zumdahl Chapter 8

  45. Chapter 8Applications of Aqueous Equilibria 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions (skip) 8.4 Buffer Capacity 8.5 Titrations and pH Curves 8.6 Acid-Base Indicators 8.7 Titration of Polyprotic Acids (skip) 8.8 Solubility Equilibria and the Solubility Product 8.9 Precipitation and Qualitative Analysis (skip) 8.10 Complex Ion Equilibria (skip) Zumdahl Chapter 8