Chem 30AP Notes
Wednesday, December 15, 2010
M. Monoprotic vs. Polyprotic Acids and Bases
 an acid capable of donating only one proton is called
monoprotic
eg) HCl(aq), HNO3(aq), HOCl(aq) etc.
 if an acid can transfer more than one proton, it is called
polyprotic (diprotic if 2 protons, triprotic if 3 protons)
eg) Label each of the following acids as monoprotic or polyprotic:
1. H2SO4(aq)
polyprotic
2. HOOCCOOH(aq)
polyprotic
3. HCOOH(aq)
monoprotic
4. CH3COOH(aq)
monoprotic

5. H2PO4 (aq)
polyprotic
+
6. NH4 (aq)
monoprotic
 a base that can accept only one proton is called a monoprotic base
 a base that can accept more than one proton is called a polyprotic
base (diprotic or triprotic)
eg) PO43(aq) can accept up to 3 H+ to form HPO42(aq),
H2PO4(aq), and H3PO4(aq) respectively
eg) Label each of the following as monoprotic or polyprotic acids,
bases or polybasic:
1. HSO4(aq)
monoprotic acid; monoprotic base

2. H2PO4 (aq)
polyprotic acid; monoprotic base
3. HPO4 (aq)
monoprotic acid; polyprotic base

4. HCO3 (aq)
monoprotic acid; monoprotic base
5. H2O(l)
monoprotic acid; monoprotic base
 reactions involving polyprotic acids or polybasic substances involve
the same principles of reaction prediction
 only one proton is transferred at a time and always from strongest
acid to strongest base
1
Example 1
Potassium hydroxide is continuously added to oxalic acid until no more
reaction occurs.
List: K+ OH HOOCCOOH H2O HOOCCOO H2O
- SB
SA
A/B
SA
A/B
1:
OH- + HOOCCOOH  H2O + HOOCCOO2:
OH- + HOOCCOO-  H2O + OOCCOO2OOCCOO2- cannot donate any more protons the rxn stops
Net Reaction: Add all reactions together (only if all quantitative), cancelling
out any species that occur in the same quantity on both the reactant and
product sides and summing any species that occur more than once on the
same side
2 OH- + HOOCCOOH  2 H2O + OOCCOO2-
Example 2
Sodium hydrogen phosphate is titrated with hydroiodic acid. If we assume
all steps are quantitative, show all steps and give the net reaction.
List: Na+
-
HPO42SB
H3O+
SA
IB
Rxn 1:
Rxn 2:
Net Rxn:
HPO42- + H3O+  H2PO4- + H2O
H2PO4- + H3O+  H3PO4 + H2O
HPO42- + 2 H3O+  H3PO4 + 2 H2O
H2O
A/B
H2PO4A/B
H3PO4
A
Your Assignment: pg 11 #1-5 “Polyprotic Acids and
Bases”
2
N. Acid-Base Stoichiometry
 use the same rules of stoich that we always have
 write the net reaction first using the acid-base reaction prediction rules
(List, identify A, B, SA, SB etc)
 if the question says “at the second endpoint”, then you know the
reaction is dealing with something that is polyprotic or
polybasic…determine all reaction steps then use the net reaction to do
the stoich calculation
Example 1
A 25.0 mL NH3(aq) solution is titrated with 15.70 mL of 0.100 mol/L
HCl(aq). Calculate the concentration of the NH3(aq).
List: NH3
SB
H3O+
SA
Cl
B
H2O
A/B
NH3
+
H3O+
⇌
NH4+
v = 0.0250 L
v = 0.01570 L
c=?
c = 0.100 mol/L
n = 0.001570 mol × 1/1 n = cv
c = n/v
= (0.100 mol/L)(0.01570L)
= (0.001570 mol)
= 0.001570 mol
(0.0250 L)
= 0.0628 mol/L
+
H2O
Example 2
What is the volume of 0.500 mol/L KOH(aq) needed to react with 10.0 mL
of 0.500 mol/L nitric acid?
List: H3O+
NO3
K+
OH
H2O
SA
B
SB
A/B
H3O+
+
v = 0.0100 L
c = 0.500 mol/L
n = cv
= (0.500 mol/L)(0.0100L)
= 0.00500 mol
OH
⇌
v=?
c = 0.500 mol/L
n = 0.00500 mol  1/1
V= n
C
3
2 H2O
= 0.00500 mol
0.500 mol/L
= 0.0100 L
Example 3
What is the pH of a mixture of 50.0 mL of 0.30 mol/L HCl(aq) with 30.0 mL
of 0.40 mol/L NaOH(aq)?
List: H3O+
ClNa+
OHH2O
SA
B
SB
A/B
OH
⇌
2 H2O
v = 0.0300L
c = 0.40 mol/L
n = cv
= (0.40 mol/L)(0.0300 L)
= 0.012 mol
limiting (used up)
nH3O+ left = 0.015 mol – 0.012 mol
= 0.0030 mol
vtotal = 0.0500 L + 0.0300 L = 0.0800L
c = n/v
= 0.0030 mol/0.0800 L
= 0.0375 mol/L
H3O+
+
v = 0.0500 L
c = 0.30 mol/L
n = cv
= (0.30 mol/L)(0.0500L)
= 0.015 mol
pH = log(0.0375 mol/L)
= 1.43
Your Assignment: pg 11 #1-8 “Acid-Base Stoichiometry”
4