Chemistry I-H
Acid/Base Problem Set #1
1. Using the Arrhenius definitions, identify the following as acids, bases or salts by checking the appropriate
column. Also, if it is an acid, determine if it is monoprotic, diprotic or triprotic.
Substance
Base
Salt
Acid
Monoprotic
HC2H3O2
Ca(OH)2
Mg(NO3)2
CoCl6
H3PO3
H2S
Al(OH)3
HF
Zn(C2H3O2)2
KOH
Diprotic
Triprotic










2. Name the following Arrhenius acids.
A. HI
hydroiodic acid
B. H2C2O4
C. HBr
hydrobromic acid
D. HClO2
E. H2CrO4
chromic acid
F. HMnO4
oxalic acid
chlorous acid
permanganic acid
3. Using the Bronsted – Lowry definitions, label each member of the following equations as
an acid, base, conjugate acid, or conjugate base.
A.
CH3COOH + H2O  H3O+
A
B
CA
B.
F- + H2O 
B
A
OHCB
+ HF
CA
C.
H2O
B
+ HOBr
A

OBrCB
D.
H2O
A
+ HS -  H2S + OHB
CA
CB
+ CH3COO –
CB
+ H3O+
CA
HNO3 + HPO4 –2  NO3- + H2PO4 –
A
B
CB
CA
4. Use the Bronsted-Lowry definition to predict the conjugate acid to the following bases.
E.
A. NH3
NH4+
B. HSO4-
H2SO4
D. ClO-
HClO
E. PO4-3
HPO42-
C. H2O
H3O+
5. Use the Bronsted - lowry definitions to predict the conjugate base to the following acids.
A. NH4+
NH3
B. H2O OH-
D. HCO3-
CO32-
E. H2C2O4 HC2O4-
C. HClO4
ClO4-
6. Predict the products of the following double replacement reactions, which are the neutralization of an
Arrhenius acid and base.
These equations have not been balanced.

A.
Ca(OH)2 + HNO3
Ca(NO3)2 +
B.
H2SO4
C.
HClO3 + LiOH

LiClO3
D.
Al(OH)3 + H2S 
Al2S3
+

+ Mg(OH)2
MgSO4
H2O
+
H2O
+ H2 O
H2O
7. Determine the pH and pOH of the following substances. FIRST determine the [H+] or [OH-] depending on
whether you are working with an Arrhenius acid or base. Remember pH + pOH = 14
ex. 0.0045 M NaOH
NaOH is a base. It releases OH- into solution. Therefore
From the [OH-], the pOH may be calculated using the formula pOH = -log [OH-].
Knowing that pH and pOH = 14 we can determine the pH
A. 0.0020 M HNO3
[ H+ ] = 0.0020
pH = 2.70
pOH = 11.3
B. 0.00035 M HCl
[ H+ ] = 0.00035
pH = 3.46
pOH = 10.5
C. 0.0782 M LiOH
[ OH-] = 0.0782
pH = 12.9
pOH = 1.11
D. 1.2 x 10 –5 HClO4
[ H+ ] = 1.2x10-5
pH = 4.92
pOH = 9.10
E. 3.7 x 10 –3 Ba(OH)2
[ OH- ] = 0.0074
pH = 11.7
pOH = 2.13