A solution containing Na_(2)CO_(3) and NaOH requires 300 mL of 0.1 NHCl using phenolphthalein as an indicator. Methyl orange is then added to above titrated solution when a further 25 mL of 0.2 M HCl is required. The amount of NaOH present in the original solution is

A solution containing Na_(2)CO_(3) and NaOH requires 300 mL of 0.1 NHC

1.	A solution containing Na_(2)CO_(3) and NaOH requires 300 mL of 0.1 NHCl using phenolphthalein as an indicator. Methyl orange is then added to above titrated solution when a further 25 mL of 0.2 M HCl is required. The amount of NaOH present in the original solution is
Answer» 0.5 G 1g 2g 4g Solution :300 ml of 0.1 N HCl will neutralize all the NAOH and HALF of `Na_(2)CO_(3)`(with phenolphthalein asindicator). The remaining half will be neutralized by 25 ml of 0.2 N HCl (with methyl orange as indicator). Thus, 0.2 N HCl required for half neutralization of `Na_(2)CO_(3) = 25 ml` or 0.1 N HCl required will be 50 ml. Hence, 0.1 N HCl USED up for NaOH = 300 - 50 = 250 ml . Gram equivalents present in 250 ml of 0.1 N HCl `=(0.1)/(1000)xx250=0.025`. It will neutralize 0.025 g eq. of NaOH, i.e., `0.025xx40g` = 1 g.