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Chapter 19

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

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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 Buffers 19.2 Acid-Base Titration Curves 19.3 Equilibria of Slightly Soluble Ionic Compounds 19.4 Equilibria Involving Complex Ions

  3. Acid-Base Buffers An acid-base buffer is a solution that lessens the impact of pH from the addition of acid or base. An acid-base buffer usually consists of a conjugate acid-base pair where both species are present in appreciable quantities in solution. An acid-base buffer is therefore a solution of a weak acid and its conjugate base, or a weak base and its conjugate acid.

  4. Figure 19.1 The effect of adding acid or base to an unbuffered solution. A 100-mL sample of dilute HCl is adjusted to pH 5.00. The addition of 1 mL of strong acid (left) or strong base (right) changes the pH by several units.

  5. Figure 19.2 The effect of adding acid or base to a buffered solution. A 100-mL sample of an acetate buffer is adjusted to pH 5.00. The addition of 1 mL of strong acid (left) or strong base (right) changes the pH very little. The acetate buffer is made by mixing 1 M CH3COOH ( a weak acid) with 1 M CH3COONa (which provides the conjugate base, CH3COO-).

  6. Buffers and the Common-ion Effect A buffer works through the common-ion effect. Acetic acid in water dissociates slightly to produce some acetate ion: CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) acetic acid acetate ion If NaCH3COO is added, it provides a source of CH3COO- ion, and the equilibrium shifts to the left. CH3COO- is common to both solutions. The addition of CH3COO- reduces the % dissociation of the acid.

  7. Table 19.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid [CH3COOH]dissoc [CH3COOH]init * % Dissociation = x 100

  8. How a Buffer Works The buffer components (HA and A-) are able to consume small amounts of added OH- or H3O+by a shift in equilibrium position. CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Added H3O+ reacts with CH3COO-, causing a shift to the left. Added OH- reacts with CH3COOH, causing a shift to the right. The shift in equilibrium position absorbs the change in [H3O+] or [OH-], and the pH changes only slightly.

  9. H3O+ OH- How a buffer works. Figure 19.3 Buffer has more HA after addition of H3O+. Buffer has equal concentrations of A- and HA. Buffer has more A- after addition of OH-. H2O + CH3COOH ← H3O+ + CH3COO- CH3COOH + OH- → CH3COO- + H2O

  10. Relative Concentrations of Buffer Components CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Since Ka is constant, the [H3O+] of the solution depends on the ratio of buffer component concentrations. [CH3COO-][H3O+] [CH3COOH] Ka = [H3O+] = Ka x If the ratio decreases, [H3O+] increases. If the ratio increases, [H3O+] decreases. [HA] [A-] [HA] [A-] [CH3COOH] [CH3COO-]

  11. PLAN: We can calculate [CH3COOH]init and [CH3COO-]init from the given information. From this we can find the starting pH. For (b) and (c) we assume that the added OH- or H3O+ reacts completely with the buffer components. We write a balanced equation in each case, set up a reaction table, and calculate the new [H3O+]. Sample Problem 19.1 Calculating the Effect of Added H3O+ or OH- on Buffer pH PROBLEM: Calculate the pH: (a) Of a buffer solution consisting of 0.50 M CH3COOH and 0.50 M CH3COONa (b) After adding 0.020 mol of solid NaOH to 1.0 L of the buffer solution (c) After adding 0.020 mol of HCl to 1.0 L of the buffer solution in (a). Ka of CH3COOH = 1.8 x 10-5. (Assume the additions cause a negligible change in volume.)

  12. CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Initial 0.50 - 0.50 0 Change −x - +x +x Equilibrium 0.50 - x - 0.50 + x x Sample Problem 19.1 SOLUTION: (a) Concentration (M) Since Ka is small, x is small, so we assume [CH3COOH] = 0.50 – x ≈ 0.50 M and [CH3COO-] = 0.50 + x ≈ 0.50 M Ka x = 1.8x10-5M x = [H3O+] = 0.50 0.50 ≈ 1.8x10-5 x 1.8x10-5M 0.50 M pH = -log(1.8x10-5) = 4.74 Checking the assumption: x 100 = 3.6x10-3% (< 5%; assumption is justified.) [CH3COOH] [CH3COO-]

  13. CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Initial 0.48 - 0.52 0 Change −x - +x +x Equilibrium 0.48 - x - 0.52 + x x Sample Problem 19.1 (b) Concentration (M) Concentration (M) CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) Setting up a reaction table for the stoichiometry: 0.020 mol [OH−]added = = 0.020 M OH− 0.020 Initial 0.50 0.50 - 1.0 L soln Change -0.020 -0.020 +0.020 - Equilibrium 0.48 0 0.52 - Setting up a reaction table for the acid dissociation, using new initial [ ]:

  14. Sample Problem 19.1 Since Ka is small, x is small, so we assume [CH3COOH] = 0.48 – x ≈ 0.48 M and [CH3COO-] = 0.52 + x ≈ 0.52 M = 1.7x10-5M pH = -log(1.7x10-5) = 4.77 Ka x x = [H3O+] = 0.48 0.52 ≈ 1.8x10-5 x [CH3COOH] [CH3COO-]

  15. CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Initial 0.52 - 0.48 0 Change −x - +x +x Equilibrium 0.52 - x - 0.48 + x x Sample Problem 19.1 (c) Concentration (M) Concentration (M) CH3COO-(aq) + H3O+(aq) → CH3COOH(aq) + H2O(l) Setting up a reaction table for the stoichiometry: 0.020 mol [H3O+]added = = 0.020 M H3O+ 0.020 Initial 0.50 0.50 - 1.0 L soln Change -0.020 -0.020 +0.020 - Equilibrium 0.48 0 0.52 - Setting up a reaction table for the acid dissociation, using new initial [ ]:

  16. Sample Problem 19.1 Since Ka is small, x is small, so we assume [CH3COOH] = 0.52 – x ≈ 0.52 M and [CH3COO-] = 0.48 + x ≈ 0.48 M = 2.0x10-5M pH = -log(2.0x10-5) = 4.70 Ka x x = [H3O+] = 0.52 0.48 ≈ 1.8x10-5 x [CH3COOH] [CH3COO-]

  17. The Henderson-Hasselbalch Equation HA(aq) + H2O(l) A-(aq) + H3O+(aq) [HA] [A-] [H3O+][A-] [HA] -log[H3O+] = -logKa – log Ka = [H3O+] = Ka x [base] [acid] pH = pKa + log [HA] [A-]

  18. Buffer Capacity The buffer capacity is a measure of the “strength” of the buffer, its ability to maintain the pH following addition of strong acid or base. The greater the concentrations of the buffer components, the greater its capacity to resist pH changes. The closer the component concentrations are to each other, the greater the buffer capacity.

  19. Figure 19.4 The relation between buffer capacity and pH change. When strong base is added, the pH increases least for the most concentrated buffer. This graph shows the final pH values for four different buffer solutions after the addition of strong base.

  20. Buffer Range The buffer range is the pH range over which the buffer is effective. Buffer range is related to the ratio of buffer component concentrations. [HA] [A-] The closer is to 1, the more effective the buffer. If one component is more than 10 times the other, buffering action is poor. Since log10 = 1, buffers have a usable range within ± 1 pH unit of the pKa of the acid component.

  21. Sample Problem 19.2 Using Molecular Scenes to Examine Buffers PROBLEM: The molecular scenes below represent samples of four HA/A- buffers. (HA is blue and green, A- is green, and other ions and water are not shown.) (a) Which buffer has the highest pH? (b) Which buffer has the greatest capacity? (c) Should we add a small amount of concentrated strong acid or strong base to convert sample 1 to sample 2 (assuming no volume changes)?

  22. Sample Problem 19.2 PLAN: Since the volumes of the solutions are equal, the scenes represent molarities as well as numbers. We count the particles of each species present in each scene and calculate the ratio of the buffer components. SOLUTION: [A-]/[HA] ratios: sample 1, 3/3 = 1; sample 2, 2/4 = 0.5; sample 3, 4/4 = 1; and sample 4, 4/2 = 2.

  23. Sample Problem 19.2 (a) As the pH rises, more HA will be converted to A-. The scene with the highest [A-]/[HA] ratio is at the highest pH. Sample 4 has the highest pH because it has the highest ratio. (b) The buffer with the greatest capacity is the one with the [A-]/[HA] closest to 1. Sample 3 has the greatest buffer capacity. (c) Sample 2 has a lower [A-]/[HA] ratio than sample 1, so we need to increase the [A-] and decrease the [HA]. This is achieved by adding strong acidto sample 1.

  24. Preparing a Buffer • Choose the conjugate acid-base pair. • The pKa of the weak acid component should be close to the desired pH. • Calculate the ratio of buffer component concentrations. • Determine the buffer concentration, and calculate the required volume of stock solutions and/or masses of components. • Mix the solution and correct the pH. [base] [acid] pH = pKa + log

  25. PROBLEM: An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of the acid rain on limsetone-rich soils. How many grams of Na2CO3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO3 to make the buffer? Ka of HCO3- is 4.7x10-11. PLAN: The conjugate pair is HCO3- (acid) and CO32- (base), and we know both the buffer volume and the concentration of HCO3-. We can calculate the ratio of components that gives a pH of 10.00, and hence the mass of Na2CO3 that must be added to make 1.5 L of solution. Sample Problem 19.3 Preparing a Buffer [CO32-][H3O+] [HCO3-] Ka = SOLUTION: [H3O+] = 10-pH = 10-10.00= 1.0x10-10M HCO3-(aq) + H2O(l) H3O+(aq) + CO32-(aq)

  26. Sample Problem 19.3 Preparing a Buffer = 0.094 M (4.7x10-11)(0.20) 1.0x10-10 = Amount (mol) of CO32- needed = 1.5 L soln x 0.094 mol CO32- 1 L soln = 0.14 mol CO32- 105.99 g Na2CO3 1 mol Na2CO3 Ka[HCO3-] [H3O+] [CO32-] = = 15 g Na2CO3 The chemist should dissolve 15 g Na2CO3 in about 1.3 L of 0.20 M NaHCO3and add more 0.20 M NaHCO3 to make 1.5 L. Using a pH meter, she can then adjust the pH to 10.00 by dropwise addition of concentrated strong acid or base. 0.14 mol Na2CO3 x

  27. Acid-Base Indicators An acid-base indicator is a weak organic acid (HIn) whose color differs from that of its conjugate base (In-). The ratio [HIn]/[In-] is governed by the [H3O+] of the solution. Indicators can therefore be used to monitor the pH change during an acid-base reaction. The color of an indicator changes over a specific, narrow pH range, a range of about 2 pH units.

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

  29. Figure 19.6 The color change of the indicator bromthymol blue. pH < 6.0 pH = 6.0-7.5 pH > 7.5

  30. Acid-Base Titrations In an acid-base titration, the concentration of an acid (or a base) is determined by neutralizing the acid (or base) with a solution of base (or acid) of known concentration. The equivalence point of the reaction occurs when the number of moles of OH- added equals the number of moles of H3O+ originally present, or vice versa. The end point occurs when the indicator changes color. - The indicator should be selected so that its color change occurs at a pH close to that of the equivalence point.

  31. Figure 19.7 Curve for a strong acid–strong base titration. The pH increases gradually when excess base has been added. The pH rises very rapidly at the equivalence point, which occurs at pH = 7.00. The initial pH is low.

  32. Calculating the pH during a strong acid–strong base titration Initial pH [H3O+] = [HA]init pH = -log[H3O+] mol H3O+remaining Vacid + Vbase pH before equivalence point initial mol H3O+ = Vacid x Macid mol OH- added = Vbase x Mbase mol H3O+remaining = (mol H3O+init) – (mol OH-added) [H3O+] = pH = -log[H3O+]

  33. Calculating the pH during a strong acid–strong base titration pH at the equivalence point pH = 7.00 for a strong acid-strong base titration. mol OH-excess Vacid + Vbase pH beyond the equivalence point initial mol H3O+ = Vacid x Macid mol OH- added = Vbase x Mbase mol OH-excess = (mol OH-added) – (mol H3O+init) [OH-] = pOH = -log[OH-] and pH = 14.00 - pOH

  34. Example: 40.00 mL of 0.1000 M HCl is titrated with 0.1000 M NaOH. The initial pH is simply the pH of the HCl solution: [H3O+] = [HCl]init = 0.1000 M and pH = -log(0.1000) = 1.00 To calculate the pH after 20.00 mL of NaOH solution has been added: OH- added = 0.02000 L NaOH x Initial mol of H3O+ = 0.04000 L HCl x = 2.000x10-3 mol OH- = 4.000x10-3 mol H3O+ 0.1000 mol 1 L 0.1000 mol 1 L The OH- ions react with an equal amount of H3O+ ions, so H3O+ remaining = 4.000x10-3 – 2.000x10-3 = 2.000x10-3 mol H3O+

  35. 2.000x10-3 mol 0.04000 L + 0.02000 L 1.000x10-3 mol 0.04000 L + 0.05000 L [H3O+] = [OH-] = = 0.03333 M = 0.01111 M pH = -log(0.03333) = 1.48 The equivalence point occurs when mol of OH- added = initial mol of HCl, so when 40.00 mL of NaOH has been added. To calculate the pH after 50.00 mL of NaOH solution has been added: OH- added = 0.05000 L NaOH x = 5.000x10-3 mol OH- OH- in excess = 5.000x10-3 – 4.000x10-3 = 1.000x10-3 mol OH- 0.1000 mol 1 L pOH = -log(0.01111) = 1.95 pH = 14.00 – 1.95 = 12.05

  36. Figure 19.8 Curve for a weak acid–strong base titration. The pH increases slowly beyond the equivalence point. The pH at the equivalent point is > 7.00 due to the reaction of the conjugate base with H2O. The curve rises gradually in the buffer region. The weak acid and its conjugate base are both present in solution. The initial pH is higher than for the strong acid solution.

  37. Calculating the pH during a weak acid–strong base titration [HA] [A-] Initial pH [H3O+] = pH = -log[H3O+] [H3O+][A-] [HA] Ka = pH before equivalence point [H3O+] = Ka x or [base] [acid] pH = pKa + log

  38. Calculating the pH during a weak acid–strong base titration Kw Ka pH at the equivalence point [OH- ] = where [A-] = and Kb = mol OH-excess Vacid + Vbase mol HAinit Vacid + Vbase A-(aq) + H2O(l) HA(aq) + OH-(aq) pH beyond the equivalence point [OH-] = pH = -log[H3O+] Kw [OH-] [H3O+] = [H3O+] ≈ and pH = -log[H3O+] Kw

  39. PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000 M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000 M NaOH: PLAN: The initial pH must be calculated using the Ka value for the weak acid. We then calculate the number of moles of HPr present initially and the number of moles of OH- added. Once we know the volume of base required to reach the equivalence point we can calculate the pH based on the species present in solution. Sample Problem 19.4 Finding the pH During a Weak Acid–Strong Base Titration (a) 0.00 mL (b) 30.00 mL (c) 40.00 mL (d) 50.00 mL SOLUTION: = 1.1x10-3M (a) [H3O+] = pH = -log(1.1x10-3) = 2.96

  40. Sample Problem 19.4 [HPr] [Pr-] [H3O+] = Ka x (b) 30.00 mL of 0.1000 M NaOH has been added. Concentration (M) HPr(aq) + OH-(aq) → Pr-(aq) + H2O(l) Initial amount of HPr = 0.04000 L x 0.1000 M = 4.000x10-3 mol HPr Amount of NaOH added = 0.03000 L x 0.1000 M = 3.000x10-3 mol OH- Each mol of OH- reacts to form 1 mol of Pr-, so - Initial 0.004000 0.003000 0 - +0.003000 −0.003000 Change -0.003000 - 0.001000 0 0.003000 Equilibrium = 4.3x10-6M 0.001000 0.003000 = (1.3x10-5) x pH = -log(4.3x10-6) = 5.37

  41. Sample Problem 19.4 (c) 40.00 mL of 0.1000 M NaOH has been added. This is the equivalence point because mol of OH- added = 0.004000 = mol of HAinit. All the OH- added has reacted with HA to form 0.004000 mol of Pr-. = 0.05000 M = 7.7x10-10 = 1.6x10-9M 0.004000 mol 0.04000 L + 0.04000 L [Pr-] = 1.0x10-14 1.3x10-5 Kw Ka pH = -log(1.6x10-9) = 8.80 Pr- is a weak base, so we calculate Kb = = 1.0x10-14 = [H3O+] ≈ Kw

  42. Sample Problem 19.4 (d) 50.00 mL of 0.1000 M NaOH has been added. Amount of OH- added = 0.05000 L x 0.1000 M = 0.005000 mol Excess OH- = OH-added – HAinit = 0.005000 – 0.004000 = 0.001000 mol 0.001000 mol 0.09000 L mol OH-excess total volume = = 0.01111 M [OH-] = 1x10-14 0.01111 Kw [OH-] = 9.0x10-13M [H3O+] = = pH = -log(9.0x10-13) = 12.05

  43. Figure 19.9 Curve for a weak base–strong acid titration. The pH decreases gradually in the buffer region. The weak base and its conjugate acid are both present in solution. The pH at the equivalence point is < 7.00 due to the reaction of the conjugate acid with H2O.

  44. Figure 19.10 Curve for the titration of a weak polyprotic acid. Titration of 40.00 mL of 0.1000 M H2SO3 with 0.1000 M NaOH pKa2 = 7.19 pKa1 = 1.85

  45. Amino Acids as Polyprotic Acids An amino acid contains a weak base (-NH2) and a weak acid (-COOH) in the same molecule. Both groups are protonated at low pH and the amino acid behaves like a polyprotic acid.

  46. Abnormal shape of red blood cells in sickle cell anemia. Figure 19.11 Several amino acids have charged R groups in addition to the NH2 and COOH group. These are essential to the normal structure of many proteins. In sickle cell anemia, the hemoglobin has two amino acids with neutral R groups instead of charged groups. The abnormal hemoglobin causes the red blood cells to have a sickle shape, as seen here.

  47. Equilibria of Slightly Soluble Ionic Compounds Any “insoluble” ionic compound is actually slightly soluble in aqueous solution. We assume that the very small amount of such a compound that dissolves will dissociate completely. For a slightly soluble ionic compound in water, equilibrium exists between solid solute and aqueous ions. Qsp = Qc[PbF2] = [Pb2+][F-]2 PbF2(s) Pb2+(aq) + 2F-(aq) [Pb2+][F-]2 [PbF2] Qc =

  48. Qsp and Ksp Qsp is called the ion-product expression for a slightly soluble ionic compound. For any slightly soluble compound MpXq, which consists of ions Mn+ and Xz-, Qsp = [Mn+]p[Xz-]q When the solution is saturated, the system is at equilibrium, and Qsp = Ksp, the solubility product constant. The Ksp value of a salt indicates how far the dissolution proceeds at equilibrium (saturation).

  49. Metal Sulfides Metal sulfides behave differently from most other slightly soluble ionic compounds, since the S2- ion is strongly basic. We can think of the dissolution of a metal sulfide as a two-step process: S2-(aq) + H2O(l) → HS-(aq) + OH-(aq) MnS(s) Mn2+(aq) + S2-(aq) MnS(s) + H2O(l) Mn2+(aq) + HS-(aq) + OH-(aq) Ksp = [Mn2+][HS-][OH-]

  50. PROBLEM: Write the ion-product expression at equilibrium for each compound: (a) magnesium carbonate (b) iron(II) hydroxide (c) calcium phosphate (d) silver sulfide PLAN: We write an equation for a saturated solution of each compound, and then write the ion-product expression at equilibrium, Ksp. Note the sulfide in part (d). Sample Problem 19.5 Writing Ion-Product Expressions SOLUTION: Ksp = [Mg2+][CO32-] Ksp = [Fe2+][OH-]2 (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) Ksp = [Ca2+]3[PO43-]2

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