Chemistry, Third edition

148 9 · CALCULATING CONCENTRATIONS

Example 9.9

If 20 cm^3 of a standard solution of magnesium ions (Mg^2 ) of

concentration 100 ppm was diluted to 200 cm^3 , what would be

the concentration of the new solution?

Answer

First, work out the factor by which you have diluted the solution by dividing the

new volume by the old volume. The 100 ppm solution of Mg^2 has been diluted

from 20 cm^3 to 200 cm^3 , i.e. by a factor of

200/20 10

The new concentration is worked out by dividing the old concentration by the

dilution factor:

concentration of the new diluted solution 

100

10 ppm of Mg^2 

10

Example 9.10

A solution of NaCl has a concentration of 0.100 mol dm^3. What

is its concentration in ppm (mg dm^3 )?

Answer

1.00 mol of NaCl has a mass of (23 + 35.5) 58.5 g. Therefore a

0.100 mol dm^3 solution contains 5.85 g dm^3 or 5850 mg dm^3 of NaCl. This is

also 5850 ppm.

Comment

When the salt dissolves in water 1 mol of NaCl produces 1 mol of Naions. A

0.1 mol dm^3 solution contains 2.3 g dm^3 of Na, or 2300 ppm (mg dm^3 ) Na,

and 3.55 g dm^3 or 3550 ppm (mg dm^3 ) Cl.

Example 9.8

If a sample of drinking water contains aluminium ions at a

concentration of 3.0 ppm, how many mol dm^3 of aluminium

ions does this represent?

Answer

The sample contains 3.0 mg of Al^3 in 1 dm^3 , or

0.0030 g of Al^3 in 1 dm^3

or

0.0030

mol of Al^3 in 1 dm^3 1.1 10 ^4 mol dm^3 of Al^3 

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