Volumetric calculations

The percentage composition can be used to determine the purity of a sample. This is done using volumetric analysis, where the concentration of the pure substance is calculated.

Example

3g of impure Na₂CO₃ are dissolved to make 250 cm³ of solution. 25 cm³ of this solution requires 21.0 cm³ of 0.105M HCl for complete neutralization.

Step i: Write the balanced equation for the reaction

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Step ii: Calculate the molarity of pure Na₂CO₃

• From the equation, $n_B = 1$ and $n_A = 2$.
• Use the molarity equation to find the molarity of Na₂CO₃:

$M_A V_A = M_B V_B$

$0.105 \times 21 = M_A \times 25$

$M_A = 0.0441 \text{ M}$

Step iii: Calculate the concentration of pure Na₂CO₃

$\text{Concentration} = M_A \times \text{molar mass} = 0.0441 \text{ M} \times 106 \text{ g/mol} = 4.67 \text{ g/dm}^3$

Step iv: Calculate % purity

$\text{Concentration of Na}_2\text{CO}_3 \text{ solution} = \frac{3.0 \text{ g}}{250 \text{ cm}^3} = 12 \text{ g/dm}^3$

$\text{Percentage purity} = \frac{4.67}{12} \times 100 = 38.9\%$

Step v: Calculate % impurity

$\text{Percentage impurity} = 100\% - 38.9\% = 61.1\%$

The dilution law describes how the concentration of a solution decreases when more solvent is added. It states that:

$M_1 V_1 = M_2 V_2$

Where:

• $M_1$ = initial concentration
• $M_2$ = final concentration
• $V_1$ = initial volume
• $V_2$ = final volume

Example

300 cm³ of 6.0M NaOH needs to be diluted to give a 0.40M solution.

Using the dilution law:

$M_1 V_1 = M_2 V_2$

$6.0 \text{ M} \times 300 \text{ cm}^3 = 0.40 \text{ M} \times V_2$

$V_2 = 4500 \text{ cm}^3$

Water of crystallization is the water trapped within the crystals of a substance. This can be determined using titration reactions.

Example 1

Determining the number of molecules of water in Na₂CO₃·XH₂O:

The relative molecular mass of Na₂CO₃·XH₂O is 286 g/mol.

The equation for Na₂CO₃·XH₂O is:

$2(23) + 12 + 3(16) + X(18) = 286$

$106 + X(18) = 286$

$X = 10$

Therefore, 10 molecules of water are bound in the crystal.

Example 2

Preparing a HBr solution:

Purity = 48%, density = 1.5 g/cm³, volume = 500 mL, and molarity = 0.6 M.

Calculate the mass required to prepare the solution:

$n = M \times V = 0.6 \text{ mol/L} \times 0.5 \text{ L} = 0.3 \text{ mol}$

$\text{Mass required} = 0.3 \text{ mol} \times 81 \text{ g/mol} = 24.3 \text{ g}$

To account for purity:

$\text{Mass of HBr required} = \frac{24.3 \text{ g}}{0.48} = 50.6 \text{ g}$

Volume of concentrated solution needed:

$\text{Volume required} = \frac{50.6 \text{ g}}{1.5 \text{ g/cm}^3} = 33.7 \text{ mL}$

Example 3

1.5g of an acid Hx was dissolved in water and neutralized by 0.115M KOH.

Use the formula:

$V_A \times M_A = V_B \times M_B$

From the equation, $n_A = n_B$ (1:1 ratio between acid and base).

Calculate the molarity of the acid (Hx):

$\text{Molarity of Hx} = 0.095 \text{ M}$

Molar mass of Hx: 63 g/mol.

The radical X is most likely HNO₃ or HCl, based on the molecular mass.