Section IV. Variations of Normality Problems

Calculate the amount of salt needed using the appropriate factors.

50.0 mEq

60.20 mg

----------X ----------X 0.100 L = 301 mg

mEq

Section IV. VARIATIONS OF NORMALITY PROBLEMS

4-10. PROBLEMS INVOLVING HYDRATES

Like molar solutions, normal solutions contain a known number of particles per

liter of solution. An one equivalent per liter (Eq/L) solution of CuSO4 will have the same

concentration as a one equivalent per liter CuSO4.5H2O solution. When solving normal

problems involving hydrates, use the gram equivalent weight of the substance being

weighed in the preparation of the solution.

Example.

How much CuSO4.5H2O is required to prepare 500 mL of a 1.50 Eq/L CuSO4 solution?

Solution. Read the problem carefully and determine the desired quantity.

Grams of CuSO4.5H2O.

Determine the GEW of the compound that is available (actually being weighed

out).

CuSO4.5H2O

63.5 X 1 = 63.5

32.1 X 1 = 32.1

16.0 X 4 = 64.0

HOH

18.0 X 5 = + 90.0

249.6 g/mol

NOTE:

When determining the TPIV of hydrates, the water (H2O) molecules are not

considered. Water has a low degree of reactivity. Also, note that copper is

ambivalent, and we must consider the charge of the anion (-) in order to

correctly determine the TPIV.

249.6 g

mol

-------- X ------ = 124.8 g/Eq

mol

2 Eq

MD0837

4-10