1. Applications of Aqueous Equilibria
McMurry & Fay ch. 15

2. Vocabulary: BUFFER SOLUTION an aqueous solution containing both an acid and its conjugate base

3. Calculate the pH in a solution prepared by dissolving 0
Calculate the pH in a solution prepared by dissolving 0.10 mol of solid NH4Cl in L of 0.40 M NH3 (Kb = 1.8 x 10-5). I C E
[NH4+][OH-]
[NH3]
Kb = = 1.8 x 10-5

4. [H+][A-] [HA] Ka = [H+][A-] [HA] - log Ka = - log [A-] [HA]
A shortcut
[H+][A-]
[HA]
Ka =
[H+][A-]
[HA]
- log Ka = - log
[A-]
[HA]
- log Ka = - log [H+] – log
= pKa
= pH

5. Henderson-Hasselbalch Equation
pKa = pH – log

6. Uses for Henderson-Hasselbalch Equation
Calculate pH of buffer solution
Calculate change in pH of buffer solution after additon of acid/base
Prepare buffer solution of given pH

7. Calculating pH of buffer solution
Calculate the pH in a solution prepared by dissolving 0.10 mol of solid NH4Cl in L of 0.40 M NH3 (Kb = 1.8 x 10-5).
pKa = pH – log
[A-]
[HA]

8. Calculating pH of buffer solution
A buffer solution is created with 0.10 M HF (Ka = 6.8 x 10-4) and 0.15 M NaF. What is the pH?
pKa = pH – log
[A-]
[HA]

9. Buffer solutions resist changes in pH: adding acid
H2CO H+ + HCO3-

10. Buffer solutions resist changes in pH: adding base
H2CO3 + OH H2O + HCO3-

11. To calculate change in pH:
Choose appropriate reaction:
HA ⇌ A- + HA if adding acid
BH+ + OH- ⇌ B + H2O
Make an ICE table that shows number of moles of each species
Calculate change using # moles of base or acid added
Use final # of moles & Henderson-Hasselbalch eqn to calculate final pH

12. Quick tip: can use either # moles or concentration in Henderson-Hasselbalch eqn
𝑝𝐻=𝑝 𝐾 𝑎 + 𝐴 − 𝐻𝐴
𝑝𝐻=𝑝 𝐾 𝑎 + 𝑚𝑜𝑙𝑒𝑠 𝐻 + 𝑣𝑜𝑙𝑢𝑚𝑒 𝑚𝑜𝑙𝑒𝑠 𝐻𝐴 𝑣𝑜𝑙𝑢𝑚𝑒
𝑝𝐻=𝑝 𝐾 𝑎 + 𝑚𝑜𝑙𝑒𝑠 𝐻 + 𝑚𝑜𝑙𝑒𝑠 𝐻𝐴

13. How much would the pH change if 100 mL of 1.0 HCl was added to water?
1.0 L of buffer solution has 0.30 M HNO2 (Ka=7.1 x 10-4) and 0.30 M NaNO mL of 1.0 M HCl are added to the solution.
How much would the pH change if 100 mL of 1.0 HCl was added to water?
How much does the pH change when the HCl is added to the buffer solution?
Choose appropriate reaction:
HA ⇌ A- + HA if adding acid
BH+ + OH- ⇌ B + H2O
Make an ICE table that shows number of moles of each species
Calculate change using # moles of base or acid added
Use final # of moles & Henderson-Hasselbalch eqn to calculate final pH

14. Preparing buffer solution of given pH
1. Pick acid with pKa within 1 pH unit of desired pH
2. Use Henderson-Hasselbalch equation to calculate ratio between acid & conjugate base
3. Multiply ratio by a factor appropriate for necessary
buffer strength

15. Pick acid with pKa near desired pHHF
3.17
CH3CO2H
4.76
Ethylenediamine
6.85
HOCl
7.53
Tris*
8.07
NH4+
9.25
*Tris(hydroxymethyl)aminomethane

16. Make a buffer with pH 7.8 [A-] pKa = pH – log [HA]
1. Pick acid with pKa within 1 pH unit of desired pH HF
3.17
CH3CO2H
4.76
Ethylenediamine
6.85
HOCl
7.53
Tris*
8.07
NH4+
9.25
2. Use Henderson-Hasselbalch equation to calculate ratio between acid & conjugate base
3. Multiply ratio by a factor appropriate for necessary
buffer strength
[A-]
[HA]
pKa = pH – log

17. Acid-Base Titrations

18. pH changes during Titration: Strong Acid-Strong Base
As you add base to an acid solution:
Some acid is neutralized; # of moles of H+ decreases
Volume of solution increases
Molarity =
so [H+] will decrease and pH will increase
Moles
Liters

19. Calculating pH changes during Strong Acid-Strong Base Titration: before equivalence point
Calculate moles of acid you start with:
# moles acid = (Molarity acid) * (volume acid)
Calculate moles of base added:
# moles base = (Molarity base) * (volume base)
Calculate moles of H+ remaining:
# moles acid at start - # moles base added
Calculate volume of sample:
Initial volume + volume of base added
Calculate [H+] (moles H+/volume) and pH

20. 20. 00 mL of 0. 400 M HCl is titrated with 0
20.00 mL of M HCl is titrated with M NaOH Calculate pH when you have added: (a) 0 mL of acid (b) 5 mL of acid (c) 7.5 mL of acid (d) 10 mL of acid (e) 15 mL of acid (f) 19 mL of acid (g) 20 mL of acid
Calculate moles of acid you start with:
# moles acid = (Molarity acid) * (volume acid)
Calculate moles of base added:
# moles base = (Molarity base) * (volume base)
Calculate moles of H+ remaining:
# moles acid at start - # moles base added
Calculate volume of sample:
Initial volume + volume of base added
Calculate [H+] (moles H+/volume) and pH

21. What happens after the equivalence point?
All acid has been used up; OH- dominates the pH
Calculate moles of base added:
# moles base = (Molarity base) * (volume base)
Calculate initial moles of acid :
# moles H+ = (Molarity acid) * (volume acid)
Calculate moles of OH-:
# moles of base – initial moles of acid
Calculate volume of sample:
Initial volume + volume of NaOH added
Calculate [OH-] (moles OH-/volume), pOH, and pH

22. Equivalence point # moles acid = # moles base

23. pH changes during Titration: Weak Acid-Strong Base
What’s different from strong acid titrations
Acid doesn’t completely dissociate so pH depends on Ka
Adding base produces the conjugate base of the weak acid
Acid + conjugate base both present = buffer solution
Use Henderson Hasselbalch to calculate pH

24. pH changes during Titration: Weak Acid-Strong Base (before equivalence point)
Calculate moles of acid you start with:
# moles acid = (Molarity acid) * (volume acid)
Calculate moles of base added:
# moles base = (Molarity base) * (volume base)
Calculate moles of conjugate base formed:
# moles A- = # moles base added
Calculate moles of acid remaining:
# moles HA left =
initial # moles acid - # moles base added
Use Henderson Hasselbalch to calculate pH

25. Example: titration of 10 mL 0.500 M benzoic acid (pKa = 4.202) with 0.200 M NaOH
What is [H+] at the start? Ka =
[H+][A-]
[HA]

26. Example: titration of 10 mL 0.500 M benzoic acid (pKa = 4.202) with 0.200 M NaOH
Where do we expect our equivalence point to be?

27. Example: titration of 10 mL M benzoic acid (pKa = 4.202) with M NaOH What is pH after you have added: 5 mL 10 mL 12.5 mL 15 mL 20 mL 25 mL
[A-]
[HA]
pKa = pH – log

28. After the equivalence point: No acid left
A- + H2O HA + OH-

29. equivalence point # moles acid = # moles base
Half equivalence point
[A-] = [HA]

30. Half-equivalence point
Point at which [A-] = [HA] Henderson-Hasselbalch equation:
[A-]
[HA]
pKa = pH – log

31. At half-equivalence point,
pH = pKa

32. pH changes during Titration: Weak Acid-Strong Base (at equivalence point)
All acid has been turned into conjugate base Solution of basic salt: calculate pH as done in ch. 14

33. pH changes during Titration: Weak Acid-Strong Base (after equivalence point)
Most OH- comes from excess NaOH
Calculate moles of acid you start with:
# moles acid = (Molarity acid) * (volume acid)
Calculate moles of base added:
# moles base = (Molarity base) * (volume base)
Calculate moles of OH- remaining:
# moles base added - # moles acid
Calculate volume of sample:
Initial volume + volume of base added
Use [H+] = 10-14/[OH-] to calculate pH

34. Indicators
Dyes that are weak acids or bases Color of acid form different from conjugate base HIn (aq)  H+ (aq) + In– (aq) Acid form base form (one color) (another color)

35. Choosing an indicator
Color change begins when two forms are equal:
Choose indicator that has pKa = pH of equiv. point!