For titration of a basic solution of
Na2CO3 and
NaHCO3 against
HCl, if phenolphthalein is used as indicator, the end point is indicated only for half neutralization of
Na2CO3, i.e., (upto
NaHCO3 ).
Na2CO3+HCl→NaHCO3+NaCl The remaining solution then contains the unreacted
NaHCO3 from this reaction plus the unreacted
NaHCO3 originally in the solution. At the phenolphthalein end point, there is no reaction between
HCl and
NaHCO3. From the equations
Mol of
HCl consumed
=mol of
Na2CO3 20 mL of 0.1 M = 20 mL of 0.1 M
∴ The concentration of
Na2CO3 in solution
X=0.1M Note that for a quantity of
Na2CO3, exactly half volume of the HCl is used at the phenolphthalein end point and the second half volume of the
HCl is required for complete neutralization of
Na2CO3 at methyl orange end point.
NaHCO3+HCl→NaCl+CO2 H2O ∴ Volume of HCl required to neutralize
Na2CO3 in original sample
=2×20mL =40mL If methyl orange is used, the end point is indicated when all the alkali is neutralized.
NaHCO3+HCl→NaCl+CO2 H2O As
40mL of
0.1MHCl is consumed in complete neutralization of
Na2CO3 at methyl orange end point, so the volume of HCl used to neutralized
NaHCO3 from the original sample would be Remaining
HCl=60−40=20mL of
0.1M As per equation
=1mol of
NaHCO3=1mol of
HCl ∴0.1mol of
NaHCO3=0.1mol of
HCl,