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Chapter 19

Chapter 19. Ionic Equilibria in Aqueous Systems. Ionic Equilibria in Aqueous Systems. 19.1 Equilibria of Acid-Base Buffer Systems. 19.2 Acid-Base Titration Curves. 19.3 Equilibria of Slightly Soluble Ionic Compounds. 19.4 Equilibria Involving Complex Ions.

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Chapter 19

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  1. Chapter 19 Ionic Equilibria in Aqueous Systems

  2. Ionic Equilibria in Aqueous Systems 19.1 Equilibria of Acid-Base Buffer Systems 19.2 Acid-Base Titration Curves 19.3 Equilibria of Slightly Soluble Ionic Compounds 19.4 Equilibria Involving Complex Ions 19.5 Application of Ionic Equilibria to Chemical Analysis

  3. Buffered solutions • A solution that resists a change in its pH when either OH- or H+ ions are added. • Ex: Blood, absorbs acid and bases without change in pH. • The components of a buffer are the conjugate acid-base pair of a weak acid. • WA and its salt-HF and NaF • WB and its salt- NH3 and NH4Cl.

  4. acid added base added an unbuffered solution acid added base added or a buffered solution Figure 19.1 The effect of addition of acid or base to … Figure 19.2

  5. Common ion effect • The shift in equilibrium position that occurs because of the addition of an ion already involved in equilibrium reaction is called common ion effect. • Consider a weak acid HF and its salt NaF. • NaF (s) _______ Na+ (aq)+ F- (aq) , Major species: Na+ F- HF H2O • HF (aq)↔ H+ (aq) + F- (aq)

  6. Common ion effect • F- is the common ion. • F- from the added NaF moves the equ position to the left according to LC principle. • If we dissolve HF in a NaF solution, the F-ion and H+ ion enter the solution. The F- ion already present combines with the H+, which lowers the H+.

  7. Buffered Solution Characteristics • Buffers contain relatively large amounts of weak acid and corresponding base. • Added H+ is consumed by A- • Added OH is consumed by HA • The pH is determined by the ratio of the concentrations of the weak acid and its conjugate base.

  8. 0.10 0.00 0.050 0.10 0.10 0.10 0.15 0.10 [CH3COOH]dissoc * % Dissociation = x 100 [CH3COOH]initial Table 19.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH3COOH]initial [CH3COO-]added % Dissociation* pH 1.3 2.89 4.44 0.036 4.74 0.018 4.92 0.012

  9. Buffer after addition of H3O+ Buffer with equal concentrations of conjugate base and acid Buffer after addition of OH- H3O+ OH- H2O + CH3COOH H3O+ + CH3COO- CH3COOH + OH- H2O + CH3COO- How a buffer works. Figure 19.3

  10. Problem P.1. A buffered solution contains 0.50 M acetic acid (Ka= 1.8 x 10-5) and 0.50 M sodium acetate. a) Calculate pH of this solution. b) Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of buffered solution described in the above problem.Compare this pH change with the one that occurs when 0.010 mol solid NaOH is added to 1.0 L of water. c) Calculate the change in pH that occurs when 0.010 mol solid HCl is added to 1.0 L of buffered solution.

  11. HA + H2O H3O+ + A- [H3O+] [A-] [HA] Ka [HA] [A-] [A-] [HA] [base] pH = pKa + log [acid] The Henderson-Hasselbalch Equation Ka = [H3O+] = - log[H3O+] = - log Ka + log

  12. Problems P. 2. Calculate the pH of a solution 0.75 M lactic acid (Ka= 1.4 x10-4 ) and 0.25 M sodium lactate. P.3. A buffer solution contains 0.25 M NH3 (Kb= 1.8 x 10-5) and 0.40 M NH4Cl. Calculate pH of this solution.

  13. Buffering Capacity It represents the amount of H+ or OH the buffer can absorb without a significant change in pH. • More concentrated the components of a buffer , the greater the buffer capacity. • pH of a buffered solution is determined by the ratio [A-]/[HA] . • If an acid or a base is added , the concentration ratio changes less when the buffer component concentrations are similar than they are different.

  14. Buffering Capacity: • Buffer has highest capacity when component concentrations are equal. • pH=pKa highest buffer capacity. • The Buffer Range is the pH range over which the buffer works effectively. • The further the buffering component concentration ratio is from 1 less effective is the buffering action.

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

  16. The relation between buffer capacity and pH change. Figure 19.4

  17. Preparing a Buffer • Choose the conjugate acid-base pair. Ratio of concentrations should be close to1. pH=pKa • Calculate ratio of buffer component concentrations. Use the Henderson Hasselbalch equation. • Determine the buffer concentration. Higher the buffer concentrations greater the buffer capacity. • Mix the solutions and adjust the pH by adding strong acid or strong base with the help of a pH probe.

  18. Preparing a Buffer 1. Choose the conjugate acid-base pair. 2. Calculate the ratio of buffer component concentrations. 3. Determine the buffer concentration. 4. Mix the solution and adjust the pH.

  19. Problem • P.4. A chemist needs a solution buffered at pH= 4.30 and can choose from the following acids and their sodium salts: • chloroacetic acid Ka= 1.35 x 10-3 • 2. propanoic acid Ka= 1.3 x 10-5 3. benzoic acid Ka= 6.4 x 10-5 4. hypochlorous acid. Ka= 3.5 x 10-8 Calculate the ratio [HA]/[A-] required for each system to yield a pH of 4.30, which system will work the best?

  20. Acid base Titration Curves A plot of pH of the solution being analyzed as a function of the amount of titrant added.

  21. Acid-Base Indicators: • It marks the end point of a titration by changing color. • Ex: phenolphthalein is colorless in its HIn form and pink in In- form, or basic form. • Indicator changes color over a range of about 2 pH units.

  22. pH Colors and approximate pH range of some common acid-base indicators. Figure 19.5

  23. change occurs over ~2pH units The color change of the indicator bromthymol blue. Figure 19.6 basic acidic

  24. Strong Acid- Strong Base Titration curves: • pH is low initially, as base is added the pH increases slowly. • The pH rises steeply when the moles of OH- nearly equals the moles of H3O+ • The additional drop of base neutralizes the tiny excess acid and introduces a tiny excess of base. • Then pH increases smoothly as more base is added.

  25. Equivalence (stoichiometric) point • Enough titrant has been added to react exactly with the solution being analyzed. • Equivalence point is nearly the vertical portion of the curve. • The point at which the number of moles of added OH- =the number of moles of H3O+ • For a SA vs SB titration pH =7(neutral ions)

  26. End point It occurs when indicator changes color. The indicator chosen should be close to the equivalence point.

  27. Figure 19.7 Curve for a strong acid-strong base titration

  28. Calculating pH: • Initial pH is the pH of the acid. • Before the equ. Point find initial moles present, moles reacted, change in volume and then the pH. • At the equ point, pH =7 • After the equ point find the excess of OH- present and then calculate pH.

  29. Problem P.5. 50.0 mL of 0.200 m HNO3 is titrated with 0.100M NaOH. Calculate the pH of the solution at selected points during the course of titration, where 0 mL, 20.0 mL, 100.0 mL, 150.0 mL of 0.100 M NaOH has been added.

  30. Weak acid-Strong Base Titrations: • Initial pH is high as weak acid dissociates slightly. • Buffer Region: A gradual rising portion of the curve appearing before the equ.point. • At midpoint of the buffer region pH=pKa. • pH at equ point is > 7. • Beyond equ point pH increases slowly as excess OH- is added.

  31. pH = 8.80 at equivalence point pKa of HPr = 4.89 methyl red [HPr] = [Pr-] Titration of 40.00mL of 0.1000M HPr with 0.1000M NaOH Figure 19.8 Curve for a weak acid-strong base titration

  32. Calculating pH • 1. When only HA present, Use ICE table calculate H+ and pH like a weak acid type calculation. • 2. As we add base, find change in moles and solve for H+ • 3. At equ. Point pH >7, Find Kb from Ka and then calculate OH- and pH. • 4. Beyond the equ. point , calculate excess of OH- present and calculate pH.

  33. Problem P.6. 50.0 mL of 0.1 M acetic acid (Ka=1.8 x 10-5) with 0.10 M NaOH. Calculate the pH at various points representing volumes of 0mL, 10.0 mL , 25.0 mL, 50.0 mL , 60.0 mL of added NaOH.

  34. Weak base –strong acid Titration curves: • Same shape curve as WA vs SB , but inverted. • Initially pH above 7 as it is a weak base. • pH decreases in buffer region. • At midpoint pH=pKa • After buffer region curve drops vertically to equ.point . • pH at equ.point is below 7 • Beyond equ point , pH decreases slowly as more acid is added.

  35. pH = 5.27 at equivalence point Titration of 40.00mL of 0.1000M NH3 with 0.1000M HCl Figure 19.9 pKa of NH4+ = 9.25 Curve for a weak base-strong acid titration

  36. Problems P.7. 100.0 mL of 0.050 M NH3 is titrated with 0.10 M HCl. Calculate the pH at various points: 1) before adding HCl 2) Before equ.point 3) At equ point (setup) P.8 If 2.00 mmol of solid acid in 100.0 mL water is titrated with 0.0500 M NaOH. After 20.0 mL NaOH has been added the pH is 6.00. What is the Ka value for acid?

  37. Titration curves of polyprotic acids: Number of curves =Number of H+ ions.

  38. Figure 19.10 Curve for the titration of a weak polyprotic acid. pKa = 7.19 pKa = 1.85 Titration of 40.00mL of 0.1000M H2SO3 with 0.1000M NaOH

  39. PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000M NaOH: PLAN: The amounts of HPr and Pr- will be changing during the titration. Remember to adjust the total volume of solution after each addition. SOLUTION: (a) Find the starting pH using the methods of Chapter 18. Amount (mol) HPr(aq) + OH-(aq) Pr-(aq) + H2O(l) Sample Problem 19.3 Calculating the pH During a Weak Acid- Strong Base Titration (a) 0.00mL (b) 30.00mL (c) 40.00mL (d) 50.00mL Ka = [Pr-][H3O+]/[HPr] [Pr-] = x = [H3O+] [Pr-] = x = [H3O+] x = 1.1x10-3 ; pH = 2.96 (b) Before addition 0.04000 - 0 - Addition - 0.03000 - - After addition 0.01000 0 0.03000 -

  40. 0.001000 mol [H3O+] = 1.3x10-5 0.003000 mol (0.004000 mol) = 0.05000M (0.004000L) + (0.004000L) [H3O+] = Kw / = 1.6x10-9M M = (0.00100) (0.0900L) Sample Problem 19.3 Calculating the pH During a Weak Acid- Strong Base Titration continued = 4.3x10-6M pH = 5.37 (c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr -] will be Ka x Kb = Kw Kb = Kw/Ka = 1.0x10-14/1.3x10-5 = 7.7x10-10 pH = 8.80 (d) 50.00mL of NaOH will produce an excess of OH-. mol XS base = (0.1000M)(0.05000L - 0.04000L) = 0.00100mol [H3O+] = 1.0x10-14/0.01111 = 9.0x10-11M M = 0.01111 pH = 12.05

  41. Sickle shape of red blood cells in sickle cell anemia. Figure 19.11

  42. Equilibria of slightly soluble ionic compounds: • An ionic substance dissociates completely in water to form hydrate cations and anions. • Both the forward and reverse reactions take place. • A stage is reached when no more solid dissolves. This is called as dynamic equilibrium. • CaF2 (s) ↔Ca2+ (aq) + 2F- (aq) • Ksp= [Ca2+ ] [F-]2 • Ksp is called as the solubility product constant or solubility product for the equilibrium expression.

  43. Equilibria of slightly soluble ionic compounds: • For the hypothetical compound, MpXq • At equilibrium Qsp = [Mn+]p [Xz-]q= Ksp • Higher the Ksp greater the solubility for formulas containing same total number of ions. • If salts being compared produce different number of ions. • Compare their solubilities, which is opposite of Ksp

  44. Ion-Product Expression (Qsp) and Solubility Product Constant (Ksp) For the hypothetical compound, MpXq At equilibrium Qsp = [Mn+]p [Xz-]q = Ksp

  45. PROBLEM: Write the ion-product expression for each of the following: PLAN: Write an equation which describes a saturated solution. Take note of the sulfide ion produced in part (d). (a) MgCO3(s) Mg2+(aq) + CO32-(aq) (b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq) (c) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq) (d) Ag2S(s) 2Ag+(aq) + S2-(aq) S2-(aq) + H2O(l) HS-(aq) + OH-(aq) Ag2S(s) + H2O(l) 2Ag+(aq) + HS-(aq) + OH-(aq) Sample Problem 19.4 Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds (a) Magnesium carbonate (b) Iron (II) hydroxide (c) Calcium phosphate (d) Silver sulfide SOLUTION: Ksp = [Mg2+][CO32-] Ksp = [Fe2+][OH-] 2 Ksp = [Ca2+]3[PO43-]2 Ksp = [Ag+]2[HS-][OH-]

  46. Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic Compounds at 250C Name, Formula Ksp Aluminum hydroxide, Al(OH)3 3 x 10-34 Cobalt (II) carbonate, CoCO3 1.0 x 10-10 Iron (II) hydroxide, Fe(OH)2 4.1 x 10-15 Lead (II) fluoride, PbF2 3.6 x 10-8 Lead (II) sulfate, PbSO4 1.6 x 10-8 Mercury (I) iodide, Hg2I2 4.7 x 10-29 Silver sulfide, Ag2S 8 x 10-48 Zinc iodate, Zn(IO3)2 3.9 x 10-6

  47. Problems • P.10. CuBr has a measured solubility of 2.0 x 10-4 mol/L at 25 C. Calculate Ksp. • P.11. Calculate Ksp value for Bi2S3 which has a solubility of 1.0 x 10-15 mol/L at 25 C. • P.12. Calculate the solubility of copper (II) iodate at 25 C, if Ksp = 1.4 x 10-7

  48. Table 19.3 Relationship Between Ksp and Solubility at 250C No. of Ions Formula Cation:Anion Ksp Solubility (M) 2 MgCO3 1:1 3.5 x 10-8 1.9 x 10-4 2 PbSO4 1:1 1.6 x 10-8 1.3 x 10-4 2 BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5 3 Ca(OH)2 1:2 5.5 x 10-6 1.2 x 10-2 3 BaF2 1:2 1.5 x 10-6 7.2 x 10-3 3 CaF2 1:2 3.2 x 10-11 2.0 x 10-4 3 Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5

  49. Common ion Effect: • Presence of a common ion decreases solubility of a slightly soluble ionic compound.

  50. CrO42- added PbCrO4(s) Pb2+(aq) + CrO42-(aq) Figure 19.12 The effect of a common ion on solubility PbCrO4(s) Pb2+(aq) + CrO42-(aq)

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