Х = 0,355 ´ A х 5. Definition of need(requirement) of an amount of lime chloride. for a chlorination of water.. SOLUTION

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Х = 0, 355 ´ A х 5

Х - residual chlorine in milligrammes ( mg) on 1 litre;

0, 355 - amount of chlorine in milligrammes, appropriate 1 ml of Sodium hyposulphitum;

A - amount 0, 01 n of a solution of Sodium hyposulphitum in a ml, use on titration 200 milliliters of water;

5 - recalculation on 1 litre of water.

Definition of need(requirement) of an amount of lime chloride

for a chlorination of water.

The calculation of a necessary amount 1 % of a solution of Lime chloride for a chlorination 1000 litres ( 1 cubic meter ) water: From three flasks of researched water to select in which the contents of residual chlorine is in limits 0, 3 - 0, 5 ml of chlorine on 1 litre.

Translate an amount of drops 1 % of a solution of the lime chloride added in the given flask, at an experienced chlorination in a ml ( 1 ml = 20 drops) contain on the average. The amount of dry lime chloride in grammes necessary for a decontamination 1 cubic meter water is determined under the formula:

В х 5 х 1000

X = ---------------------, where:

100

Х - amount of lime chloride in grammes on 1 м3 of water.

В - amount 1 % of a solution of lime chloride ( in a ml ) added in a flask at experience chlorination;

1000 - recalculation on 1 м3 of water;

100 - recalculation in grammes of dry lime chloride.

Conclusion: 1 gramme of dry lime chloride on 1 м3 of water is necessary for a decontamination of researched water.

Example 1:

It was titrated 200 milliliters of water in 3 flasks. In 1 flask we were adding 2 drops 0, 01 n Solution of Sodium hyposulphitum, in 2 flask - 4 drops and in 3 flask - 8 drops. What is necessary quantity of dry lime chloride for a decontamination of water ( in grammes on 1 м3 of water )?

SOLUTION

1). 1 ml 0, 01 n Solution of Sodium hyposulphitum contain 20 drops. In 1 flask we were add 2 drops

1 ml - 20 drops

Х ml - 2 drops

1 х 2

Х = -------- = 0, 1 mls of Sodium hyposulphitum

2). 1 ml 0, 01 n Solution of Sodium hyposulphitum contain 20 drops. In 2 flask we were add 4 drops

1 ml - 20 drops

Х ml - 4 drops

1 х 4

Х = -------- = 0, 2 mls of Sodium hyposulphitum

3). 1 ml 0, 01 n Solution of Sodium hyposulphitum contain 20 drops. In 3 flask we were add 8 drops

1 ml - 20 drops

Х ml - 8 drops

1 х 8

Х = -------- = 0, 4 mls of Sodium hyposulphitum

4). Residual quantity of chlorine in each flask determine under the formula:

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