Quantitative Analysis IV
Percentage Purity

The percentage purity of a sample can be calculated from an impure sample of a titrant. The percentage purity is given as:

$$ \text{% purity} = \frac{\text{mass conc of pure sample in g/dm³}}{\text{mass conc of impure sample in g/dm³}} × 100 \tag{1}$$

Scroll right to view formula $$ \text{% purity }= \frac{MC_p}{MC_i} × 100 $$ $$\text{% impurity} =\text{ 100 - % purity }$$

Example 1: 25cm³ of a solution containing 1.0g impure sodium hydroxide pellets per 250cm³ required 20.0cm³ of 0.050 mol/dm³ tetraoxosulphate(VI) acid for complete neutralization. Determine the percentage purity of the sodium hydroxide pellets. (Na = 23, O = 16, H = 1, S = 32)

Solution

$$\text{First, calculate the conc of pure NaoH} $$ $$\text{The equation of reaction is} $$
$${H_2SO_4} + {2NaOH} ——> {Na_2SO_4} + {2H_2O} $$
$$ {C_A} = 0.050 mol/dm³ $$ $${V_A} = 20cm³ $$ $${V_B} = 25cm³ $$ $${C_B} = ? $$ $${a:b} = 1:2 $$ $$\frac{0.05 × 20}{C_B × 25} = \frac{1}{2} $$ $${C_B} = \frac{2 × 0.05 × 20}{25} $$ $$\text{Conc of pure NaOH} = 0.08 mol/dm³ $$ $$\text{Get the mass conc. of pure NaOH} $$ $$ \text{Mass conc.} = \text{Molar conc × M.m} $$ $$ \text{Molar mass of NaOH} = 23 + 16 + 1 $$ $$ = 40 g/mol $$ $$\text{Mass conc. of pure NaOH} = 0.08 × 40 $$ $$ = 3.2 g/dm³ $$ $$\text{Get mass conc of impure in g/dm³} $$ $$ \text{If 1g of NaOH is in 250cm³} $$ $$\text{xg would be in 1000cm³} $$ $$\text{mass conc of impure} = \frac{1000 × 1}{250} $$ $$ = 4g/dm³ $$ $$\text{ % purity }= \frac{MC_p}{MC_i} × 100 $$ $$ \text{% purity }= \frac{3.2}{4} × 100 $$ $$ \text{% purity }= 80\text{%}$$

Example 2: 25cm³ of a solution containing 6.50g/dm³ of impure KOH required 21cm³ of 5.0g/dm³ solution of HCl for complete neutralization. Calculate the :

  1. Concentration of A in mol/dm³
  2. Concentration of B in mol/dm³
  3. Percentage purity of KOH in B
    4. (H = 1, Cl = 35.5, KOH = 56.0g/mol)

Solution

  1. Concentration of A in mol/dm³
$$\text{Equation of reaction } $$ $${HCl} + {NaOH} ——> {NaCl} + {H_2O} $$ $$ \text{Molar conc. of A} = \frac{\text{mass Conc. of A}}{\text{M.m of A}} $$ $$ = \frac{5}{1 + 35.5} $$ $$ = 0.137 mol/dm³ $$
  1. Concentration of B in mol/dm³
$$\frac{C_AV_A}{C_BV_B} = \frac{a}{b} $$ $$ \frac{0.137 × 21}{C_B × 25} = \frac{1}{1} $$ $${C_B} = \frac{0.137 × 21}{25} $$ $$ {C_B} = 0.115 mol/dm³ $$
  1. Percentage purity of KOH in B
$$ \text{mass conc. of pure KOH} = 0.115 × 56 $$ $$ = 6.44g/dm³ $$ $$\text{% purity} = \frac{MC_p}{MC_i} × 100 $$ $$ =\frac{6.44}{6.50} × 100 $$ $$ \text{% purity of KOH in B} = 99.07\text{%} $$

Summary