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Aqueous Equilibria / Buffers / Ksp

Aqueous Equilibria / Buffers / Ksp. AP Chemistry Pages 681-736. Solutions of Acids or Bases Containing a Common Ion. Solutions that contain weak acid (HA) and its salt ( NaA ); for example, a solution of hydrofluoric acid (HF, K a = 7.2 x 10 -4 ) and its salt sodium fluoride.

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Aqueous Equilibria / Buffers / Ksp

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  1. Aqueous Equilibria/ Buffers/ Ksp AP Chemistry Pages 681-736

  2. Solutions of Acids or Bases Containing a Common Ion • Solutions that contain weak acid (HA) and its salt (NaA); for example, a solution of hydrofluoric acid (HF, Ka = 7.2 x 10-4) and its salt sodium fluoride. NaF(s)  Na+(aq) + F-(aq) strong electrolyte common ion is F-, produced by both HF and NaF • Compare 1.0 M HF solution to solution of 1.0 M HF and 1.0 MNaF HF(aq) ↔ H+(aq) + F-(aq) • Common Ion Effect: We would expect equilibrium of second solution to be driven to the left because of the addition of F- ions from the NaF. Therefore a solution containing both NaF and HF is less acidic than a solution of HF alone. • Other examples: adding solid NH4Cl to a 1.0 M NH3 solution. NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)

  3. Example Problem • Calculate [H+], pH and the percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF. (note: a 1.0 M solution of only HF has [H+] = 2.7 x 10-2 (pH = 1.57) and a percent dissociation of 2.7%.)

  4. Practice Problem • Calculate the pH of each of the following solutions. • A. 0.100 M HONH2 (Kb= 1.1 x 10-8) • B. 0.100M HONH3Cl • C. Pure H2O • D. A mixture containing 0.100 M HONH2 and 0.100M HONH3Cl • Compare the percent ionization of the base in part A with the percent ionization of the base in part D. Explain any differences.

  5. Buffered Solutions • A buffer is a solution that contains a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid.) A buffered solution resists drastic changes in its pH when either OH- ions or H+ ions are added. • Our blood is buffered (pH = 7.42) • The most important buffering system in the blood involves HCO3- and H2CO3. • so is Country Time Lemonade

  6. Example 1 (pH of a buffered solution) • A buffered solution contains 0.50 M acetic acid (Ka = 1.8 x 10-5) and 0.50 M sodium acetate. Calculate the pH of this solution.

  7. Example 2 (a) Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the buffered solution in the previous example.(b) Calculate the pH after the addition of 0.020 molHCl to 1.0 L of the buffered solution. Note: when a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations.

  8. Examples of Buffers: • HC2H3O2 and… • H2CO3 and… • HF and… • NH4+ and… • H3PO4 and… • H2PO4- and… • Why do HCl and Cl-not make a buffer?

  9. The Henderson-Hasselbalch Equation • For any weak acid (HA + H2O  H3O+ + A-) • Ka= • [H3O+] =   If the amounts of HA and A- are large compared to the amount of OH- added, the change in the ratio [HA] / [A-] will be small. The pH will remain essentially constant.

  10. The Henderson-Hasselbalch Equation • -log[H3O+] = -logKa – log[HA]/[A-] • pH = pKa + log[A-]/[HA] Thus, pH = pKa + log([base]/[acid]) The Henderson-Hasselbalch Equation

  11. Example Problem(pH of a buffered solution) • Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 x 10-4) and 0.25 M sodium lactate. Lactic acid (HC3H5O3) is a common constituent of biologic systems. For example, it is found in milk and is present in human muscle tissue during exertion.

  12. Example Problem • A buffered solution contains 0.25 M NH3(Kb = 1.8 x 10-5) and 0.40 M NH4Cl. Calculate the pH of this solution.

  13. Example Problem(Adding a strong acid to a buffered solution) • Calculate the pH of the solution that results when 0.10 mol gaseous HCl is added to 1.0 L of the buffered solution from the previous example.

  14. Individual Practice Problems • Beginning on page 740 • # 21,23, 25, 27, 31, 33, 35, 37, and 38

  15. Buffering Capacity • The buffering capacity of a buffered solution represents the amount of H+ or OH- ions the buffer can absorb without a significant change in pH. • A buffer is most effective when the concentrations of the acid and its conjugate base are equal; if the concentrations of the acid and conj. base are different by more than a factor of 10, the buffer will not be reasonably effective. • The more concentrated the components of a buffer are, the greater the buffer capacity. (A buffer made from 1.0 M acetic acid and 1.0 M sodium acetate has the same pH as a buffer made from 0.1 M solutions of each, but has a much greater ability to resist pH changes.)

  16. Example Problem(Adding Strong Acid to a Buffered Solution) • Calculate the change in pH that occurs when 0.010 mol gaseous HCl is added to 1.0 L of each of the following solutions: • Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2 • Solution B: 0.050 M HC2H3O2and 0.050 M NaC2H3O2 For acetic acid, Ka = 1.8 x 10-5

  17. Preparing a Buffer • Large changes in the ratio [A-] / [HA] will produce large changes in pH. We want to avoid this situation. • Optimal buffering occurs when [A-] = [HA]. • So pH = pKa. (Most effective range for a buffer is within ±1 pH units of the pKa of the weak acid.)

  18. Steps to preparing a buffer: 1.Choose the conjugate acid-base pair. • It is best to use a conjugate acid-base pair whose pKa of the weak acid is within ±1 unit of the desired pH. • For a buffer at a pH of 3.90, formic acid, HCO2H (pKa = 3.74) would be a good choice. 2.Calculate the ratio of buffer components that gives the desired pH (using the Henderson-Hasselbalch equation).

  19. Steps to preparing a buffer continued… 3.Determine the buffer concentration. • For many laboratory applications, concentrations of 0.50 M are suitable. • From the concentration of the acid solutions in the available solutions, calculate the mass of conjugate base necessary to make the buffer. (e.g. we have 0.40 M formic acid and we want to make 1.0 liters of the buffer solution.) 4.Mix the solution and adjust the pH by adding small amounts of strong acid or base.

  20. Example Problem • A chemist needs a solution buffered at pH 4.30 and can choose from the following acids (and their sodium salts). Which system would work best, and why? 1. 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)

  21. Titrations and pH Curves • Equivalence Point: (stoichiometric point) • The point in the titration where an amount of base has been added to exactly react with all acid originally present in the solution. • End Point: • Point at which the indicator changes color.

  22. Strong Acid-Strong Base Titrations (H+ + OH- H2O) Case Study – titrating 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH. a. No NaOH has been added b. 10.0 mL of 0.100 M NaOH has been added. c. 20.0 mL (total) of 0.100 M NaOH has been added. d. 50.0 mL (total) of 0.100 M NaOH has been added. e. 100.0 mL (total) of 0.100 M NaOH has been added f. 200.0 mL (total) of 0.100 M NaOH has been added.

  23. Titration Curve

  24. Titrations of Weak Acids with Strong Bases • Calculating the pH curve for a weak acid-strong base titration is a two-step procedure: • 1. A stoichiometry problem: The reaction of hydroxide ion with the weak acid is assumed to run to completion, and the concentration of acid remaining and the conjugate base formed are determined. • 2. An equilibrium problem: The position of the weak acid equilibrium is determined, and the pH is calculated.

  25. Case Study – titrating 50.0 mL of 0.10 M acetic acid (Ka = 1.8 x 10-5) with 0.10 M NaOH. a. No NaOH has been added. b. 10.0 mL of 0.100 M NaOH has been added. c. 25.0 mL (total) of 0.100 M NaOH has been added. d. 40.0 mL (total) of 0.100 M NaOH has been added. e. 50.0 mL (total) of 0.100 M NaOH has been added. f. 60.0 mL (total) of 0.100 M NaOH has been added. g. 75.0 mL (total) of 0.100 M NaOH has been added.

  26. Titration Curve

  27. Example Problem(Titration of a Weak Acid) • Hydrogen cyanide gas (HCN) is highly toxic. It is a very weak acid (Ka = 6.2 x 10-10) when dissolved in water. If a 50.0-mL sample of 0.100 HCN is titrated with 0.100 M NaOH, calculate the pH of the solution a. after 8.00 mL of 0.100 M NaOH has been added. b. at the halfway point of the titration c. at the equivalence point of the titration.

  28. Example Problem(Calculating Ka) • A chemist has synthesized a monoprotic weak acid and wants to determine its Ka value. To do so, the chemist dissolves 2.00 mmol of the solid acid in 100.0 mL water and titrates the resulting solution with 0.0500 M NaOH. After 20.0 mL NaOH has been added, the pH is 6.00. What is the Ka value for the acid?

  29. Titrations of Weak Bases with Strong Acids Case Study – titrating 100.0 mL of 0.050 M NH3 with 0.10 M HCl. (Kb= 1.8 x 10-5) Calculate the pH… • a. Before the addition of any HCl • b. Before the equivalence point. (half-way) • c. At the equivalence point. • d. Beyond the equivalence point.

  30. Titration Curve

  31. KHP is a COMMON standardizing agent. • Moles KHP = moles of H+

  32. Titration Curve for Polyprotic Acids • Titration curve for Phosphoric Acid (H3PO4) with NaOH

  33. Acid-Base Indicators Two ways to determine the equivalence point of an acid-base titration: • 1.Use a pH meter to monitor pH and plot the titration curve. The center of the vertical region represents the equivalence point. • 2.Use an acid-base indicator, which marks the end point in a titration by changing color. The equivalence point of a titration defined by stoichiometry is not necessarily the same as the end point, but careful selection of the indicator will ensure the error is negligible. We choose an indicator with an end point close to the equivalence point, so that the color change in the indicator corresponds to the point at which the acid has been neutralized.

  34. Acid-Base Indicators Continued… Phenolphthalein: HIn (colorless)  H+ + In-(pink) • How does a molecule function as an indicator? Hypothetical indicator HIn, a weak acid with Ka = 1.0 x 10-8. HIn(aq)  H+(aq) + In-(aq) Red Blue • For most indicators, about 1/10 of the initial form must be converted to the other form before a new color is apparent.

  35. Example Problem • Bromothymol blue, an indicator with a Ka value of 1.0 x 10-7, is yellow in its HIn form. Suppose we put a few drops of this indicator in a strongly acidic solution. If this solution is titrated with NaOH, at what pH will the indicator color change first be visible?

  36. Example Problem What indicator should be used for the following titrations: (page 715) a.100 mL of 0.100 M HCl with 0.100 M NaOH (equivalence point at pH 7.00) b.100 mL of 0.100 M HC2H3O2 with 0.100 M NaOH (equivalence point at pH 8.7)

  37. Individual Practice • Beginning on page 741 • #39, 41, 49, 51, 53, 55, 57, 59, 61, 63, 65, 69, 71, and 73

  38. Solubility Equilibria and the Solubility Product Constant (Ksp) PbCl2(s) ↔ Pb2+(aq) + 2 Cl-(aq) what does it mean that PbCl2 is “insoluble”? Solubility Product Constant (Ksp) PbCl2 = 1.17 x 10-5

  39. Example Problem • Write the solubility product expression for the following species: • magnesium carbonate • iron (II) hydroxide • silver carbonate • calcium phosphate

  40. Example Problems • Copper (I) bromide has a measured solubility of 2.0 x 10-4mol/L at 25°C. Calculate the Ksp value. • Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0 x 10-15mol/L at 25°C. • The Ksp value for copper (II) iodate is 1.4 x 10-7 at 25°C. Calculate its molar solubility at 25°C.

  41. CAUTION: Be careful comparing relative solubilities • AgI(Ksp = 1.5 x 10-16); CuI (Ksp = 5.0 x 10-12); CaSO4 (Ksp = 6.1 x 10-5) • CuS(Ksp = 8.5 x 10-45); Ag2S (Ksp = 1.6 x 10-49); Bi2S3 (Ksp = 1.1 x 10-73)

  42. Common Ion Effect • What is the solubility of silver chromate (Ksp = 9.0 x 10-12) in a 0.100 M solution of silver nitrate? • Calculate the solubility of solid CaF2 (Ksp = 4.0 x 10-11) in a 0.025 M NaF solution.

  43. pH and Solubility The pH of a solution can greatly affect a salt’s solubility. • Milk of Magnesia: Mg(OH)2 ↔ Mg2+(aq) + 2 OH-(aq) Ksp = 1.8 x 10-11 • How would the addition of OH- ions (increase in pH) affect the solubility? • How would the addition of H+ ions (decrease in pH) affect the solubility?

  44. Example Problem Saturated solution of Mg(OH)2. (Ksp = 1.8 x 10-11) • What is [Mg2+]? • What is pH? • Now suppose that solid Mg(OH)2 is equilibrated with a solution buffered at a more acidic pH of 9.0. What is [Mg2+]?

  45. pH and Solubility Ag3PO4(s) ↔ 3 Ag+(aq) + PO43-(aq) • Why more soluble in acidic solutions? • Is PO43- a strong or weak base? (H+ + PO43- HPO42-) AgCl(s) ↔Ag+(aq) + Cl-(aq) • Is Cl- a strong or weak base? • How does this affect the solubility of AgCl in acidic solutions? General Rule – If the anion X- is an effective base (HX is a weak acid), the salt MX will show an increased solubility in acidic solutions.

  46. Precipitation and Qualitative Analysis • Ion Product (Qsp) – just like Ksp, but use initial concentrations (Q = [Ca2+]0[F-]02) • If Q > Ksp, precipitation occurs and will continue until the concentrations are reduced to the point that they satisfy Ksp. • If Q < Ksp, no precipitation occurs. • If Q = Ksp, it is a saturated solution.

  47. Example Problems • A solution is prepared by adding 750.0 mL of 4.00 x 10-3 M Ce(NO3)3 to 300.0 mL of 2.00 x 10-2 M KIO3. Will Ce(IO3)3 (Ksp = 1.9 x 10-10) precipitate from this solution? • Suppose we mix 100.0 mL of 0.0500 M lead (II) nitrate and 200.0 mL of 0.100 M sodium iodide. Determine whether a precipitate occurs and then determine the equilibrium concentrations of Pb2+ and I- ions.(Ksp = 1.4 x 10-8)

  48. Example Problem • A solution is prepared by mixing 150.0 mL of 1.00 x 10-2 M Mg(NO3)2 and 250.0 mL of 1.00 x 10-1 M NaF. Calculate the concentrations of Mg2+ and F- at equilibrium with solid MgF2 (Ksp = 6.4 x 10-9).

  49. Example Problem(Selective Precipitation) • A solution contains 1.0 x 10-4 M Cu+ and 2.0 x 10-3 M Pb2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp = 1.4 x 10-8) or CuI (Ksp = 5.3 x 10-12) precipitate first? Specify the concentration of I- necessary to begin precipitation of each salt.

  50. Selective Precipitation • 1.Since metal sulfide salts differ dramatically in their solubilities, the sulfide ion is often used to separate metal ions by selective precipitation. For example, consider a solution containing a mixture of Fe2+ (Ksp of FeS = 3.7 x 10-19) and Mn2+ (Ksp of MnS = 2.3 x 10-13). Which will precipitate first as you carefully add the S2-? • 2.Another advantage of the sulfide ion is that it is basic, so its concentration can be controlled by regulating the pH of the solution. H2S ↔ H+ + HS- Ka1 = 1.0 x 10-7 HS- ↔ H+ + S2-Ka2= 1.0 x 10-19 • In an acidic solution [S2-] will be small, but in basic solutions [S2-] will be relatively large. This means that the most insoluble salts (for example, CuS, Ksp = 8.5 x 10-45) can be precipitated from an acidic solution, leaving the more soluble ones still dissolved. The remaining cations can be precipitated by making the solution slightly basic.

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