Page 233 - Theory and Problems of BEGINNING CHEMISTRY

222 SOLUTIONS [CHAP. 15

EXAMPLE 15.11. What is the number of equivalents in 4.00 mol H 2 SO 3 ?
Ans. Assuming that the H 2 SO 3 will react with a base to replace both hydrogen atoms, we have reaction (b) from
Example 15.9, and there are 2 equiv per mole. In 4.00 mol H 2 SO 3 ,

2 equiv
4.00 mol H 2 SO 3 = 8.00 equiv
1 mol
The major use of equivalents stems from its definition. Once you define the number of equivalents in a
certain mass of a substance, you do not need to write the equation for its reaction. That equation has already
been used in defining the number of equivalents. Thus, a chemist can calculate the number of equivalents in a
certain mass of substance, and technicians can subsequently use that definition without knowing the details of
the reaction.
One equivalent of one substance in a reaction always reacts with one equivalent of each of the other
substances in that reaction.

EXAMPLE 15.12. How many equivalents of NaOH does the 82.0 g of H 2 SO 3 react with in reaction (b) of Example 15.9?
Ans. Since there are 2 equiv of H 2 SO 3 , they must react with 2 equiv of NaOH. Checking, we see that 2 equiv of NaOH
liberate 2 mol of OH and also react with 2 mol H , and thus there are 2 equiv by definition also.
−
+
15.5. NORMALITY
Analogous to molarity, normality is defined as the number of equivalents of solute per liter of solution. Its
unit is normal with the symbol N. Thus the normality of a solution may be 3.0 normal, denoted 3.0 N.

EXAMPLE 15.13. What is the normality of a solution containing 1.75 equiv in 2.50 L of solution?
Ans. 1.75 equiv
= 0.700 N
2.50 L
The normality of the solution is 0.700 normal, or, stated another way, the solution is 0.700 normal.

Normality is some integral multiple (1, 2, 3, ...) of molarity, since there are always some integral number
of equivalents per mole.

EXAMPLE 15.14. What is the normality of 2.25 M H 2 SO 3 in each of the following reactions?
(a)H 2 SO 3 + NaOH −→ NaHSO 3 + H 2 O (b)H 2 SO 3 + 2 NaOH −→ Na 2 SO 3 + 2H 2 O
Ans. We found in Example 15.9 that there were 1 equiv per mol in the first reaction and 2 equiv per mol in the second.
We use these factors to solve for normality:
2.25 mol 1 equiv 2.25 equiv

(a)2.25 M = = = 2.25 N
1L 1 mol 1L

2.25 mol 2 equiv 4.50 equiv
(b)2.25 M = = = 4.50 N
1L 1 mol 1L
EXAMPLE 15.15. What is the molarity of 1.25 N H 2 SO 3 in each of the following reactions?
(a)H 2 SO 3 + NaOH −→ NaHSO 3 + H 2 O (c) 6 H + H 2 SO 3 + 6 e −→ H 2 S + 3H 2 O
+
−
(b)H 2 SO 3 + 2 NaOH −→ Na 2 SO 3 + 2H 2 O
Ans. Again we use the factors found in Sec. 15.4:
1.25 equiv 1 mol 1.25 mol
(a)1.25 N = = = 1.25 M
1L 1 equiv 1L
1.25 equiv 1 mol 0.625 mol
(b)1.25 N = = = 0.625 M
1L 2 equiv 1L