Question

A solution contains both NaHCO3 and Na2CO3. Titration of a 50.00mL portion to a phenolphtalein end point requires 20.19 mL of 0.1029 M HCl. A second 50.00mL aliquot requires 47.28 mL of the HCl solution when titrated to a bromocresol green end point. Calculate the molar concentration of NaHCO3 in the solution.

Answer 1

For the first end-point,
mol of HCl required for titration = 0.1029 mol/L*(20.19/1000)L = 2.0775*10^-3mol
it is therefore clear that 2.0775*10^-3mol of HCl gives 2.0775*10^-3mol of Na₂CO₃
For the second end point,
mol of HCl required for titration = 0.1029mol/L *(47.28/1000)L = 4.8651*10^-3mol
4.8651*10^-3mol of HCl will react with  4.8651*10^-3mol of NaHCO₃
but you should know that 1 mol of Na₂CO₃ will be converted into 1 mol of NaHCO₃
mol of NaHCO₃ in sample = ( 4.8651*10^-3) - (2.0775*10^-3) = 2.7876*10^-3 mol
Molar concentration of NaHCO₃ in the solution = (2.7876*10^-3 mol) / (50/1000)L
Molar concentration of NaHCO₃ in the solution = 0.05575 mol/L or M