Question

1.25 g of impure Na2CO3 is dissolved in water and the
solution is made up to 250 mL. To 50 mL of this
solution, 50 mL of 0.1 N-HCl is added and the mixture
is shaken well. It further required 10 mL of 0.16 N
solution for complete neutralization. Calculate the
percentage purity of Na2CO3 sample.​

Answer 1

Given:- 12.5g of Impure Na2Co3

50 ml of 0.1 N HCl

10 ml of 0.16 N solution

To Find:- Percentage purity of Na2Co3

Solution:-

- The formula of percentage purity is

= mass of pure sample obtained/ mass of given sample × 100

- Milli equivalent of HCl added to the Na2Co3 solution

= 0.1 × 50

= 5

- According to the question,

Milli equivalent of HCl used to hydrolysis NaoH

= 0.16 × 10

= 1.6

- Hence, Milli equivalent of HCl used to neutralize 50 ml of Na2Co3 solution

= 5- 1.6

=3.5

- Milli equivalent of HCl used to neutralize 250 ml of Na2Co3 solution

= 3.4/50 × 250

= 17

- Here, Formula used to find number of equivalent is

No. of equivalent = weight of sample/ equivalent weight of sample

So, No. of equivalent

= molecular weight of Na2Co3/equivalent wt of Na2Co3

=17 equivalent /1000= wt/106/2

- Hence, The weight of Na2Co3 is

= wt = 0.901 g if Na2Co3

- Percentage purity

= 0.901/1 × 100

= 90.10 %

- So, The Answer is 90.10%