Question

A mixture of NaHCO₃ and Na₂CO₃ is neutralised by x ml of 1M HCl by using phenolphthalein indicator. The above mixture is neutralised by y ml of 1M HCl by using methyl orange indicator. The mole of CO₂ evolved on heating the above mixture at lower temperature is

Answer 1

In case of phenolphthalein indicator only Na₂ CO₃ will react with HCl and from NaHCO₃

Na₂ CO₃ + Hcl →NaHCO₃ + NaCl

it means, Number of moles of HCl is equal to the Number of moles of HCl 

= x ml x 1M

= x millimole

Therefore, Number of moles of Na₂ CO₃ in mixture is = x millimole

In case of methyl orange indicator both NaHCO₃ and Na₂CO₃ will react with HCl.

NaHCO₃ + Na₂CO₃ + 3HCl → 2H₂ CO₃ + 3 Na Cl

Number of millimoles of HCl are used to complete Neutralization of NaHCO₃ + Na₂CO₃ 

= y ml x 1M

= y millimoles. 

We know that, to complete Neutralization of one mole of Na₂CO₃ we have to require 2 mole of HCl.

Therefore, Number of millimoles of HCl that neutralized 

NaHCO₃ = (y-2x) millimole.

Hence, Number of millimoles at NaHCO₃ in mixture = (y-2x) millimole at lower temperature - Na₂CO₃ is stable but NaHCO3 convert into Na₂CO₃

2 NaHCO₃ → Na₂CO₃ + H₂O + CO₂

In the above equation, 2 moles of NaHCO₃ on heating gives 1 mole of CO₂

∴ (y-2x) millimole NaHCO₃ gives = \(\frac {y-2x}{2}\) millimole of CO₂

Hence, on Heating the mixture at lower temperature gives \(\frac {y-2x}{2}\) millimoles or (\(\frac {y-2x}{2000}\)) mole of CO₂