A solution containing Na2CO, and NaOH requires 300 ml of 0.1 N HCl using phenolphthalein as an indicator. Methyl orange is then added to the above titrated solution when a further 25 ml of 0.2 N HCl is required. The amount of NaOH present in solution is (NaOH - 40,Na2CO3 = 106)

(I) Phenolphthalein indicate partial neutralization of Na2CO3→NaHCO3Meq. Na2CO3+Meq. of NaOH = Meq. of HClWE×1000+WE×1000=NV Suppose Na2CO3=ag,NaOH=bg ) a106×1000+b40×1000=300×0.1      . . . (1)(II) Methyl orange indicate complete neutralization of HCIN1V1=N2V2,25×0.2=0.1×V2 so V2=50ml excess ∴ a53×1000+b40×1000=350×0.1 . . . . (2)From Eq. (1) and Eq.(2), b = 1 g.