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Dilution

Module by: Sunil Kumar Singh. E-mail the author

Dilution means adding solvent (A) to a solution (S). The amount of solute (B) remains constant, but the amount of solution increases. Dilution of solution is needed as solutions are purchased and stored in concentrated form; whereas we use solution of different concentration in accordance with the requirement of the application.

Concentration of solution is expressed in different concentration format as mass percent, mole fraction, molarity, normality or any other format. The underlying principle, here, is that quantity of solute remains the same before and after dilution.

Dilution of a solution of a given chemical entity can also be realized by mixing solution of higher concentration with solution of lower concentration. This mixing is equivalent to increasing amount of solvent. This aspect has already been dealt in the module titled “Gram equivalent concept”.

Mass percentage and dilution

Mass percentage is equal to percentage of mass of solute with respect to solution and is given by :

y = Mass of solute(B) Mass of solution(S) = W B W S X 100 = g B g S X 100 y = Mass of solute(B) Mass of solution(S) = W B W S X 100 = g B g S X 100

Let 1 and 2 subscripts denote before and after states, then mass of solute before and after dilution is same :

W B 1 = W B 2 W B 1 = W B 2

y 1 W S 1 = y 2 W S 2 y 1 W S 1 = y 2 W S 2

Similarly,

y 1 g S 1 = y 2 g S 2 y 1 g S 1 = y 2 g S 2

Example 1

Problem : How much water would you need to prepare 600 grams of a 4% (w/w) Hydrogen Peroxide ( H 2 O 2 H 2 O 2 ) solution using a solution of 20% (w/w) ?

Solution : Here,

g S 1 = 600 g m g S 1 = 600 g m

y 1 = 4 y 1 = 4

y 2 = 20 y 2 = 20

y 1 g S 1 = y 2 g S 2 y 1 g S 1 = y 2 g S 2

4 X 600 = 20 X g S 2 4 X 600 = 20 X g S 2

g S 2 = 4 X 600 20 = 120 g m g S 2 = 4 X 600 20 = 120 g m

Total mass of 4% = mass of 20% solution + mass of water added Total mass of 4% = mass of 20% solution + mass of water added

mass of water added = 600 120 = 480 g m mass of water added = 600 120 = 480 g m

Molarity and dilution

Molarity is equal to ratio of moles of solute and volume of solution and is given by :

M = moles of solute V L = = mill−moles of solute V C C M = moles of solute V L = = mill−moles of solute V C C

Let 1 and 2 subscripts denote before and after dilution, then :

moles before dilution = moles after dilution moles before dilution = moles after dilution

M 1 V L 1 = M 2 V L 2 M 1 V L 1 = M 2 V L 2

Similarly,

M 1 V C C 1 = M 2 V C C 2 M 1 V C C 1 = M 2 V C C 2

Example 2

Problem : What volume of water should be added to 100 ml nitric acid having specific gravity 1.4 and 70% strength to give 1 M solution?

Solution : We first need to convert the given strength of concentration to molarity. Molarity is given by :

M = 10 y d M O M = 10 y d M O

The specific gravity is 1.4 (density compared to water). Hence, density of solution is 1.4 gm/ml.

M = 10 X 70 X 1.4 1 + 14 + 3 X 16 = 15.55 M M = 10 X 70 X 1.4 1 + 14 + 3 X 16 = 15.55 M

Applying dilution equation,

M 1 V C C 1 = M 2 V C C 2 M 1 V C C 1 = M 2 V C C 2

15.55 X 100 = 1 X V 2 15.55 X 100 = 1 X V 2

V 2 = 1555 m l V 2 = 1555 m l

The volume of water to be added = 1555 100 = 1455 m l The volume of water to be added = 1555 100 = 1455 m l

Example 3

Problem : In an experiment, 300 ml of 6.0 M nitric acid is required. Only 50 ml of pure nitric acid is available. Pure anhydrous nitric acid (100%) is a colorless liquid with a density of 1522 k g / m 3 k g / m 3 . Would it be possible to carry out the experiment?

Solution : Here,

M 1 = 6 M ; M 2 = ? ; V C C 1 = 300 m l ; V C C 2 = ? M 1 = 6 M ; M 2 = ? ; V C C 1 = 300 m l ; V C C 2 = ?

In order to apply dilution equation, we need to know the molarity of concentrated nitric acid. Now 1 litre pure concentrate weighs 1000 X 1.522 = 1522 gm. The number of moles in 1 litre of nitric acid is :

n B = 1522 1 + 14 + 3 X 16 = 24.16 n B = 1522 1 + 14 + 3 X 16 = 24.16

Hence, molarity of the nitric acid concentrate is 24.16 M.

Applying dilution equation,

M 1 V C C 1 = M 2 V C C 2 M 1 V C C 1 = M 2 V C C 2

V C C 2 = M 1 V C C 1 M 2 = 6 X 300 24.16 = 74.5 m l V C C 2 = M 1 V C C 1 M 2 = 6 X 300 24.16 = 74.5 m l

It means requirement of concentrate (74.5 ml) is more than what is available (50 ml). As such, experiment can not be completed.

Example 4

Problem : A 100 ml, 0.5 M calcium nitrate solution is mixed with 200 ml of 1.25M calcium nitrate solution. Calculate the concentration of the final solution. Assume volumes to be additive.

Solution : We first calculate milli-moles for each solution. Then, add them up to get the total moles present in the solution. Finally, we divide that total moles by the total volume.

Milli-moles in the first solution = M 1 V C C 1 = 100 X 0.5 = 50 Milli-moles in the first solution = M 1 V C C 1 = 100 X 0.5 = 50

Milli-moles in the second solution = M 2 V C C 2 = 200 X 1.25 = 250 Milli-moles in the second solution = M 2 V C C 2 = 200 X 1.25 = 250

Milli-moles in the final solution = 50 + 250 = 300 Milli-moles in the final solution = 50 + 250 = 300

Total volume of the solution = 100 + 200 = 300 m l Total volume of the solution = 100 + 200 = 300 m l

Hence, molarity of the final solution is :

M = Milli-moles of solute V C C = 300 300 = 1 M M = Milli-moles of solute V C C = 300 300 = 1 M

Alternative method

Molarity of the first solution when diluted to final volume of 300 ml is obtained as :

M 1 V C C 1 = M 2 V C C 2 M 1 V C C 1 = M 2 V C C 2

0.5 X 100 = M 2 X 300 0.5 X 100 = M 2 X 300

M 2 = 50 300 = 1 6 M 2 = 50 300 = 1 6

Molarity of the second solution when diluted to final volume of 300 ml is obtained as :

M 1 V C C 1 = M 2 V C C 2 M 1 V C C 1 = M 2 V C C 2

1.25 X 200 = M 2 X 300 1.25 X 200 = M 2 X 300

M 2 = 250 300 = 5 6 M 2 = 250 300 = 5 6

Hence, molarity of the final combined solution is :

M = 1 6 + 5 6 = 1 M = 1 6 + 5 6 = 1

Normality and dilution

As there is no change in the amount of solute due to dilution, the amount expressed as gram equivalents should also remain same before and after dilution. Like in the case of molarity, we state the condition of dilution in terms of normality as :

Gram equivalents in the solution = N 1 V 1 = N 2 V 2 Gram equivalents in the solution = N 1 V 1 = N 2 V 2

Example 5

Problem : What volume of water should be added to 100 ml sulphuric acid having specific gravity 1.1 and 80% strength to give 1 N solution?

Solution : We first need to convert the given strength of concentration to Normality. For that we first calculate molarity and then use the formula : :

N = x M N = x M

Now, molarity is given by

M = 10 y d M O M = 10 y d M O

The specific gravity is 1.4 (density compared to water). Hence, density of solution is 1.4 gm/ml.

M = 10 X 80 X 1.1 2 X 1 + 32 + 4 X 16 = 8.98 M M = 10 X 80 X 1.1 2 X 1 + 32 + 4 X 16 = 8.98 M

The valence factor of H 2 S O 4 H 2 S O 4 is 2. Hence,

N = x M = 2 X 8.98 = 17.96 N N = x M = 2 X 8.98 = 17.96 N

Applying dilution equation,

N 1 V C C 1 = N 2 V C C 2 N 1 V C C 1 = N 2 V C C 2

17.96 X 100 = 1 X V 2 17.96 X 100 = 1 X V 2

V 2 = 1796 m l V 2 = 1796 m l

The volume of water to be added = 1796 100 = 1696 m l The volume of water to be added = 1796 100 = 1696 m l

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