Chemistry 51
ANSWRE KEY
REVIEW QUESTIONS
Chapter 10
1. For each reaction shown below, determine the Brønsted-Lowry acid and base and
their conjugates:
A)
H3BO3
+
acid
HCO3–
B)
base
+
acid
CN–
C)
base
2.
3.
H3O+
H2O
acid
OH–
base
H2O
base
+
H2BO3–
+
+
acid
H2O
base
HCN
acid
CO32–
+
acid
OH–
base
Identify the Brønsted-Lowry acid for each base shown below:
a)
NO3–
HNO3
b)
OH–
H2 O
c)
SO42–
HSO4–
d)
CH3O –
CH3OH
e)
O 2–
OH–
f)
HPO42–
H2PO4–
Identify the Brønsted-Lowry base for each acid shown below:
a)
NH3
NH2–
b)
H2SO4
HSO4–
c)
HS–
S2–
d)
HClO
ClO–
e)
HNO2
NO2–
f)
H2O
OH–
1
4. Each diagram below represents an acid solution with the formula HX. For each solution,
determine if it is a strong acid or a weak acid.
A) Weak acid (mostly molecules)
B) Strong acid (only ions)
C) Weak acid (mostly molecules)
5. Complete the missing information in the table below:
[H3O+]
[OH-]
Acidic/Basic
1.0 x 10 –12
1.0 x 10–2
Basic
3.8 x 10 –4
2.6 x 10–11
Acidic
2.4 x 10–11
4.2 x 10 –4
Basic
1.0 x 10–5
1.0 x 10 –9
Acidic
6.5 x 10 –8
1.5 x 10–7
Basic
Sample calculations:
[OH - ] =
Kw
1.0 x 10-14
=
= 2.6 x 10-11
+
-4
[H 3O ] 3.8 x 10
[H 3O + ] =
Kw
1.0 x 10-14
=
= 2.4 x 10-11
-4
[OH ] 4.2 x 10
6. Identify each of the substances below as strong electrolyte, weak electrolyte or nonelectrolyte:
a) KCl
c) CH3OH
e) H3PO4
strong electrolytes
non-electrolytes
weak electrolytes
2
b) HNO3
strong electrolyte
d) HF
weak electrolyte
7. What are the [H3O+] and [OH–] for a solution with the following pH values?
a) 4.10
[H3O+] = antilog (–4.10) = 7.9 x 10–5 M
[OH ] =
Kw
1.0 x 10
=
+
[H 3O ] 7.9 x 10
14
5
= 1.3 x 10
10
M
b) 9.80
[H3O+] = antilog (–9.80) = 1.6 x 10–10 M
Kw
1.0 x 10
[OH ] =
=
+
[H 3O ] 1.6 x 10
8.
14
10
= 6.3 x 10 5M
What is the pH of a solution that contains 1.54 g of HNO3 in 0.500 L of solution?
1 mol
1
x
= 0.0489 M
63.01 g 0.500 L
pH = - log (0.0489) = 1.311 (3 sig figs)
[H 3O + ] = [HNO 3 ] = 1.54 g x
9. What is the pH of a solution prepared by dissolving 1.00 g of Ca(OH)2 in enough
water to make 875 mL of solution?
[Ca(OH)2 ] = 1.00 g x
1 mol
1
x
= 0.0154 M
74.03 g 0.875 L
[OH ] = 2 [Ca(OH)2 ] = 2 (0.0154) = 0.0308 M
Kw
1.0 x 10 14
[H 3O ] =
=
= 3.25 x 10
[OH ]
0.0308
+
pH = - log (3.25 x 10
13
) = 12.488
3
13
M
(3 sig figs)