Page 232 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 15] SOLUTIONS 221

1 mol H 2 O

25.0gH O = 1.39 mol H 2 O
2
18.0gH O
2
0.375 mol CH 3 OH
X CH 3 OH = = 0.212
0.375 mol CH 3 OH + 1.39 mol H 2 O
Since the mole fraction is a ratio of moles (of one substance) to moles (total), the units cancel and mole fraction has
no units.
EXAMPLE 15.8. Show that the total of both mole fractions in a solution of two compounds is equal to 1.
x mol A y mol B
Ans. X A = X B =
x mol A + y mol B x mol A + y mol B
x mol A y mol B x mol A + y mol B
X A + X B = + = = 1
x mol A + y mol B x mol A + y mol B x mol A + y mol B

15.4. EQUIVALENTS
Equivalents are measures of the quantity of a substance present, analogous to moles. The equivalent is
defined in terms of a chemical reaction. It is defined in one of two different ways, depending on whether an
oxidation-reduction reaction or an acid-base reaction is under discussion. For an oxidation-reduction reaction,
one equivalent is the quantity of a substance that will react with or yield 1 mol of electrons. For an acid-base
reaction, one equivalent is the quantity of a substance that will react with or yield 1 mol of hydrogen ions or
hydroxide ions. Note that the equivalent is defined in terms of a reaction, not merely in terms of the formula
of a compound. Thus, the same mass of the same compound undergoing different reactions can correspond to
different numbers of equivalents. The ability to determine the number of equivalents per mole is the key to
calculations in this section.

EXAMPLE 15.9. How many equivalents are there in 82.0 g of 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. Since 82.0 g of H 2 SO 3 is 1.00 mol of H 2 SO 3 , we will concentrate on that quantity of H 2 SO 3 . Both of these reactions
are acid-base reactions, and so we define the number of equivalents of H 2 SO 3 in terms of the number of moles of
−
hydroxide ion with which it reacts. In the first equation, 1 mol of H 2 SO 3 reacts with 1 mol of OH .Bydefinition,
that quantity of H 2 SO 3 is 1 equiv. For that equation, 1 equiv of H 2 SO 3 is equal to 1 mol of H 2 SO 3 . In the second
−
equation, 1 mol of H 2 SO 3 reacts with 2 mol OH .Bydefinition, that quantity of H 2 SO 3 is equal to 2 equiv. Thus,
in that equation, 2 equiv of H 2 SO 3 is 1 mol of H 2 SO 3 .
(a) 1 equiv = 1 mol (b) 2 equiv = 1 mol
We can use these equalities as factors to change moles to equivalents or equivalents to moles, especially to calculate
normalities (Sec. 15.5) or equivalent masses (Sec. 15.6).
EXAMPLE 15.10. How many equivalents are there in 82.0 g of H 2 SO 3 in the following reaction?
+
6H + H 2 SO 3 + 3Zn −→ H 2 S + 3Zn 2+ + 3H 2 O
Ans. This reaction is a redox reaction (Chap. 14), and so we define the number of equivalents of H 2 SO 3 in terms of the
number of moles of electrons with which it reacts. Since no electrons appear explicitly in an overall equation, we
will write the half-reaction in which the H 2 SO 3 appears:

−
+
6H + H 2 SO 3 + 6 e −→ H 2 S + 3H 2 O
It is now apparent that 1 mol of H 2 SO 3 reacts with 6 mol e , and by definition 6 equiv of H 2 SO 3 react with
−
6 mol e . Thus, 6 equiv equal 1 mol in this reaction. Since 82.0 g H 2 SO 3 is 1 mol, there are 6 equiv in 82.0 g H 2 SO 3 .
−
Some instructors and texts ask for the number of equivalents per mole of an acid or base without specifying a
particular reaction. In that case, merely assume that the substance undergoes an acid-base reaction as completely
as possible. State that assumption in your answers on examinations.