1. Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 10 (Chp 17): Buffers & Acid-Base Titrations (more Ka, Kb, Kw, pH, pOH)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

2. Buffers: H+ + HCO3– → H2CO3 3 HCl 1 free H+
weak conjugate acid-base pair.
resist pH changes by reacting with added acid/base. 3
3 HCl
How many H+’s added?
How many H+’s remain? 1
H+ + HCO3– → H2CO3
Cl–
H2CO3
HCO3–
HCO3– H+
Na+
H2CO3
Na+
H2CO3  HCO3–
buffer solution
1 free H+

3. Buffers:
H2CO3 H+ +
HCO3–
H2CO3 H+
HCO3–

4. Buffers:
animation
Animation:

5. HW p. 762 #17
Buffers:
If acid is added, the F− reacts to form HF and water.

6. Common-Ion Effect Consider a solution of acetic acid:
If NaC2H3O2 (acetate ion) is added, the equilibrium will shift to the left. (Le Châtelier)
HC2H3O2(aq) 
H+(aq) + C2H3O2−(aq)
“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has a common ion with the weak electrolyte.” OR
adding common ion shifts left (less ionized)

7. Common-Ion Effect First a review of what you can do already:
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF.
Ka for HF is 6.8  10−4
[H+] [F−]
[HF]
Ka =
= 6.8  10–4

8. Common-Ion Effect
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF.
HF(aq) 
H+(aq) + F−(aq)
[HF], M
[H+], M
[F−], M
Initial
0.20
Change −x +x
Equilibrium
0.20 − x
 0.20 x x2
(0.20)
6.8  10−4 =
[H+] = M
pH = 1.92
x = 0.012

9. Common-Ion Effect
NOW, calculate the [F–] and pH of M HF and 0.10 M NaF.
NaF (strong electrolyte) so [F–]init ≠ 0.
HF(aq) 
H+(aq) + F−(aq)
[HF], M
[H+], M
[F−], M
Initial
0.20
0.10
Change −x +x
Equilibrium
0.20 − x
 0.20 x x
 0.10
(x)(0.10)
(0.20)
6.8  10−4 =
[H+] = M
pH = 2.85
x =

10. Common-Ion Effect
“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has a common ion with the weak electrolyte.”
NaF
HF(aq) 
H+(aq) + F−(aq)
Previously
0.20 M HF:
[H+] = M
pH = 1.92
With NaF (common ion F–)
0.20 M HF and 0.10 M NaF:
[H+] = M
pH = 2.85
HW p. 761 #11

11. pH of Buffers
HW p. 762 #19a
What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.11 M in sodium lactate (NaC3H5O3)?
Ka for lactic acid is 1.4  10−4
HC3H5O3 + H2O  H3O+ + C3H5O3–
[H3O+] [C3H5O3−]
[HC3H5O3]
Ka =
ICE gives:
(x)(0.11)
(0.12)
1.4  10−4 =
pH = 3.82
x = 1.5  10−4 M H3O+
pH = –log(1.5  10−4)

12. Buffer Capacity & pH Range
buffer capacity: the amount of acid or base neutralized before significant pH changes.
pH range: the range of pH values over which a buffer works effectively.
Best Capacity and Range:
- weak acid with a pKa near desired pH
- [HA] = [A–] (so that pKa = pH)
if Ka =
[H+][A–]
[HA]
, then pKa =
pH[A–]
[HA]

13. When Strong Acids or Bases Are Added to a Buffer…
…it is safe to assume that all of the strong acid or base is consumed in the reaction.

14. Adding Strong Acid or Base to Buffer
Add OH– consumes [HA] and produces [A–].
Add H+ consumes [A–] and produces [HA].
Use M’s in K exp. to find new [H+] & pH.
add mol OH–
add mol H+
HW p. 762 #21, 24, 26

15. Titration
(video clip)
The analytical technique used to calculate the moles in an unknown soln.
mass % (g /gtotal)
molarity (mol/L)
molar mass (g/mol)
buret
titrant
known vol. (V)
known conc. (M)
Safari Montage (2 min) Titration
analyte
known vol. (V)
unknown conc. (M)
(or moles)

16. equivalence point,(Veq):
end point:
indicator color change persists
equivalence point,(Veq):
equal stoichiometric amounts react completely
HBr + NaOH 
mol Acid =
mol Base
2 HA + Ca(OH)2 
H2SO KOH 

17. Titration Standard: solution of known conc. (M) Vanalyte (known)
animation
Standard: solution of known conc. (M)
Vanalyte (known)
Vtitrant (known)
Mtitrant (known)
Animation (2 min):
Manalyte (unknown) 17

18. Indicators:
Pink in BASE
Phenolphthalein
Colorless in ACID

19. Add H3O+ (titrate w/ acid)
Indicators:
weak acids with conj. bases of diff. color
Example: Phenolphthalein
HIn + H2O   H3O+  + In–
Add H3O+ (titrate w/ acid)
HIn + H2O  H3O+  + In–
Demo: Magic Pitcher
colorless
Remove H3O+ (titrate w/ base)
HIn + H2O   H3O+  + In–
pink

20. From start to near the equivalence point (Veq), the pH goes up very slowly.SA
with SB
p. 733

21. Just before and after the Veq, the pH increases rapidly (steep jump).SA
with SB

22. SA with SB 7 At Veq, pH = ? At Veq,
moles acid = moles base, the solution contains only water and salt. SA
with SB
At Veq,
pH = ? 7

23. SA
with SB
After Veq,
pH levels off.

24. WA with SB p. 736 At Veq, pH > 7
Conjugate base of acid raises pH of H2O by…
p. 736
Animation: Titration
A– + H2O ↔ HA + OH–
Animation: (

25. Indicators & pH Ranges
animation
Choose indicator that changes color (has pKa) near pH of equivalence point (Veq) of titration.

26. (stoich. calc. with mol-to-mol)WA
with SB
Before Veq, the pH is determined from the amounts of the acid and its conjugate base present at that particular time.
(stoich. calc. with mol-to-mol)

27. WA with SB Weak acids: higher initial pH
gradual pH rise (not flat then jump)
subtle pH change at equiv. point (less steep)
Weak bases: (same
but drop not rise)

28. WB
with SA
At Veq,
pH < 7
conjugate acid of base lowers pH of H2O by…
HA + H2O ↔ H3O+ + A–

29. In a solution of 0.10 M H3PO4, which species is in the least amount?
Polyprotic Acids
In a solution of 0.10 M H3PO4, which species is in the least amount?
There’s a Ka and pKa for each dissociation, with a distinct Veq.
H3PO4
H2PO4–
HPO42–
PO43–

30. Calculating pH for 4 parts of Titration Curve:
pH BEFORE titration has begun:
pH = –log [H+] for SA b/c [HA] = [H+] 1
STRONG A with Strong B
pH DURING titration:
Titrn Rxn: H+ + OH–  H2O + conj.
RICE in moles, ÷L’s total = [H+] left
pH = –log [H+] left 2 4 3 1 2
EQUIVALENCE (Veq):
no work (pH = 7.00, only H2O , Kw) 3
pH AFTER equivalence (Veq):
Titrn Rxn: H+ + OH–  H2O + conj.
RICE in moles, ÷L’s total = [OH–] excess
pOH = –log [OH–] excess 4
What’s in the flask?
(write a rxn)

31. Calculating pH for 4 parts of Titration Curve:
pH BEFORE titration has begun:
Equil Rxn: HA + H2O ⇄ H3O+ + A–
RICE in M’s & Ka w/ x2 for [H+]
pH = –log [H+] 1
WEAK A with Strong B 2
pH DURING titration: (buffer region)
Titrn Rxn: HA + OH–  H2O + A–
RICE in moles, ÷L’s total = [HA],[A–]
Equil Rxn: HA + H2O ⇄ H3O+ + A–
RICE in M’s & Ka w/ x (not x2) for [H+]
pH = –log [H+] 2 1
(buffer)
What’s in the flask?
(write a rxn)
@ ½Veq: pH = pKa
Ka =
[H+][A–]
[HA]
b/c ½Veq
and on next slide 3 4

32. Calculating pH for 4 parts of Titration Curve:
and continued 3 4
WEAK A with Strong B
EQUIVALENCE (Veq):
Titrn Rxn: HA + OH–  H2O + A–
RICE in moles, ÷L’s total = [A–] only
Equil Rxn: A– + H2O ⇄ HA + OH–
RICE in M’s & Kb w/ x2 for [OH–]
pOH = –log [OH–] 3 4 3 2 1
(buffer)
What’s in the flask?
(write a rxn)
pH AFTER equivalence (Veq):
Titrn Rxn: HA + OH–  H2O + A–
RICE in moles, ÷L’s total = [OH–] excess
pOH = –log [OH–] excess ( [A–] produces negligible [OH–] ) 4