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Question

25mL of a mixed solution of Na2CO3 and NaHCO3 required 12mL of N/20 HCl when titrated using phenolphthalein as an indicator. But 25mL of the same, when titrated separately with methyl orange required 30mL of N/20 HCl. Calculate the amount of Na2CO3 and NaHCO3 in grams per litre.

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Solution

In the case of phenolphthalein as indicator only Na2CO3 reacts with HCl to give NaHCO3

Na2CO3+HClNaHCO3+NaCl
25ml 12ml and 120N

mmole of Na2CO3 in 25ml= mmole of HCl=12×120=0.6

wt106×1000=0.6

Wt of Na2CO3 in 25ml=0.6×1061000=0.0636g

Wt of Na2CO3 in 1 lit =0.063625×1000=2.544g

In the presence of methyl orange as indicator:

Na2CO3+2HCl2NaCl+H2O+CO2

NaHCO3+HClNaCl+H2O+CO2

meq of Na2CO3+ meq. of NaHCO3= meq of HCl

2×mmole of Na2CO3+mmole of NaHCO3=mmole of HCl

2×0.6+ mmole of NaHCO3=30×120×1

mmole of NaHCO3=1.51.2=0.3

wt.ofNaHCO384×1000=0.3

wt of NaHCO3 in 25ml=0.3×841000=0.0252

wt of NaHCO3 in 1 lit =0.025225×1000=1.008g

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