Determination of Percentage Purity

When an impure chemical substance is used in titration, only the pure part of that substance takes part in the chemical reaction and is consumed. The impurity will not react. 

Example 1

The equation for the reaction is: 

2HNO_3(aq) + Na₂CO_3(aq) → 2NaNO_3(aq) + H₂O_(l) + CO_(g)

Results and Calculations

Indicator used – methyl orange

Volume of pipette – 25 cm³

Burette reading (cm3)	Rough	1st	2nd
Final burette reading (cm3)	33.70	33.40	33.50
Initial burette reading (cm3)	0.00	0.00	0.00
Volume of acid used (cm3)	33.70	33.40	33.50

The average titre value = \( \frac{3.30 \; + \; 33.50}{2} \scriptsize cm^3 \\ = \scriptsize 33.45cm^3 \)

(i). The mole of acid used = \( \frac{C \;\times \; V}{100} \\ = \frac{0.100 \;\times \; 33.45}{100} \\ = \scriptsize 0.00335 mol\)

From the equation of the reaction;

2 mol of HNO₃ requires 1mol of Na₂CO₃

Hence, 0.00335 mol of HNO₃ will require

\( \frac{1\;\times \; 0.00335}{2} \)

= 0.00168 mol of Na₂CO₃

25 cm³ of B contains 0.00168 mol of Na₂CO₃

1000 cm³ of B will contain

\( \frac{0.00168 \;\times \; 1000}{25} \)

 = 0.0672 mol dm^-3

Alternatively,

\( \frac{C_A V_A}{C_B V_B} = \frac{n_A}{n_B}\)

Substituting the values, we get:

\( \frac{0.1 \; \times \; 33.45}{C_B \; \times \; 25} = \frac{2}{1}\\ \scriptsize C_B = \normalsize \frac{0.1 \; \times \; 33.45 \; \times \; 1}{25 \; \times \; 2} \\ = \scriptsize 0.00672moldm^{-3}\)

(ii). The molar concentration of Na₂CO₃ in the solution of Bis 0.0672 moldm^-3

Molar mass of Na₂CO₃ = 106gmol^-1

∴Mass concentration of Na₂CO₃ = 0.0672 moldm^-3 x 106 gmol^-1 = 7.12g dm^-3.

(iii). 500 cm³ of B contains 5.00g of the mixture

1000 cm³ will contain

\( \frac{1000 \; \times \; 5}{500}\\ \scriptsize = 10.0gdm^{-3} \)

Therefore %purity =

\( \frac{concentration \; of \; pure}{concentration \; of \; impure} \; \times \; \frac{100}{1}\)

= \( \frac{7.12}{10} \; \times \; \frac{100}{1} \\ = \scriptsize 71.2 \%\)

Now, %impurity = 100 – %purity

= 100 – 71.2 = 28.8%