1. Unit 10 SUMMARY: Buffers & Acid-Base Titrations
weak conjugate acid-base pair.
resist pH changes (reacts added acid/base)
Buffer Capacity: more moles (L & M) of buffer react more acid/base before significant pH changes.
pH range: buffer works near it’s pKa
Optimum Capacity and Range:
- weak acid with a pKa near desired pH
- [HA] = [A–] (so pKa = pH)
Ka =
[H+][A–]
[HA]

2. Common Ion Effect:
adding common ion shifts left causing the weak acid/base to be less ionized.
pH of Buffers:
0.12 M lactic acid, HC3H5O (Ka = 1.4  10−4)
0.11 M sodium lactate, NaC3H5O3
HC3H5O3 + H2O  H3O+ + C3H5O3–
[H3O+] [C3H5O3−]
[HC3H5O3]
Ka =
I M 0 M M
C –x + x x
E – x x x
≈ 0.12 M ≈ 0.11 M
(x)(0.11)
(0.12)
1.4  10−4 =
x = 1.5  10−4 M H3O+
pH = –log(1.5  10−4) = 3.82

3. Titration
The analytical technique used to calculate the moles in an unknown soln.
mass % (g /gtotal)
molarity (mol/L)
molar mass (g/mol)
buret
titrant
end point:
indicator color change
known vol. (V)
known conc. (M)
equivalence point:
equal stoich. amounts (mols) react completely
Safari Montage (2 min) Titration
analyte
known vol. (V)
unknown conc. (M)
(or moles)

4. SA with SB At Veq, pH = 7 At Veq… moles added moles reacted
(stoichiometrically =)
moles reacted
…the solution contains only water & salt.
(irrelevant conjugate) SA
with SB
At Veq,
pH = 7
H2O + H2O ↔ H3O+ + OH–

5. WA with SB At Veq, pH > 7 (only water & conj. base)
A– + H2O ↔ HA + OH–
Animation: (

6. WB
with SA
At Veq,
pH < 7
(only water & conj. acid)
HA + H2O ↔ H3O+ + A–
Indicators: weak acids with diff color conjugates
choose one that changes color (has pKa) near pH of equivalence point (Veq) of titration.

7. WA with SB Verbally describe the visual curve: 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”)

8. Titration Calculations
Calculate unknown moles of the ANALYTE.
(then calculate: M OR molar mass OR mass %)
Stoich: __ L T x mol T x mol A = mol A
1 L T mol T
Calculate unknown VOLUME of TITRANT (mL) added to reach equivalence (all reacted).
Stoich: __ L A x mol A x mol T x 1 L T = L T
1 L A mol A mol T
Calculate unknown pH at any point in the titration (especially at equivalence).
Stoich: What’s in the sol’n? Find M of H+ or OH–

9. Calculating pH for 4 parts of Titration Curve:
pH BEFORE titration has begun:
pH = –log [H+] for SA b/c [HA] = [H+] 1
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+] 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

10. Calculating pH for 4 parts of Titration Curve:
pH BEFORE titration has begun:
Equil Rxn: HA + H2O ⇄ H3O+ + A–
RICE: M
pH: –log [H+] 1
WEAK A with Strong B
0 M 0 M
–x x x
K = x2
[HA]
x x 2 1
pH DURING titration: (buffer region)
MREP (More RICE Eq. Please!)
Moles: L x M = mol HA & mol OH–
RICE: HA + OH–  H2O + A–
mol÷L’s total = [HA] & [A–]
Equil Rxn: HA + H2O ⇄ H3O+ + A–
RICE in M’s & Ka w/ x for [H+]
pH: –log [H+]
(buffer) 2
What’s in the flask?
(write a rxn)
Ka =
[H+][A–]
[HA]
@ ½Veq: pH = pKa
b/c ½Veq
K = x[A–]
[HA]

11. Calculating pH for 4 parts of Titration Curve:
WEAK A with Strong B
and continued 3 4
EQUIVALENCE (Veq):
MREP (More RICE Eq. Please!)
Moles: L x M = mol HA & mol OH–
RICE: HA + OH–  H2O + A–
mol÷L’s total = ONLY [A–]
Equil Rxn: A– + H2O ⇄ HA + OH–
RICE in M’s & Kb w/ x2 for [OH–]
pH: pOH = –log [OH–] & 14 – pOH = 3 4 3 2 1
(buffer)
Kb = x2
[A–]
What’s in the flask?
(write a rxn)
pH AFTER equivalence (Veq):
MRP (M & R give [OH–] excess)
pOH = –log [OH–] & 14 – pOH = 4
MREP (NO Eq)
b/c [A–] produces
negligible [OH–]