2012 topic 18 2   buffer solutions
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2012 topic 18 2 buffer solutions

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2012 topic 18 2   buffer solutions 2012 topic 18 2 buffer solutions Presentation Transcript

  • IB Chemistry Power Points Topic 18 Acids and Baseswww.pedagogics.ca Buffer Solutions
  • Buffer Solutions DEFINITION: A buffer solution contains a weak acid mixed with its conjugate base (or weak base and conjugate acid) Buffers resist changes in pH when a small amount of a strong acid or base is added to it. HA ∏ H+ + A-
  •  If a small amount of a strong acid (H+) is added eqm shifts to the left as [H+] increases so system adjusts to increase [HA] and reduce [H+] again. HA ∏ H+ + A-
  •  A small amount of a strong base will react with H+ to form H2O and eqm will shift to the right to increase [H+] again. HA ∏ H+ + A-
  • Making Buffer Solutions An example of a weak acid is ethanoic acid. This could be mixed with sodium ethanoate which will provide ethanoate ions (conjugate base). CH3COOH(aq) ∏ CH3COO-(aq)+H+(aq) An example of a weak base is ammonia. This could be mixed with ammonium chloride to provide ammonium ions (conjugate acid). NH3(aq) + H2O(aq) ∏ NH4+(aq) +OH-(aq) In order for a buffer to work well the concentration of the acid/base and its salt must be much higher than the strong acid/base added.
  • Optimum Buffer A buffer is most effective when the concentration of weak acid and its salt (the conjugate base) are equal and the pH is equal to pKa. In practice it will work reasonably well with similar concentrations and the effective buffer range of any weak acid/ base is pKa 1.
  • Blood has buffering capacity Blood must maintain a pH of 7.4 so its enzymes can work. If 0.01 mol of H+ or OH- is added to blood it only changes pH by 0.1 unit. The eqm is: CO2(aq) + H2O(l) ) ∏ H+(aq) + HCO3-(aq)
  • Buffer Calculation #1Calculate the pH of a 1.00 dm3 buffer solution made bydissolving 0.50 mol of sodium ethanoate into a 0.075 mol dm-3ethanoic acid solution. 1. Identify the weak acid / conjugate base OR weak base / conjugate acid pair. Determine their concentrations.
  • Buffer Calculation #1Calculate the pH of a 1.00 dm3 buffer solution made bydissolving 0.50 mol of sodium ethanoate into a 0.075 mol dm-3ethanoic acid solution. 1. Identify the weak acid / conjugate base OR weak base / conjugate acid pair. Determine their concentrations. weak acid [CH3COOH] = 0.075 mol dm-3 conjugate base [CH3COO-] = 0.50 mol dm-3
  • Buffer Calculation #1Calculate the pH of a 1.00 dm3 buffer solution made bydissolving 0.50 mol of sodium ethanoate into a 0.075 mol dm-3ethanoic acid solution. 2. Write the equilibrium equation and expression
  • Buffer Calculation #1Calculate the pH of a 1.00 dm3 buffer solution made bydissolving 0.50 mol of sodium ethanoate into a 0.075 mol dm-3ethanoic acid solution. 2. Write the equilibrium equation and expression CH3COOH ∏ H+ + CH3COO- [H + ][CH 3COO - ] - pK Ka = = 10 a [CH 3COOH ]
  • Buffer Calculation #1Calculate the pH of a 1.00 dm3 buffer solution made bydissolving 0.50 mol of sodium ethanoate into a 0.075 mol dm-3ethanoic acid solution. ICE table for clarity CH3COOH ∏ H+ + CH3COO- I C E HA 0.075 -x 0.075 - x H+ 0 +x x A- 0.50 +x 0.50 + x
  • Buffer Calculation #1Calculate the pH of a 1.00 dm3 buffer solution made bydissolving 0.50 mol of sodium ethanoate into a 0.075 mol dm-3ethanoic acid solution. 3. In this case, assume the equilibrium concentrations of the weak acid and the salt anion are assumed to be the same as the given information (very little change when equilibrium is established.
  • Buffer Calculation #1Calculate the pH of a 1.00 dm3 buffer solution made bydissolving 0.50 mol of sodium ethanoate into a 0.075 mol dm-3ethanoic acid solution. 3. In this case, assume the equilibrium concentrations of the weak acid and the salt anion are assumed to be the same as the given information (very little change when equilibrium is established. [H + ][0.50] Ka = = 10-4.76 [0.075] æ 0.075´10-4.76 ö pH = -log[H + ] = -log ç ÷ è 0.50 ø
  • Buffer Calculation #1Calculate the pH of a 1.00 dm3 buffer solution made bydissolving 0.50 mol of sodium ethanoate into a 0.075 mol dm-3ethanoic acid solution. pH = 5.6
  • Henderson-Hasselbalch Equation
  • Henderson-Hasselbalch Equation æ [A- ] ö pH = pK a + log ç ÷ è [HA] ø æ 0.50 ö pH = 4.76 + log ç ÷ è 0.075 ø pH = 5.58
  • Buffer Calculation #2 Calculate the mass of ammonium chloride that would need to be dissolved into 1.00 dm3 of 0.100 mol dm-3 NH3 solution to create a buffer with a pH of 9.00. Assume no change in overall volume. 1. Identify the weak acid / conjugate base OR weak base / conjugate acid pair. Determine their concentrations.
  • Buffer Calculation #2 Calculate the mass of ammonium chloride that would need to be dissolved into 1.00 dm3 of 0.100 mol dm-3 NH3 solution to create a buffer with a pH of 9.00. Assume no change in overall volume. 1. Identify the weak acid / conjugate base OR weak base / conjugate acid pair. Determine their concentrations. weak base [NH3] = 0.100 mol dm-3 conjugate acid [NH4+] = ?
  • Buffer Calculation #2 Calculate the mass of ammonium chloride that would need to be dissolved into 1.00 dm3 of 0.100 mol dm-3 NH3 solution to create a buffer with a pH of 9.00. Assume no change in overall volume. 2. Write the equilibrium equation and expression
  • Buffer Calculation #2 Calculate the mass of ammonium chloride that would need to be dissolved into 1.00 dm3 of 0.100 mol dm-3 NH3 solution to create a buffer with a pH of 9.00. Assume no change in overall volume. 2. Write the equilibrium equation and expression NH3+ H2O ∏ NH4+ + OH- [NH 4+ ][OH - ] - pKb Kb = = 10 [NH 3 ]
  • Buffer Calculation #2 Calculate the mass of ammonium chloride that would need to be dissolved into 1.00 dm3 of 0.100 mol dm-3 NH3 solution to create a buffer with a pH of 9.00. Assume no change in overall volume. 3. In this case, assume the equilibrium concentrations of the weak base is the same as the given information (very little change when equilibrium is established. Calculate [OH-] from desired pH (14 pH ) [ NH 4 ][10 ] 4.75 kb 10 [0.100] 0.100 10 4.75 [ NH 4 ] 0.1778 mol dm-3 10 (14 9)
  • Buffer Calculation #2 Calculate the mass of ammonium chloride that would need to be dissolved into 1.00 dm3 of 0.100 mol dm-3 NH3 solution to create a buffer with a pH of 9.00. Assume no change in overall volume. 3. In this case, assume the equilibrium concentrations of the weak base is the same as the given information (very little change when equilibrium is established. Calculate [OH-] from desired pH
  • Buffer Calculation #2 Calculate the mass of ammonium chloride that would need to be dissolved into 1.00 dm3 of 0.100 mol dm-3 NH3 solution to create a buffer with a pH of 9.00. Assume no change in overall volume. mass = 9.50 g
  • Buffer Calculation #3 A buffer can also be made by mixing excess weak acid/base with a lesser amount of strong base/acid. For example, calculate the pH of a buffer formed when 25 cm3 of 0.075 mol dm-3 HCl is added to 40 cm3 of a 0.150 mol dm-3 ammonia solution. 1. Do the stoichiometry to determine what the concentrations are AFTER neutralization. -
  • Buffer Calculation #3 A buffer can also be made by mixing excess weak acid/base with a lesser amount of strong base/acid. For example, calculate the pH of a buffer formed when 25 cm3 of 0.075 mol dm-3 HCl is added to 40 cm3 of a 0.150 mol dm-3 ammonia solution. 1. Do the stoichiometry to determine what the concentrations are AFTER neutralization. HCl + NH3  NH4+ + Cl-
  • Buffer Calculation #3 A buffer can also be made by mixing excess weak acid/base with a lesser amount of strong base/acid. For example, calculate the pH of a buffer formed when 25 cm3 of 0.075 mol dm-3 HCl is added to 40 cm3 of a 0.150 mol dm-3 ammonia solution. 1. Do the stoichiometry to determine what the concentrations are AFTER neutralization. HCl + NH3  NH4+ + Cl- (0.025 0.075) NH 4 HCl limiting (why?) 0.065 NH 4 0.02885 (0.040 0.150) (0.025 0.075) NH 3 0.065 NH 3 0.06346
  • Buffer Calculation #3 A buffer can also be made by mixing excess weak acid/base with a lesser amount of strong base/acid. For example, calculate the pH of a buffer formed when 25 cm3 of 0.075 mol dm-3 HCl is added to 40 cm3 of a 0.150 mol dm-3 ammonia solution.2. Write the equilibrium equation and expression
  • Buffer Calculation #3 A buffer can also be made by mixing excess weak acid/base with a lesser amount of strong base/acid. For example, calculate the pH of a buffer formed when 25 cm3 of 0.075 mol dm-3 HCl is added to 40 cm3 of a 0.150 mol dm-3 ammonia solution.2. Write the equilibrium equation and expression NH3+ H2O ∏ NH4+ + OH- [NH 4+ ][OH - ] - pKb Kb = = 10 [NH 3 ]
  • Buffer Calculation #3 A buffer can also be made by mixing excess weak acid/base with a lesser amount of strong base/acid. For example, calculate the pH of a buffer formed when 25 cm3 of 0.075 mol dm-3 HCl is added to 40 cm3 of a 0.150 mol dm-3 ammonia solution. 3. In this case, assume the equilibrium concentrations of the weak base and cation are the same as determined by stoichiometry (very little change when equilibrium is established.
  • Buffer Calculation #3 A buffer can also be made by mixing excess weak acid/base with a lesser amount of strong base/acid. For example, calculate the pH of a buffer formed when 25 cm3 of 0.075 mol dm-3 HCl is added to 40 cm3 of a 0.150 mol dm-3 ammonia solution. 3. In this case, assume the equilibrium concentrations of the weak base and cation are the same as determined by stoichiometry (very little change when equilibrium is established. [0.02885][OH - ] Kb = = 10-4.75 [0.06346] æ 0.06346 ´10-4.75 ö pOH = -log ç ÷ è 0.02885 ø
  • Buffer Calculation #3 A buffer can also be made by mixing excess weak acid/base with a lesser amount of strong base/acid. For example, calculate the pH of a buffer formed when 25 cm3 of 0.075 mol dm-3 HCl is added to 40 cm3 of a 0.150 mol dm-3 ammonia solution. pH= 9.6