0 . 4 g mixture of NaOH , Na 2 CO 3 and some inert impurities was first titrated with N 10 HCl using phenolphthalein as an indicator, 17 . 5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1 . 5 mL of same HCl was required for the next end point. The weight percentage of Na 2 CO 3 in the mixture is (Rounded-off to the nearest integer)

Question

0 . 4 g mixture of NaOH , Na 2 CO 3 and some inert impurities was first titrated with N 10 HCl using phenolphthalein as an indicator, 17 . 5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1 . 5 mL of same HCl was required for the next end point. The weight percentage of Na 2 CO 3 in the mixture is (Rounded-off to the nearest integer),

MARKS App · Accepted Answer

Upto first end point gm equi. of NaOH + Na 2 CO 3 = HCl x + y × 1 = 1 10 × 17 . 5 x + y = 1 . 75 . . . . 1 Upto second end point NaOH + Na 2 CO 3 ≡ HCl x + y × 2 = 1 10 × 19 x + 2 y = 1 . 9 . . . . 2 y = 0 . 15 % Na 2 CO 3 = 0 . 15 × 10 - 3 × 106 0 . 4 × 100 = 3 . 975 % = 4 % Hence answer is 4

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