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Chemistry 14CL AcidBase Chemistry Review Key
1.
Barium hydroxide furnishes two moles of hydroxide in solution for each mole of barium
hydroxide that you start with, and HCl donates one mole of protons/mole of acid.
Thus the
concentration of [OH

] = 0.0800 M.
Since
equivalents of acid = equivalents of base,
(X mL) (0.0800 M) = (15.0 mL)(0.100 M)
and X = 18.75 mL.
2.
Sulfuric acid furnishes two moles of protons for each mole of acid. Thus the concentration of
[H
+
] = 0.500 M. Since moles of [H
+
] = moles of [OH],
(X mL)(0.500 M) = (38.40 mL)(0.200 M) and X = 15.4 mL.
3.
moles of HCl = moles of NaOH = 0.0125 moles
Molecular weight of NaOH = 40.0g, Weight of NaOH required = 0.50 g
4.
This is a strong base which dissociates completely in solution.
[OH

] = 0.037M;
pOH = 1.43;
pH = 12.57
5. [H+] = 0.030,
pH = 1.5
pOH = (141.5) = 12.5
[OH] =3.3 x 10
13
6.
(a) [H+] =
0.05 M
pH
=
1.3
(b) [H+] = 1.2 x 10
4
pH
=
3.9
7.
(a) pOH = 1.3, pH = 12.7, [H+] = 2 x 10
13
(b) pOH = 7, pH
= 7, [H+] = 1 x 10
 7
8. (a)
[H+] = 3.0 10
3
M,
pH =2.5
(b)
[H+] = 8.9 x 10
5
,
pH = 4.0
9.
This is a limiting reagent problem only. The total moles of nitric acid involved in the titration
is 2.00 x 10
3
.
Therefore, half way through the titration, there are 1.00 x 1
3
moles left
untitrated.
The volume of the solution at this point in the titration can be calculated from the
original 10 mL of nitric acid and the added 10 mL of base solution giving a solution volume
of 20 mL.
Therefore, the [H
+
] of the solution at this point in the titration is
1.00 x 10
3
μολ
0.020Λ
= 5.0ξ10
2
Μ
and the pH is 1.3
10.
The calculations differs from #9 only in the fact that you have a different amount of
unreacted nitric acid and solution volume.
1.60 x 10
 3
mol
0.014L
= 0.114M
and the theoretical pH is 0.94
11.
This is also a limiting reagent problem.
However, since it also involves a weak acid, the
equilibrium expression must be used to calculate
[H
+
] after the stoichiometry is
determined from the limiting reagent calculations as was done in the problem above.
Like problem 9, half way through the titration the reaction has produced 1.00 x 10
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