1. Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chemistry, The Central Science, 10th edition
Unit 6 (Chp 16,17): Acid-Base Equilibria including Buffers & Titrations (Ka , Kb , Kw , pH )
St. Charles Community College
John D. Bookstaver
 2006, Prentice Hall
St. Peters, MO

2. WEAK Acids & Bases have Ka & Kb and RICE
HCl , NH4+
CH3COOH
Acid: proton (H+) donor
Base: proton (H+) acceptor
(forms conjugate base)
Cl– , :NH3
CH3COO–
(forms conjugate acid)
6 Strong Acids: (completely ionize)
HNO3 , H2SO4 , HCl, HBr, HI, HClO4
Strong Bases: (completely dissociate)
the soluble hydroxides (OH–) of…
Group 1 and Ca2+, Sr2+, Ba2+
WEAK Acids & Bases have Ka & Kb and RICE

3. Factors Affecting Acid Strength
The more O’s the more polar (weaker) the H-O bond the more acidic (more likely to lose H+)

4. Factors Affecting Acid Strength
larger X in H–X
stronger acid
(bond is longer, weaker)
stronger acid
(greater ∆EN)
more polar H–X bond
(weaker, likely to break and lose H+)

5. pH pH = −log [H+] or pH = −log [H3O+] H2O(l) + H2O(l) 
H3O+(aq) + OH−(aq)
In pure water, [H3O+] = [OH−]
Kw = [H3O+] [OH−] = 1.0  10−14
x2 = 1.0  10−14
[H3O+] = 1.0  10−7 M
pH = −log(1.0  10−7) = 7.00
Acids: higher [H3O+], 1 x 10–(<7), so pH <7
Bases: lower [H3O+], 1 x 10–(>7) , so pH >7
(in pure water)

6. pH pOH [H+] [OH–] Calculations
pH = –log[H+]
pOH = –log[OH–]
10–pH = [H+]
10–pOH = [OH–]
Kw = [H+] [OH–] = 1.0 x 10–14
pH + pOH = 14
Ka  Kb = Kw
(for conjugates)
pKa = −log Ka
Given any 1 of these, you know all 4 values. H+
OH– pH
pOH

7. GIVEN pH, Calculate Ka pH = −log [H3O+] 2.38 = −log [H3O+] (0.0042)2
[HCOOH]
[H3O+]
[HCOO−] I
0.10 M
0 M C E
– M M M
= 0.10 M M M
pH = −log [H3O+]
2.38 = −log [H3O+]
−2.38 = log [H3O+]
10−2.38 = [H3O+]
(0.0042)2
(0.10)
Ka =
[H3O+]eq = M

8. HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO–(aq)
Percent Ionization
[H3O+]eq
[HA]in
NOT on equation sheet
% ionization =  100%
HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO–(aq)
[HCOOH]
[H3O+]
[HCOO−] I
0.10 M
0 M C
−0.0042 E
= 0.10 M M
0.0042
0.10
% ionization =  100% =
4.2%
4.2% OF 0.10 M HCOOH IS ionized.

9. HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2–(aq)
GIVEN Ka , Calculate pH
HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2–(aq)
[H3O+] [C2H3O2−]
[HC2H3O2]
Ka =
= 1.8  10−5
[HC2H3O2]
[H3O+]
[C2H3O2−] I
0.30 M C E –x +x +x
0.30 – x x x
 0.30 M
(b/c K <<<1)
(x)2
(0.30)
1.8  10−5 =
x = 2.3  10−3 M H+
pH = −log [H+]

10. Basic & Acidic Ions M(OH)3 + H+ M3+ H ANIONS are bases (raise pH).
react with water in hydrolysis reaction to form OH− and the conjugate acid:
X− + H2O 
HX + OH−
anion base
conj. acid
CATIONS with acidic protons (like NH4+) will lower pH.
NH4+ + H2O 
NH3 + H3O+
Metal Cations (Fe3+,Cu2+,Ag+) ↓pH (acidic).
↑charge , ↓size = more acidic
M(OH)3 + H+
M3+ H

11. Cations: Anions: NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq) (acidic, ↓pH)
(donates H+)
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
Cations:
(acidic, ↓pH)
Cu+(aq) + H2O(l)  CuOH(s) + H+(aq)
Most metal cations will lower pH (acidic).
(Fe3+, Zn2+, …) (take OH– from H2O to lose H+)
But, metal cations of strong base CAN’T affect pH.
(Li+,Na+,K+,Ca2+,Ba2+) (can’t take OH– from H2O)
Anions:
(basic, ↑pH)
F−(aq) + H2O(l)  HF(aq) + OH−(aq)
Most anions will increase pH (basic) (F–, HCO3–, NO2–) (take H+ from H2O)
But, anions from conj. base of a strong acid CAN’T affect pH. (Cl–, HSO4–, NO3–) (can’t take H+ from H2O)

12. Optimum Capacity and Range: - weak acid with a pKa near desired pH
Buffer:
weak conjugate acid-base pair.
resist pH changes (reacts with added acid/base)
Buffer Capacity: more moles (M & L) 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)
if Ka =
[H+][A–]
[HA]
, then pKa =
pH[A–]
[HA]
when
[HA] = [A–]

13. Common Ion Effect: [H+] & pH change
common ion shifts left causing the weak electrolyte to be less ionized.
[H+] & pH change
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
(x)(0.11)
(0.12)
1.4  10−4 =
≈ 0.12 M ≈ 0.11 M
b/c K << 1
x = 1.5  10−4 M H3O+
WS #1-4
pH = –log(1.5  10−4) = 3.82

14. 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)

15. Titration Standard: solution of known conc. (M) Vanalyte (known)
Vtitrant (known)
Mtitrant (known)
Animation (2 min):
Manalyte (unknown) 15

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

17. Indicators
Colorless in ACID
Phenolphthalein
Pink in BASE

18. 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
H+ + OH– ↔ H2O

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

20. 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.

21. 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”)

22. 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–

23. 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–

24. 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

25. 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]

26. Calculating pH for 4 parts of Titration Curve:
WEAK A with Strong B
and continued 3 4
WS #7
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–]