Question

). In a titration, a student obtained an average titre value of 3.9 cm3 of 0.3 M HCl. If the volume of Na2CO3 solution used is 10 cm3 and the indicator used is phenolphthalein, calculate; (i). the molarity of the Na2CO3 (4 marks) (ii). the mass concentration of the Na2 theCO3. (4 marks)

Answer 1

Answer:

I) 0.0585 M

ii)6.2 g dm-3

Explanation:

The reaction equation is given as;

Na2CO3(aq) +2HCl(aq)------> 2NaCl(aq) + CO2(g) +H2O(l)

Concentration of acid CA= 0.3 M

Volume of acid VA= 3.9 cm^3

Concentration of base CB= the unknown

Volume of base VB= 10 cm^3

Number of moles of acid NA= 2

Number of moles of base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB=CBVBNA

CB= CAVANB/VBNA

substituting values;

CB= 0.3 × 3.9 × 1/ 10.0 × 2

CB= 0.0585 M

ii) mass concentration= molar concentration × molar mass

Molar mass of Na2CO3= 106 gmol-1

Mass concentration= 0.0585 × 106 = 6.2 g dm-3