Question

When 10g mixture of Na₂CO₃ and NaOH was titrated against HCl by using phenolphthalein followed by adding methyl orange after first equivalence point,

Answer 1

Here given reaction is Na₂CO₃(aq) + HCl(aq) → NaHCO₃(aq) + NaCl(aq)

Equivalent of  HCl in step II = (10 -x)/(106) x (84)/(84) is because of 1 mole of NaHCO₃ is formed and molecular mass of NaHCO₃ is 84. So for II step which is done by using methyl orange we get 1 mole of NaHCO₃