Quantitative Analysis III
Dilution

Dilution can be defined as the process of reducing the concentration of a solute in a solution by dissolving it in more solvent e.g water. Before titration, the concentration of the stock solution is diluted. The equation for dilution is given by: $${C_1}{V_1} = {C_2}{V_2} $$ $$ where $$ $${C_1} = \text{Initial concentration} $$ $$ {C_2} = \text{Final concentration} $$ $$ {V_1} = \text{Initial volume of solution} $$ $${V_2} = \text{final volume of solution} $$ $${V_2} - {V_1} = \text{volume of water added }$$ $$\text{Dilution factor} = \frac{V_2}{V_1} $$

Example 1: 5cm³ of 1.0 mol/dm³ HCl solution is diluted to 1dm³. What is the concentration of the diluted solution?

Solution

$$ {V_1} = 5cm³ = 0.005dm³ $$ $$ {C_1} = 1mol/dm³ $$ $$ {V_2} = 1dm³ $$ $$ {C_2} = ? $$ $$ {C_1}{V_1} = {C_2}{V_2} $$ $$ {C_2} = \frac{1 × 0.005}{1} $$ $$ {C_2} = 0.005mol/dm³ $$

Example 2: How much water will be added to 100cm³ of a solution containing 5.3g of Na2CO3 per 250cm³ so as to prepare 0.05mol/dm³ solution?

Solution

$$ \text{To get }{C_1} $$ $$ \text{If 5.3g of }{Na_2CO_3 } ——> 250cm³ $$ $$\text{xg } ——> 1000cm³ $$ $$ \text{Conc in g/dm³} = \frac{5.3 × 1000}{250} $$ $$ = 21.2g/dm³ $$ $$ \text{Conc. in mol/dm³} = \frac{21.2}{106} $$ $$ {C_1} = 0.2 mol/dm³ $$ $$ {C_2} = 0.05 mol/dm³ $$ $${V_1} = 100cm³ $$ $${V_2} = ? $$ $${V_2} = \frac{0.2 × 100}{0.05} $$ $$ {V_2} = 400cm³ $$ $$\text{Volume of water added} = 400-100 $$ $$ = 300cm³ $$

Example 3 : Calculate the volume of distilled water that should be added to 20cm³ of 0.200 mol/dm³ NaOH solution in order to dilute it ten times.

Solution

$$\text{Dilution factor(DF) } = 10 $$ $${V_2} = ? $$ $${V_1} = 20cm³ $$ $$ Df = \frac{V_2}{V_1} $$ $$ 10 = \frac{V_2}{20} $$ $$ {V_2} = 20 × 10 = 200cm³ $$ $$\text{Vol. of water added} = 200 - 20 $$ $$ = 180cm³ $$
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