The Foundations of Chemistry

Solution

First we calculate the number of equivalents of Na 2 CO 3 in the sample.

no. eq Na 2 CO 3 0.364 g Na 2 CO 3 6.87 10 ^3 eq Na 2 CO 3

Because no. eq H 2 SO 4 no. eq Na 2 CO 3 , we can write

LH 2 SO 4 NH 2 SO 4 6.87 10 ^3 eq H 2 SO 4

NH 2 SO 4 0.172 NH 2 SO 4

You should now work Exercise 50.

OXIDATION–REDUCTION REACTIONS

Our rules for assigning oxidation numbers are constructed so that in all redox reactions

the total increase in oxidation numbers must equal the total decrease in oxidation

numbers.

This equivalence provides the basis for balancing redox equations. Although there is no
single “best method” for balancing all redox equations, two methods are particularly useful:
(1) the half-reaction method, which is used extensively in electrochemistry (Chapter 21)

6.87 10 ^3 eq H 2 SO 4



0.040 L

6.87 10 ^3 eq H 2 SO 4



LH 2 SO 4

1 eq Na 2 CO 3



53.0 g Na 2 CO 3

Automatic titrators are used in modern

analytical laboratories. Such titrators rely

on electrical properties of the solutions.

Methyl red indicator changes from yellow

to red at the end point of this titration.

11-4 Equivalent Weights and Normality 415

The starting values in this example are

the same as those in Example 11-7.

The normality of this H 2 SO 4 solution

is twice the molarity obtained in

Example 11-7 because 1 mol of H 2 SO 4

is 2 eq.