CHEM 5013 - Applied Chemical Principles
Suggested problems for Quiz 5 in to be given in class on Thursday October 17
1. Classify each of the following as an acid or a base. Then classify each as a strong or weak acid
or base. What ions are produced when each is dissolved in water?
a. Mg(OH)2
STRONG BASE
Mg2+ and 2 OH -
b. H2SO4
STRONG ACID
2 H+ and SO4 2-
c. LiOH
STRONG BASE
Li+ and OH -
d. H3PO4
WEAK ACID
3 H+ and PO4 3-
2. Write the formula of the conjugate acid of each of the following bases:
a. OHH2O
b. NH3
NH4 +
c. HPO4 -2
H2PO4 -1
d. CO3 -2
HCO3 -1
3. Write the formula of the conjugate base of each of the following acids:
a. HNO3
NO3 -1
b. H2O
OH -1
c. HSO4 –
SO4 -2
d. H2CO3
HCO3 -1
4. In each of the acid/base reactions below, identify all acids, all bases, and all conjugate
acid/base pairs.
a. CN – (aq) + H2O (l)  HCN (aq) + OH – (aq)
BASE
ACID
ACID
BASE
b. CH3COOH (aq) + HS – (aq)  CH3COO – (aq) + H2S (aq)
ACID
BASE
BASE
ACID
5. Write the formulas of the two products, balance the equation, and write out the complete ionic
and net ionic equations for the neutralization reaction shown below:
2
H3PO4 (aq) +
6
NaOH (aq) 
2
Na3PO4 (aq)
+
6
H2O (l)
CIE: 6 H+ + 2 PO4-3 + 6 Na+ + 6 OH-  6 Na+ + 2 PO4 3- + 6 H2O (l)
NIE: 6 H+ + 6 OH-  6 H2O
OR
H+ + OH-  H2O
6. Given a 1.0 M solution of HNO3, calculate the:
a. [H3O+] :
Strong acid so 1.0 M
b. [OH-] :
1x10-14 = [H3O+][OH-] or [OH-] = 1.0x10-14 / [H3O+]
1.0x10-14 / 1.0 = 1.0 x 10-14 M
c. pH :
-log [H3O+] = -log 1.0 = 0
d. pOH :
pH + pOH = 14
7. Given a 0.5 M solution of LiOH, calculate the:
a. [H3O+] :
1x10-14 = [H3O+][OH-]
so
pOH = 14 - pH = 14 - 0 = 14
or [H3O+] = 1.0x10-14 / [OH-]
1.0x10-14 / 0.5 = 2.0 x 10-14 M
b. [OH-] :
Strong base so [OH-] = 0.5 M
c. pH :
pH + pOH = 14
d. pOH :
-log [OH-] = -log 0.5 = 0.3
so
pH = 14 - pOH = 14 – 0.3 = 13.7