Question

A sample containing 25.14 g of neutral salts, glucose and a sodium carbonate/bicarbonate

buffer was dissolved in 100 ml of water. A 25 ml aliquot of the resultant solution required

20.35 ml of 0.0987 M HCl when titrated to the PP end-point. A second 25 ml aliquot was

titrated to the MO end-point and required 56.75 ml of the acid. Calculate the percentage of

Na2CO3 (molecular weight 106) and NaHCO3 (molecular weight 84) in the sample

Answer 1

In strongly alkaline media, pH> 11 PP (reversibly) discolors

But titration end-ppint for 8.3 pH

Titration with phenolphthalein leads to titration of only half of the carbonate according to the equation.

Na₂CO₃+HCl=>NaCl+NaHCO₃ (1)

It's half. 20.35ml.

40.7ml - titration (1) for end.

0.004017 moles (1)

OH^- + H⁺ = H₂O(2)

Methyl Orange titration for 3.8 pH point end NaHCO₃+HCl=>H₂O+CO₂+NaCl (1)+(3)

[H+]=(2)=10^-8.3-(-3.8)=10^-4.5 Moles/litres

0.0003204 volume HCl(2)=0.3204 ml

56.75-40.7-0.3204-=15.73ml

0.0015525moles HCl(3)=NaHCO3

n(HCO3):n(Na₂CO3)=0.3864:1

In aliquote.

In solution *4

In sample:

Weight NaHCO3=0.52164 g

Na2CO3=1.7032 g

0.3063:1