Question

Titration question:

How do I calculate the mass of Na2CO3 and NaHCO3 in the initial (250mL) solution?

-The titrant is .09522 M HCl

-An unknown solution containing a Na2CO3 & NaHCO3 mixture is being titrated

-The unknown solution was made using 2.3051g of the mixture in 250mL of CO2-free water and a 25mL aliquot of this solution is what is being titrated.

-Bromocresol green indicator was used

-The first equivalence point occurred when 19.81mL titrant was added

-The second equivalence point was added when 42.84mL of titrant was added

Answer 1

When a mixture of Na₂CO₃ & NaHCO₃ is titrated against HCl the following reactions take place:

1st -

This reaction gives the first equivalence point.

2nd -

This reaction gives the second equivalence point.

The volume of HCl required to reach the 1st equivalence point is 19.81ml (0.09522M)

Therefore moles of HCl used = Molarity x Volume = 0.09522 x 19.81 = 1.8863 mmoles

Since one mole of HCl requires one mole Na₂CO₃, moles of Na₂CO₃ present in 25ml of aliquot = 1.8863 mmoles

Moles of Na₂CO₃ present in 250ml = 18.863 mmoles

Mass of Na₂CO₃ in 250ml = 18.863 x 106 = 1999.5 mg = 1.9995 gms

You can calculate mass of NaHCO₃ in the foloowing ways:

1. 2nd equivalence point occured at 42.84ml of HCl, hence volume of HCl added after 1st equivalence point = 42.84 - 19.81 = 23.03ml    

This volume of HCl is used to neutralize NaHCO₃ (from Na₂CO₃ and the one initially present)

Moles of HCl used = Molarity x Volume = 0.09522 x 23.03 = 2.1929 mmoles

Hence moles of NaHCO₃ present = 2.1929 moles

Of this 1.8863 mmoles is obtained from Na₂CO₃ during 1st reaction, therefore NaHCO₃ present in 25ml aliquot = 2.1929 - 1.8863 = 0.3066 mmoles

Hence NaHCO₃ in 250ml solution = 3.066 mmoles

Mass of NaHCO₃ = 3.066 x 84 = 257.54 mg = 0.2575 gms

2. Since mass of the mixture is 2.3051 gms and mass of Na₂CO₃ calculated is 1.9995 gms, remaining amount should be NaHCO₃

Mass of NaHCO₃ = 2.3051 - 1.9995 = 0.3056 gms

P.S:- The slight error in mass of NaHCO₃ from the two methods may be attributed to calculation error due to approximation.