Solution " X " contains Na2CO3 and NaHCO3. 20mL of X when titrated using methyl orange indicator consumed 60mL of 0.1M HCl solution. In another experiment, 20mL of X solution when titrated using phenolphthalein consumed 20mL of 0.1M HCl solution. The concentrations (in molL−1 ) of Na2CO3 and NaHCO3 in X are respectively
For titration of a basic solution of Na2CO3 andNaHCO3 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,