Page 238 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 15] SOLUTIONS 227

Ans. By definition, normality is the number of equivalents per liter. Then
x equiv 1000 mequiv 1L x mequiv

=
1L 1 equiv 1000 mL 1mL
15.20. Explain why volume times normality of one reagent is equal to volume times normality of each of the
other reagents in the reaction.
Ans. Volume times normality is the number of equivalents, and by definition, the number of equivalents of one
reagent is the same as the number of equivalents of any other reagent in a given reaction.

15.21. What volume of 1.50 N H 2 SO 4 is completely neutralized by 35.8 mL of (a) 2.50 N NaOH, (b) 2.50 M
NaOH, (c) 2.50 M Ba(OH) 2 , and (d) 2.50 N Ba(OH) 2 ?
Ans. (a) The number of equivalents of NaOH is 0.0358 L × 2.50 N = 0.0895 equiv. That also is the number of
equivalents of H 2 SO 4 .

1L
0.0895 equiv = 0.0597 L = 59.7mL
1.50 equiv
(b) Since 2.50 M NaOH is 2.50 N NaOH, the answer is the same as that in part a.

2.50 mol Ba(OH) 2 2 equiv
(c) 2.50 M Ba(OH) 2 = = 5.00 N Ba(OH) 2
1L 1 mol
5.00 N(0.0358 L) = 0.179 equiv
The quantity of Ba(OH) 2 is 0.179 equiv. The volume of H 2 SO 4 is therefore
1L

0.179 equiv H SO 4 = 0.119 L = 119 mL
2
1.50 equiv
(d) 2.50 N Ba(OH) 2 has the same neutralizing ability as 2.50 N NaOH, and the answer is the same as that
in part a.
15.22. Calculate the number of milliequivalents per milliliter in 6.11 N H 2 SO 4 .

6.11 equiv 1000 mequiv 1L 6.11 mequiv
Ans. 6.11 N = =
1L 1 equiv 1000 mL 1mL
Normality may be defined as the number of milliequivalents per milliliter as well as the number of equivalents
per liter.

15.23. What volume of 2.25 N H 3 PO 4 will 14.7 equivalents of a base neutralize completely? Explain why you
did not need to know the formula of the base to answer this question.

Ans. 14.7 equiv of base reacts with 14.7 equiv of acid, no matter what the base.

1L
14.7 equiv H PO 4 = 6.53 L
3
2.25 equiv
(The identity of the base was used previously to determine the number of equivalents of base; it is not
necessary to know its identity now.)
15.24. A bottle is marked 1.00 N H 2 SO 4 . What is its probable molarity?

Ans. Assuming that the normality is for complete neutralization, the molarity is given by
1.00 equiv 1 mol
= 0.500 M
1L 2 equiv

15.25. A bottle is marked 1.00 N H 2 SO 4 for production of NaHSO 4 . What is its molarity?
Ans. H 2 SO 4 + NaOH −→ NaHSO 4 + H 2 O
1.00 equiv 1 mol
= 1.00 M
1L 1 equiv