A Getting-It-On Review and Self-Test
Acids, Bases and Compounds in Solution
According to Arrhenius, an acid contains (1)
and a base
contains (2)
. However, Brönsted defines an acid as a
(3)
and a base as a (4)
.
Lewis defines an acid as an (5)
and a base
as an (6)
.
In the Brönsted system, a species may gain a proton to become a
(7)
or lose a proton to become a (8)
.
Solutions of acid taste (9)
and solutions of base taste
(10)
. Compounds which are sensitive to acid or base and change
color are called (11)
. Reactions of acids and bases in solution
are (12)
and an indicator is used to tell the (13)
.
At the end-point, the (14)
of acid equals the (15)
of
base.
The concentration of solutes may be given in (16)
,
(17)
, (18)
, and (19)
. The definition of
molarity (M) is (20)
of solution and normality (N) is (21)
of
solution. Molarity is always (22)
to normality.
23.
Complete and balance the following equations:
H2SO4 + BaO →
Zn(OH)2 + NaOH →
HCl + Al →
NaOH + NaHCO3 →
MnCl2 + KOH →
24.
Complete the table:
Normality
Molarity
1.0 N Mg(OH)2
0.125 M HNO3
0.75 M H2SO3
Equivalent weight
2
25.
Calculate the volume or weight percents of the following solutions:
Solvent
Solute
45 g H2O
15 g NaOH
1 liter H2O
300 ml C2H5OH
500 ml C2H5OH
300 ml C6H6
300 ml H2O
60 g KCl
26.
Percent
Fill-in the Conjugate table:
Conjugate acid
NH4
species
Conjugate base
H2O
CO32
H2S
27.
Calculate the weight of C2H5OH required to make 200 ml of a
6.5 M solution.
28.
The density of 30.0% by weight sulfuric acid is 1.32 g/ml. Find the
molarity and normality.
29.
Find the volume of 0.105 N H2SO4 required to neutralize 0.192 g
KOH.
30.
It requires 25.86 ml of 0.1005 N NaOH to neutralize 30.09 ml of
nitric acid. Find the normality of the acid.
31.
Calculate the weight of calcium chloride formed in the
neutralization of 0.560 g CaO with HCl.
3
32.
To what volume must you dilute 100 ml of 10 M H2SO4 to make a
2 N solution?
33.
Determine the pH and the pOH of 0.005 M Ba(OH)2.
34.
Calculate the normality and molarity of an H2SO4 solution where
the pH is 3.76
35.
Calculate the pH of these:
a.
b.
c.
0.015 N H3PO4
0.00075 M H3AsO4
0.0000607 M HCl
ANSWERS
1. proton (H+)
2. hydroxide (OH–)
3. proton donor
4. proton acceptor
5. electron pair acceptor
6. electron pair donor
7. conjugate acid
8. conjugate base
9. sour
10. bitter
11. indicators
12. neutralizations
13. end-point
14. equivalents
15. equivalents
16. volume %
17. weight %
18. molarity
19. normality
20. moles/liter
21. equivalents/liter
22. 
4
23.
H2SO4 + BaO →
BaSO4 + H2O
Zn(OH)2 + 2 NaOH → Na2ZnO2 + 2 H2O
6 HCl + 2 Al →
2 AlCl3 + 3 H2
NaOH + NaHCO3 →
Na2CO3 + H2O
MnCl2 + 2 KOH →
Mn(OH)2 + 2 KCl
24.
Normality
Molarity
Equivalent weight
1.0 N Mg(OH)2
0.5 M
29.15 g
0.125 N
0.125 M HNO3
63 g
1.5 N
0.75 M H2SO3
41.04 g
25.
Solvent
Solute
Percent
45 g H2O
15 g NaOH
25% (weight)
1 liter H2O
300 ml C2H5OH
23.1% (volume)
500 ml C2H5OH
300 ml C6H6
37.5% (volume)
300 ml H2O
60 g KCl
16.7% (weight)
5
26.
27.
28.
Conjugate acid
NH4
species
NH3
Conjugate base
NH2
H3O
H2O
OH
H2CO3
HCO3
CO32
H3S
H2S
HS
0.2 liters 
1 liter 
6.5 mole
46 g

 59.8 g C2H5OH
1 liter
1 mole
1000 ml 1.32 g

 1,320 g solution
1 liter
ml
1,320 g solution 
396 g H2SO4 
30 g H2SO4
 396 g H2SO4
100 g solution
1 mole H2SO4 4.04 mole

or 4.04 M
98 g H2SO4
liter
4.04 mole 2 equiv 8.08 equiv


or 8.08 N
liter
1 mole
liter
29.
0.192 g KOH 
30.
1 liter H2SO4
1 mole
1 equiv


 0.0326 liters or 32.6 ml
56.1 g KOH 1 mole 0.105 equiv H2SO4
0.1005 N 
25.86 ml
 0.0864 N
30.09 ml
31.
0.560 g CaO 
1 mole CaO 1 mole CaCl2 111 g CaCl2


 1.11 g CaCl2
56 g CaO
1 mole CaO 1 mole CaCl2
6
32.
10 M H2SO4 = 20 N H2SO4
100 ml 20 N
2N
total volume
33.
OH  
Total volume  100 ml 
0.005 mole Ba(OH)2
2 mole OH
0.01 mole OH


liter
1 mole Ba(OH)2
liter
pOH   log (0.01)   log102  (2)  2
34.
pH   log H   3.76
M
35.
20 N
 1000 ml
2N
pH = 14 – 2 = 12
H   103.76  1.74  104  N
1.74  104 mole H 1 mole H2SO4

 8.69  105 M

liter
2 mole H
Hint: N  H 
a.
pH   log (0.015)  1.82
b.
N  3  M  3  (7.5 104 )  2.25 103
pH   log (2.25  103 )  2.65
c.
H   6.07  105
pH   log (6.07  105 )  4.22