Now determine the grams of hydrate needed using the appropriate factors.

1.50 Eq 24.8 g

--------X ------X 0.500 L = 93.6 g

L

Eq

When preparing solutions using compounds where only part of the molecule will

be involved in the reaction, a factor must be used to account for the reactivity of the ion

of interest per mole of compound.

TPIV expresses the number of equivalents per mole of substance.

Example**. **How much Na2HPO4 is required to prepare 2.00 liters of a 1.00 Eq/L

sodium standard?

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

Grams of Na2HPO4.

Calculate the GEW of the compound using the TPIV of the ion of interest.

Na2HPO4

Na 23.0 X 2 = 46.0

H 1.0 X 1 =

1.0

P 31.0 X 1 = 31.0

O 16.0 X 4 = + 64.0

142.0 g/mol Na2HPO4

142.0 g Na2HPO4

mol Na2HPO4

---------------- X ----------------= 71.00 g Na2HPO4/Eq Na

mol Na2HPO4

2 Eq Na

Determine the amount of salt (hydrate) needed in the usual manner using the

appropriate factors.

1.00 Eq Na 71.00 g Na2HPO4

---------- X ----------------X 2.00 L = 142 g Na2HPO4

L

Eq Na

MD0837

4-11

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