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 hydrogen 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. (stoichiometrically =)
end point:
indicator color change persists
equivalence point,(Veq):
equal stoichiometric amounts react completely
HBr + NaOH 
mol titrant
(stoichiometrically =)
mol analyte
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. p. 733 SA
with SB
very low initial pH
level pH rise

21. SA
with SB
rapid, steep jump in pH at Veq

22. SA with SB 7 at Veq, pH = ___ At Veq… moles added moles reacted
(stoichiometrically =)
moles reacted
…the solution contains only water and salt.
(irrelevant conjugate) SA
with SB
at Veq, pH = ___ 7

23. SA
with SB
after Veq, very high pH levels.

24. Titration Calculations
Calculate unknown moles of the ANALYTE.
(then calculate: M OR molar mass OR mass %)
Stoich: L X  mol X  mol Y = mol Y
L X mol X
Calculate unknown VOLUME of TITRANT (mL) added to reach equivalence (all reacted).
Stoich: L X  mol X  mol Y  L Y = L Y
L X mol X mol Y
Calculate unknown pH at any point in the titration (especially at equivalence).
Stoich: What’s in the sol’n? Find excess H+, OH–
HW p. 763 #38

25. Calculating pH for 4 parts of Titration Curve:
pH BEFORE titration has begun:
pH = –log [H+] for SA b/c [HA] = [H+] 1
HW p. 763 #40
STRONG with Strong
pH DURING titration:
MRP (More RICE Please!)
Moles: L x M = mol H+ & mol OH–
RICE: H+ + OH–  H2O + conj.
mol÷L’s total = [H+] or [OH–]
pH: –log [H+] or –log [OH–] 2 4 3 1 2
EQUIVALENCE (Veq):
pH = 7.00 (only H2O & salt, Kw) 3
What’s in the flask?
(write a rxn)
pH AFTER equivalence (Veq):
MRP (More RICE Please!) 4

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

27. 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: (

28. WB with SA HW p. 763 #33 At Veq, pH < 7
conjugate acid of base lowers pH of H2O by…
HA + H2O ↔ H3O+ + A–

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

30. (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)

31. In a solution of 0.10 M H3PO4, which species is in the least amount?
HW p. 763
#36
Polyprotic Acids
There’s a Ka and pKa for each dissociation, with a distinct Veq.
In a solution of 0.10 M H3PO4, which species is in the least amount?
H3PO4
H2PO4–
HPO42–
PO43–

32. Calculating pH for 4 parts of Titration Curve:
pH BEFORE titration has begun:
pH = –log [H+] for SA b/c [HA] = [H+] 1
HW p. 763 #40
STRONG with Strong
pH DURING titration:
MRP (More RICE Please!)
Moles: L x M = mol H+ & mol OH–
RICE: H+ + OH–  H2O + conj.
mol÷L’s total = [H+] or [OH–]
pH: –log [H+] or –log [OH–] 2 4 3 1 2
EQUIVALENCE (Veq):
pH = 7.00 (only H2O & salt, Kw) 3
What’s in the flask?
(write a rxn)
pH AFTER equivalence (Veq):
MRP (More RICE Please!) 4

33. 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
pH DURING titration: (buffer region)
MREP
(More RICE Equil. Please!)
Moles: L x M = mol H+ & mol OH–
RICE: H+ + OH–  H2O + conj.
mol÷L’s total = [H+] or [OH–]
Equil Rxn: HA + H2O ⇄ H3O+ + A–
RICE in M’s & Ka w/ x (not x2) for [H+]
pH = –log [H+] 2
What’s in the flask?
(write a rxn) 2 1
(buffer)
@ ½Veq: pH = pKa
Ka =
[H+][A–]
[HA]
b/c ½Veq

34. Calculating pH for 4 parts of Titration Curve:
and continued 3 4
HW p. 763 #42
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