Page 234 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 15] SOLUTIONS 223

1.25 equiv 1 mol 0.208 mol
(c) 1.25 N = = = 0.208 M
1L 6 equiv 1L
We can do concentration problems with normality just as we did them with molarity (Chap. 11). The
quantities are expressed in equivalents, of course.

EXAMPLE 15.16. How many equivalents are present in 3.0 L of 2.0 N solution?

2.0 equiv
Ans. 3.0L = 6.0 equiv
1L
EXAMPLE 15.17. How many liters of 2.00 N H 3 PO 4 does it take to hold 1.80 equiv?

1L
Ans. 1.80 equiv = 0.900 L
2.00 equiv
EXAMPLE 15.18. What is the final concentration of HCl if 2.0 L of 2.2 N HCl and 1.5 L of 3.4 N HCl are mixed and
diluted to 5.0 L?
Ans. (2.0L)(2.2 N) = 4.4 equiv
(1.5L)(3.4 N) = 5.1 equiv
total = 9.5 equiv
9.5 equiv
= 1.9 N
5.0L
Equivalents are especially useful in dealing with stoichiometry problems in solution. Since 1 equiv of one
thing reacts with 1 equiv of any other thing in the reaction, it is also true that the volume times the normality of
the first thing is equal to the volume times the normality of the second.

EXAMPLE 15.19. What volume of 2.00 N NaOH is required to neutralize 25.00 mL of 2.70 N H 2 SO 4 ?
Ans. N 1 V 1 = N 2 V 2
N 1 V 1 (2.70 N H 2 SO 4 )(25.00 mL)
V 2 = = = 33.8mL
N 2 2.00 N NaOH
Note that less volume of H 2 SO 4 is required than of NaOH because its normality is greater.

15.6. EQUIVALENT MASS
The equivalent mass of a substance is the mass in grams of 1 equiv of the substance. The equivalent mass
is often a useful property to characterize a substance.

EXAMPLE 15.20. A new solid acid was prepared in a laboratory; its molar mass was not known. It was titrated with
standard base, and the number of moles of base was calculated. Without knowing the formula of the acid, can you tell how
many moles of the acid were present in a certain mass of acid? Can you tell how many equivalents of acid were present?

Ans. Without knowing the formula, you cannot tell the number of moles, but you can tell the number of equivalents. For
example, HX and H 2 X 2 would both neutralize the same number of moles of NaOH per gram of acid. Suppose X had
a molar mass of 35 g/mol. HX would have a molar mass of 36 g/mL; H 2 X 2 would have a molar mass of 72 g/mol.
To neutralize 1.00 mol NaOH would take 1.00 mol HX or 0.500 mol H 2 X 2 :
HX + NaOH −→ NaX + H 2 O
or H 2 X 2 + 2 NaOH −→ Na 2 X 2 + 2H 2 O
One mole of HX is 36 g; 0.500 mol of H 2 X 2 is also 36 g. You cannot tell from the mass which is the formula of the
acid. In either case, the equivalent mass of the acid is 36 g per equivalent. The equivalent mass of H 2 X 2 is given by
72 g 1 mol 36 g
=
1 mol 2 equiv 1 equiv
To convert from molar mass to equivalent mass, use the same factors as were introduced in Sec. 15.4.