Page 240 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 15] SOLUTIONS 229

The number of millimoles of base is determined from the2:1 ratio in the balanced chemical equation:

2 mmol NaOH
84.33 mmol H 2 SO 4 = 168.7 mmol NaOH
1 mmol H 2 SO 4
168.7 mmol NaOH
= 6.748 M NaOH = 6.748 N NaOH
25.00 mL NaOH
15.32. Calculate the equivalent mass of each of the following acids toward complete neutralization. (a)H 3 PO 3 ,(b)H 2 SO 4 ,
(c) HBr, (d)HC 2 H 3 O 2 , and (e)H 2 C 2 O 4 .
Ans. (a) 27.3 g/equiv (b) 49.0 g/equiv (c) 80.9 g/equiv (d) 60.0 g/equiv (e) 45.0 g/equiv.

15.33. A sample of 5.00 g of a solid acid was treated with 50.00 mL of 2.500 N NaOH, which dissolved it completely (by
reacting with it). There was enough excess NaOH to require 9.13 mL of 1.000 N HCl to neutralize the excess base.
What is the equivalent mass of the acid?
Ans. The numbers of equivalents of base and HCl are as follows:
2.500 mequiv

50.00 mL = 125.0 mequiv NaOH
1mL
1.000 mequiv

9.13 mL = 9.13 mequiv HCl
1mL
The difference is the number of milliequivalents of the solid acid:
125.0 mequiv − 9.13 mequiv = 115.9 mequiv
The equivalent mass is
5000 mg 43.1mg 43.1g
= =
115.9 mequiv 1 mequiv 1 equiv
15.34. Write an equation for the half-reaction in which H 2 C 2 O 4 is oxidized to CO 2 . What is the equivalent mass of
H 2 C 2 O 4 ?
+
Ans. H 2 C 2 O 4 −→ 2CO 2 + 2H + 2 e −
90.0g 1 mol 45.0g
=
1 mol 2 equiv 1 equiv
15.35. What is the equivalent mass of H 2 C 2 O 4 in the following reaction?
−
+
6H + 2 MnO 4 + 5H 2 C 2 O 4 −→ 10 CO 2 + 8H 2 O + 2Mn 2+
Ans. The half-reaction and thus the equivalent mass are given in the prior problem. The 1 mol of acid liberates
2 mol of electrons, and so there are 2 equiv of acid per mole of acid. If we wrote the half-reaction for the
equation given, we would find that there was 10 mol of electrons involved in reducing 5 mol of acid, and so
again we get
10 equiv/5 mol = 2 equiv/mol
Still another method to balance the equation is by the oxidation state change method:

H 2 C 2 O 4 −→ 2CO 2
| |
—————-
2(3 → 4) =+2
The change in oxidation state is 2 per molecule of H 2 C 2 O 4 , and so there are 2 equiv per mole.