1. 1 How Much Does the pH of a Buffer Change When an Acid or Base Is Added?  though buffers do resist change in pH when acid or base are added to them, their pH does change  calculating the new pH after adding acid or base requires breaking the problem into 2 parts a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other added acid reacts with the A − to make more HA added base reacts with the HA to make more A − an equilibrium calculation of [H 3 O + ] using the new initial values of [HA] and [A − ]  though buffers do resist change in pH when acid or base are added to them, their pH does change  calculating the new pH after adding acid or base requires breaking the problem into 2 parts a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other added acid reacts with the A − to make more HA added base reacts with the HA to make more A − an equilibrium calculation of [H 3 O + ] using the new initial values of [HA] and [A − ]

2. 2 Compare the effect on pH of adding 0.010 mol NaOH to a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L to adding 0.010 mol NaOH to 1.00 L of pure water? HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 - + H 3 O + pK a for HC 2 H 3 O 2 = 4.745 Buffer Pure Water

3. 3 Basic Buffers  buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, (H:B + )Cl − H 2 O (l) + NH 3 (aq) NH 4 + (aq) + OH − (aq) B: (aq) + H 2 O (l) H:B + (aq) + OH − (aq)

4. 4 What is the pH of a buffer that is 0.50 M NH 3 (pK b = 4.75) and 0.20 M NH 4 Cl? find the pK a of the conjugate acid (NH 4 + ) from the given K b Assume the [B] and [HB + ] equilibrium concentrations are the same as the initial Substitute into the Henderson-Hasselbalch Equation Check the “x is small” approximation NH 3 + H 2 O NH 4 + + OH −

5. 5 Buffering Effectiveness  a good buffer should be able to neutralize moderate amounts of added acid or base  however, there is a limit to how much can be added before the pH changes significantly  the buffering capacity is the amount of acid or base a buffer can neutralize  the buffering range is the pH range the buffer can be effective  the effectiveness of a buffer depends on two factors  (1) the relative amounts of acid and base  (2) the absolute concentrations of acid and base  a good buffer should be able to neutralize moderate amounts of added acid or base  however, there is a limit to how much can be added before the pH changes significantly  the buffering capacity is the amount of acid or base a buffer can neutralize  the buffering range is the pH range the buffer can be effective  the effectiveness of a buffer depends on two factors  (1) the relative amounts of acid and base  (2) the absolute concentrations of acid and base

6. HAA-A- OH − mols Before0.180.0200 mols added --0.010 mols After 0.170.030≈ 0 Effect of Relative Amounts of Acid and Conjugate Base Buffer 1 0.100 mol HA & 0.100 mol A - Initial pH = 5.00 Buffer 2 0.18 mol HA & 0.020 mol A - Initial pH = 4.05 pK a (HA) = 5.00 after adding 0.010 mol NaOH pH = 5.09 HA + OH − A - + H 2 O HAA-A- OH − mols Before0.100 0 mols added --0.010 mols After 0.0900.110≈ 0 after adding 0.010 mol NaOH pH = 4.25 a buffer is most effective with equal concentrations of acid and base pH change after adding 0.010M NaOH

7. HAA-A- OH − mols Before0.500.5000 mols added --0.010 mols After 0.490.51≈ 0 HAA-A- OH − mols Before0.050 0 mols added --0.010 mols After 0.0400.060≈ 0 Effect of Absolute Concentrations of Acid and Conjugate Base Buffer 1 0.50 mol HA & 0.50 mol A - Initial pH = 5.00 Buffer 2 0.050 mol HA & 0.050 mol A - Initial pH = 5.00 pK a (HA) = 5.00 after adding 0.010 mol NaOH pH = 5.02 HA + OH − A - + H 2 O after adding 0.010 mol NaOH pH = 5.18 a buffer is most effective when the concentrations of acid and base are largest pH change after adding 0.010M NaOH

8. 8 Effectiveness of Buffers  a buffer will be most effective when the [base]:[acid] = 1  equal concentrations of acid and base  effective when 0.1 < [base]/[acid] < 10  a buffer will be most effective when the [acid] and the [base] are large  a buffer will be most effective when the [base]:[acid] = 1  equal concentrations of acid and base  effective when 0.1 < [base]/[acid] < 10  a buffer will be most effective when the [acid] and the [base] are large

9. 9 Buffering Range  A buffer will be effective when 0.1 < [base]:[acid] < 10  substituting into the Henderson-Hasselbalch we can calculate the maximum and minimum pH at which the buffer will be effective  A buffer will be effective when 0.1 < [base]:[acid] < 10  substituting into the Henderson-Hasselbalch we can calculate the maximum and minimum pH at which the buffer will be effective Lowest pH Highest pH therefore, the effective pH range of a buffer is pK a ± 1 when choosing an acid to make a buffer, choose one whose is pK a is closest to the pH of the buffer

10. 10 Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25? Chlorous Acid, HClO 2 pK a = 1.95 Nitrous Acid, HNO 2 pK a = 3.34 Formic Acid, HCHO 2 pK a = 3.74 Hypochlorous Acid, HClOpK a = 7.54

11. 11 The pK a of HCHO 2 is closest to the desired pH of the buffer, so it would give the most effective buffering range. Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25? Chlorous Acid, HClO 2 pK a = 1.95 Nitrous Acid, HNO 2 pK a = 3.34 Formic Acid, HCHO 2 pK a = 3.74 Hypochlorous Acid, HClOpK a = 7.54

12. 12 What ratio of NaCHO 2 : HCHO 2 would be required to make a buffer with pH 4.25? Formic Acid, HCHO 2, pK a = 3.74 to make the buffer with pH 4.25, you would use 3.24 times as much NaCHO 2 as HCHO 2

13. 13 Buffering Capacity  buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness  the buffering capacity increases with increasing absolute concentration of the buffer components  as the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves  buffers that need to work mainly with added acid generally have [base] > [acid]  buffers that need to work mainly with added base generally have [acid] > [base]  buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness  the buffering capacity increases with increasing absolute concentration of the buffer components  as the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves  buffers that need to work mainly with added acid generally have [base] > [acid]  buffers that need to work mainly with added base generally have [acid] > [base]

14. 14 Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer

15. 15 Titration  in an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete  when the reaction is complete we have reached the endpoint of the titration  an indicator may be added to determine the endpoint  an indicator is a chemical that changes color when the pH changes  when the moles of H 3 O + = moles of OH −, the titration has reached its equivalence point  in an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete  when the reaction is complete we have reached the endpoint of the titration  an indicator may be added to determine the endpoint  an indicator is a chemical that changes color when the pH changes  when the moles of H 3 O + = moles of OH −, the titration has reached its equivalence point

16. 16 Titration

17. 17 Titration Curve  a plot of pH vs. amount of added titrant  the inflection point of the curve is the equivalence point of the titration  prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH  the pH of the equivalence point depends on the pH of the salt solution  equivalence point of neutral salt, pH = 7  equivalence point of acidic salt, pH < 7  equivalence point of basic salt, pH > 7  beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH  a plot of pH vs. amount of added titrant  the inflection point of the curve is the equivalence point of the titration  prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH  the pH of the equivalence point depends on the pH of the salt solution  equivalence point of neutral salt, pH = 7  equivalence point of acidic salt, pH < 7  equivalence point of basic salt, pH > 7  beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH

18. 18 Titration Curve: Unknown Strong Base Added to Strong Acid

19. 19 added 30.0 mL NaOH 0.00050 mol NaOH pH = 11.96 added 35.0 mL NaOH 0.00100 mol NaOH pH = 12.22 Adding NaOH to HCl 25.0 mL 0.100 M HCl 0.00250 mol HCl pH = 1.00 added 5.0 mL NaOH 0.00200 mol HCl pH = 1.18 added 10.0 mL NaOH 0.00150 mol HCl pH = 1.37 added 15.0 mL NaOH 0.00100 mol HCl pH = 1.60 added 20.0 mL NaOH 0.00050 mol HCl pH = 1.95 added 25.0 mL NaOH equivalence point pH = 7.00 added 40.0 mL NaOH 0.00150 mol NaOH pH = 12.36 added 50.0 mL NaOH 0.00250 mol NaOH pH = 12.52

20. 20 Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH  HCHO 2(aq) + NaOH (aq)  NaCHO 2(aq) + H 2 O (aq)  What is the Initial pH before adding any NaOH?  HCHO 2(aq) + NaOH (aq)  NaCHO 2(aq) + H 2 O (aq)  What is the Initial pH before adding any NaOH? [HCHO 2 ][CHO 2 - ][H 3 O + ] initial 0.1000.000≈ 0 change -x-x+x+x+x+x equilibrium 0.100 - xxx K a = 1.8 x 10 -4

21. 21 Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH  HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq)  initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 -3  What is the pH after adding 5.0 mL 0.100M NaOH?  HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq)  initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 -3  What is the pH after adding 5.0 mL 0.100M NaOH? HCHO 2 CHO 2 - OH − mols Before2.50E-300 mols added --5.0E-4 mols After 2.00E-35.0E-4≈ 0

22. 22 Titration of 25 mL of 0.100 M HCHO 2 with 0.100 M NaOH  HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq) : K b = 5.6 x 10 -11  initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 -3  at equivalence (25 mL of NaOH added  HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq) : K b = 5.6 x 10 -11  initial mol of HCHO 2 = 0.0250 L x 0.100 mol/L = 2.50 x 10 -3  at equivalence (25 mL of NaOH added HAA-A- OH − mols Before2.50E-300 mols added --2.50E-3 mols After 02.50E-3≈ 0 [HCHO 2 ][CHO 2 - ][OH − ] initial 00.0500≈ 0 change +x+x-x-x+x+x equilibrium x5.00E-2-xx [OH - ] = 1.7 x 10 -6 M

23. 23 Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH  HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq)  Added 30 mL NaOH (after equivalence point)  HCHO 2(aq) + NaOH (aq)  NaCHO 2 (aq) + H 2 O (aq)  Added 30 mL NaOH (after equivalence point) 5.0 x 10 -4 mol NaOH xs

24. 24 added 30.0 mL NaOH 0.00050 mol NaOH xs pH = 11.96 added 35.0 mL NaOH 0.00100 mol NaOH xs pH = 12.22 Adding NaOH to HCHO 2 added 12.5 mL NaOH 0.00125 mol HCHO 2 pH = 3.74 = pK a half-neutralization initial HCHO 2 solution 0.00250 mol HCHO 2 pH = 2.37 added 5.0 mL NaOH 0.00200 mol HCHO 2 pH = 3.14 added 10.0 mL NaOH 0.00150 mol HCHO 2 pH = 3.56 added 15.0 mL NaOH 0.00100 mol HCHO 2 pH = 3.92 added 20.0 mL NaOH 0.00050 mol HCHO 2 pH = 4.34 added 40.0 mL NaOH 0.00150 mol NaOH xs pH = 12.36 added 25.0 mL NaOH equivalence point 0.00250 mol CHO 2 − [CHO 2 − ] init = 0.0500 M [OH − ] eq = 1.7 x 10 -6 pH = 8.23 added 50.0 mL NaOH 0.00250 mol NaOH xs pH = 12.52

25. 25 Titrating Weak Acid with a Strong Base  the initial pH is that of the weak acid solution  calculate like a weak acid equilibrium problem  before the equivalence point, the solution becomes a buffer  calculate mol HA init and mol A − init using reaction stoichiometry  calculate pH with Henderson-Hasselbalch using mol HA init and mol A − init  half-neutralization pH = pK a  the initial pH is that of the weak acid solution  calculate like a weak acid equilibrium problem  before the equivalence point, the solution becomes a buffer  calculate mol HA init and mol A − init using reaction stoichiometry  calculate pH with Henderson-Hasselbalch using mol HA init and mol A − init  half-neutralization pH = pK a

26. 26 Titrating Weak Acid with a Strong Base  at the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established  mol A − = original mole HA  [A − ] init = mol A − /total liters  calculate like a weak base equilibrium problem  beyond equivalence point, the OH is in excess  [OH − ] = mol MOH xs/total liters  [H 3 O + ][OH − ]=1 x 10 -14  at the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established  mol A − = original mole HA  [A − ] init = mol A − /total liters  calculate like a weak base equilibrium problem  beyond equivalence point, the OH is in excess  [OH − ] = mol MOH xs/total liters  [H 3 O + ][OH − ]=1 x 10 -14

27. A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the volume of KOH at the equivalence point Write an equation for the reaction for B with HA. Use Stoichiometry to determine the volume of added B HNO 2 + KOH  NO 2 - + H 2 O

28. 28 A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH Write an equation for the reaction for B with HA. Determine the moles of HA before & moles of added B Make a stoichiometry table and determine the moles of HA in excess and moles A - made HNO 2 + KOH  NO 2 - + H 2 O HNO 2 NO 2 - OH − mols Before0.004000≈ 0 mols added --0.00100 mols After ≈ 0 0.003000.00100

29. 29 A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH Write an equation for the reaction of HA with H 2 O Determine K a and pK a for HA Use the Henderson- Hasselbalch Equation to determine the pH HNO 2 + H 2 O  NO 2 - + H 3 O + HNO 2 NO 2 - OH − mols Before0.004000≈ 0 mols added --0.00100 mols After 0.003000.00100≈ 0 Look up K a = 4.6 x 10 -4

30. 30 A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point Write an equation for the reaction for B with HA. Determine the moles of HA before & moles of added B Make a stoichiometry table and determine the moles of HA in excess and moles A  made HNO 2 NO 2 - OH − mols Before0.004000≈ 0 mols added --0.00200 mols After ≈ 0 0.00200 at half-equivalence, moles KOH = ½ mole HNO 2 HNO 2 + H 2 O  NO 2 - + H 3 O +

31. 31 A 40.0 mL sample of 0.100 M HNO 2 is titrated with 0.200 M KOH. Calculate the pH at the half-equivalence point Write an equation for the reaction of HA with H 2 O Determine K a and pK a for HA Use the Henderson- Hasselbalch Equation to determine the pH HNO 2 NO 2 - OH − mols Before0.004000≈ 0 mols added --0.00200 mols After 0.00200 ≈ 0 Look up K a = 4.6 x 10 -4 HNO 2 + H 2 O  NO 2 - + H 3 O +

32. 32 Titration Curve of a Weak Base with a Strong Acid

33. 33 Titration of a Polyprotic Acid  if K a1 >> K a2, there will be two equivalence points in the titration  the closer the K a ’s are to each other, the less distinguishable the equivalence points are  if K a1 >> K a2, there will be two equivalence points in the titration  the closer the K a ’s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of 0.100 M H 2 SO 3 with 0.100 M NaOH

34. 34 Monitoring pH During a Titration  the general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H 3 O + ]  using a probe that specifically measures just H 3 O +  the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve  if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator  the general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H 3 O + ]  using a probe that specifically measures just H 3 O +  the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve  if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator

35. 35 Monitoring pH During a Titration

36. 36 Indicators  many dyes change color depending on the pH of the solution  these dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution HInd (aq) + H 2 O (l) Ind - (aq) + H 3 O + (aq)  the color of the solution depends on the relative concentrations of Ind - :HInd ≈ 1, the color will be mix of the colors of Ind - and HInd  when Ind - :HInd > 10, the color will be mix of the colors of Ind -  when Ind - :HInd < 0.1, the color will be mix of the colors of HInd  many dyes change color depending on the pH of the solution  these dyes are weak acids, establishing an equilibrium with the H 2 O and H 3 O + in the solution HInd (aq) + H 2 O (l) Ind - (aq) + H 3 O + (aq)  the color of the solution depends on the relative concentrations of Ind - :HInd ≈ 1, the color will be mix of the colors of Ind - and HInd  when Ind - :HInd > 10, the color will be mix of the colors of Ind -  when Ind - :HInd < 0.1, the color will be mix of the colors of HInd

37. 37 Phenolphthalein

38. 38 Methyl Red

39. 39 Monitoring a Titration with an Indicator  for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point  an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH  pK a of HInd ≈ pH at equivalence point  for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point  an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH  pK a of HInd ≈ pH at equivalence point

40. 40 Acid-Base Indicators