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Applications of Aqueous Equilibria

Applications of Aqueous Equilibria. Buffers, Acid-Base Titrations and Precipitation Reactions. Buffers. Buffers are solutions of a weak acid and its conjugate base.

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Applications of Aqueous Equilibria

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  1. Applications of Aqueous Equilibria Buffers, Acid-Base Titrations and Precipitation Reactions

  2. Buffers Buffers are solutions of a weak acid and its conjugate base. Buffers resist changes in pH when small amounts of strong acid or base are added. This is because there is both an acid and its conjugate base present initially. Buffers typically have the acid and its conjugate base (or salt) present in roughly equal concentrations.

  3. Buffers Consider a buffer of weak acid HA and NaA in equimolar concentrations. The predominant reaction is: HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) If a small amount of strong acid is added, there is enough A- available to become protonated. The equilibrium shifts to the left, and the [H3O+] and pH do not change much.

  4. Buffers HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) Likewise, if a small amount of strong base is added, some of the hydronium will react with it. H3O+(aq) + OH-(aq)  2 H2O(l) As hydronium ion reacts, more HA will dissociate so as to replenish the [H3O+]. As a result, the pH remains fairly constant.

  5. Buffers Because buffers contain both a weak acid and its conjugate base, they can react with strong acids or bases and maintain their pH.

  6. Buffers

  7. Problem: Buffers • Determine the pH of a solution that contains 0.50M HClO and 0.60M NaClO. (Ka for HClO = 3.5 x 10-8) - Write the major reactions: NaClO(aq)  Na+(aq) + ClO-(aq) The NaClO serves as a source of ClO- ions. The sodium ion is a spectator.

  8. Problem: Buffers • Determine the pH of a solution that contains 0.50M HClO and 0.60M NaClO. Even though both HClO and ClO- are present initially, it’s important to realize that they are in equilibrium with each other, and go on opposite sides of the equation. HClO(aq) + H2O(l) ↔ H3O+(aq) + ClO-(aq)

  9. Problem: Buffers • Determine the pH of a solution that contains 0.50M HClO and 0.60M NaClO. (Ka for HClO = 3.5 x 10-8) HClO(aq) + H2O(l) ↔ H3O+(aq) + ClO-(aq) Ka = [H3O+][ClO-] = 3.5 x 10-8 [HClO]

  10. Problem: Buffers • Determine the pH of a solution that contains 0.50M HClO and 0.60M NaClO. (Ka for HClO = 3.5 x 10-8) HClO(aq) + H2O(l) ↔ H3O+(aq) + ClO-(aq)

  11. The Henderson-Hasselbalch Equation For buffers, regardless of the value of Ka, you can always assume that x will be small compared to the concentrations of the acid or its conjugate base. This is because the reaction will proceed only slightly to the right due to the presence of the conjugate base initially. HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq)

  12. The Henderson-Hasselbalch Equation As a result, the Ka expression can be written in logarithmic form. Ka = [H3O+][A-] [HA] [H3O+] = Ka [HA] [A-] pH = pKa + log [A-] [HA]

  13. The Henderson-Hasselbalch Equation In more general terms, the equation is written as: pH = pKa + log This equation is for buffers only, and is a quick way to calculate pH. [base] [acid]

  14. Problem: Adding Acid to a Buffer • Calculate the change in pH if 5.00 mL of 0.010M HNO3 is added to 50.0 mL of the HClO/NaClO buffer. The nitric acid will protonate some of the ClO- to form additional HClO. HNO3(aq) + H2O(l)  H3O+(aq) + NO3-(aq) H3O+(aq) + ClO-(aq) HClO(aq) + H2O(l)

  15. Problem: Adding Acid to a Buffer HNO3(aq) + H2O(l)  H3O+(aq) + NO3-(aq) H3O+(aq) + ClO-(aq) HClO(aq) + H2O(l) Before you can use the Henderson-Hasselbalch equation, you need to calculate the new concentration of HClO and ClO- that results from adding the nitric acid.

  16. Problem: Adding Acid to a Buffer HNO3(aq) + H2O(l)  H3O+(aq) + NO3-(aq) H3O+(aq) + ClO-(aq) HClO(aq) + H2O(l) For each mole of nitric acid added, you form a mole of HClO and react a mole of ClO-.

  17. Problem: Adding Acid to a Buffer 1. You will need to calculate the moles of HNO3 added (5.00 mL of 0.010M HNO3). 2. For each mole of nitric acid added, you will produce a mole of HClO and lose a mole of ClO-. 3. Since you need concentrations, remember that the new volume of the solution has changed to 55.0 mL to 50.0 mL.

  18. Problem: Compare to Water • Calculate the change in pH when 5.00 mLs of 0.0100M HNO3 is added to 50.0 mL of water.

  19. Preparation of Buffer Solutions Scientists often need to make a solution that is buffered to a specific pH. They may need to mimic biological conditions, test the corrosiveness of metal parts at a specific pH, etc. Conjugate acid/base pairs have a pH range at which they can effectively serve as buffers.

  20. Choosing the Acid and Salt For optimum buffering, choose an acid with a Ka value close to the desired [H3O+] of the buffer, or with a pKa value near the desired pH. You can then use the Henderson-Hasselbalch equation (or Ka expression) to determine the relative concentration of base to acid needed to prepare the buffer.

  21. Preparing a Buffer • Choose an appropriate acid and base to make a buffer with a pH of 6.50. Calculate the relative concentration of acid and base needed.

  22. Acid-Base Titrations Titration is a laboratory technique in which the amount and concentration of one reactant is known, and the concentration of the other reactant is determined. Titrations can be applied to redox reactions, and precipitation reactions, but they are most commonly used for analyzing solutions of acids or bases.

  23. Acid-Base Titrations Acid-base titrations are based on neutralization reactions, in which an acid is completely neutralized by a base. The progress of the reaction may be monitored using a pH meter, or pH indicators.

  24. Titrations In either method, base is added to the acid (or vice versa) until the equivalence point is reached. The equivalence point has been reached when the moles of acid exactly equals the moles of base. It is sometime called the stoichiometric point.

  25. Strong Acid/Strong Base Titration

  26. Titrations Not all titrations reach the equivalence point at a pH of 7. The nature of the acid and base (weak or strong) will influence the pH at the equivalence point. As a result, indicators much be carefully chosen to change color at the desired pH.

  27. Strong Acid & Strong Base When a strong acid is titrated with a strong base, a “neutral” salt and water result. Since the conjugate base of a strong acid has no tendency to accept protons, and the conjugate acid of a strong base has no tendency to donate protons, the pH at the equivalence point will be 7.

  28. Strong Acid & Strong Base If the pH is monitored during the titration using a pH meter, a graphical presentation of the pH versus volume of titrant can be obtained.

  29. Strong Acid & Strong Base The graph has certain features characteristic of a strong acid-strong base titration.

  30. Strong Acid & Strong Base Note the equivalence point of 7, and the nearly vertical rise in pH from 4-10.

  31. Strong Acid & Strong Base Because the rise in pH is vertical over such a broad range, there are several indicators that may be used with accurate results.

  32. Strong Acid & Strong Base Phenolphthalein, which changes color at a pH of around 9, and methyl red, which changes color at a pH of approximately 5 will both give highly accurate results.

  33. Strong Acid & Strong Base Phenolphthalein is usually used because it goes from colorless to pink, and our eyes are better at detecting the presence of color than a change in color.

  34. Weak Acid & Strong Base When a weak acid is titrated with a strong base, the salt that is produced will contain the conjugate base of the weak acid. As a result, the pH will be greater than 7 at the equivalence point.

  35. Weak Acid & Strong Base When a weak acid is titrated with a strong base, the salt that is produced will contain the conjugate base of the weak acid. As a result, the pH will be greater than 7 at the equivalence point.

  36. Weak Acid & Strong Base The titration curve for acetic acid with NaOH shows an equivalence point at a pH of 9. The lack of a steep vertical rise in pH means that selection of the proper indicator is crucial.

  37. Weak Acid & Strong Base Another feature of the curve is the flattening out or leveling off of pH near the half-way point. This is not observed in strong acid-strong base titrations.

  38. Weak Acid & Strong Base Another feature of the curve is the flattening out or leveling off of pH near the half-way point. This is not observed in strong acid-strong base titrations.

  39. Weak Acid & Strong Base The pH levels off due to the formation of a buffer mid-way through the titration. Half of the acid has been de-protonated to form its conjugate base.

  40. Titration of Benzoic Acid • 25.0 mL of 0.10M benzoic acid (Ka = 6.4 x 10-5) is titrated with 0.10M NaOH. Calculate the pH: - Initially - After 12.5 mL of NaOH have been added - At the equivalence point Choose an appropriate indicator for the titration.

  41. Titration of Benzoic Acid • 25.0 mL of 0.10M benzoic acid (Ka = 6.4 x 10-5) is titrated with 0.10M NaOH. We will need to know how much NaOH is needed for complete neutralization. In this case, since the concentrations of acid and base are the same, 25.0 mL of 0.10M acid will require an equal volume of 0.10M base.

  42. Calculating Volume for Neutralization In less obvious cases, the following relationship can be used. We are assuming complete neutralization of a monoprotic acid by a monobasic base. moles acid = moles base MacidVacid = MbaseVbase

  43. Titration of Benzoic Acid • 25.0 mL of 0.10M benzoic acid (Ka = 6.4 x 10-5) is titrated with 0.10M NaOH. Calculate the pH: - Initially Major Reaction: HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq) init: 0.10 0 0 chg: -x +x +x equ. .10-x x x

  44. Titration of Benzoic Acid • 25.0 mL of 0.10M benzoic acid (Ka = 6.4 x 10-5) is titrated with 0.10M NaOH. Calculate the pH: - After 12.5 mL of NaOH have been added Titration Reaction: HA(aq) + OH-(aq)  H2O(l) + A-(aq) For each mole of NaOH added, a mole of HA has been converted to A-.

  45. Titration of Benzoic Acid • 25.0 mL of 0.10M benzoic acid (Ka = 6.4 x 10-5) is titrated with 0.10M NaOH. Calculate the pH: - After 12.5 mL of NaOH have been added This problem can be greatly simplified if you realize that this is the mid-point of the titration. Since complete neutralization will require 25.0 mLs of NaOH, at 12.5 mL base, half of the acid has been converted to its conjugate base.

  46. Titration of Benzoic Acid • 25.0 mL of 0.10M benzoic acid (Ka = 6.4 x 10-5) is titrated with 0.10M NaOH. Calculate the pH: - After 12.5 mL of NaOH have been added In essence, a buffer has been formed, with [HA] = [A-]. (This is the place in the titration curve where the pH begins to level off.)

  47. Titration of Benzoic Acid • 25.0 mL of 0.10M benzoic acid (Ka = 6.4 x 10-5) is titrated with 0.10M NaOH. Calculate the pH: - After 12.5 mL of NaOH have been added We can use the Henderson-Hasselbalch equation, because a buffer is present. Since [HA] = [A-], we need not calculate the actual concentration, as they will cancel out.

  48. Titration of Benzoic Acid • 25.0 mL of 0.10M benzoic acid (Ka = 6.4 x 10-5) is titrated with 0.10M NaOH. Calculate the pH: - After 12.5 mL of NaOH have been added pH = pKa + log ([A-]/[HA]) pH = -log (6.4 x 10-5) + log (1.0) pH = 4.19

  49. Titration of Benzoic Acid • Calculate the pH at the equivalence point. At the equivalence point, enough NaOH has been added (25.0 mL) to completely neutralize the benzoic acid. There are two reactions to consider. 1. The neutralization reaction. 2. The subsequent reaction of the benzoate ion with water.

  50. Titration of Benzoic Acid • At the equivalence point, benzoate ion will react with water, since it is the conjugate base of a weak acid. Major reactions: HA(aq) + OH-(aq)  H2O(l) + A-(aq) A-(aq) + H2O(l) ↔ OH-(aq) + HA(aq)

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