A solution containing Na2CO3 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 HCI is required. The amount of NaOH present in solution is (NaOH = 40, Na2CO3 = 10)

Question

A solution containing Na2CO3 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 HCI is required. The amount of NaOH present in solution is (NaOH = 40, Na2CO3 = 10) (A) 0.6 g (B) 1.0 g (C) 1.5 g (D) 2.0 g

Tardigrade Admin · Accepted Answer

(I) Phenolphthalein indicate partial neutralisation of Na2O3 → NaHCO3 Meq. of Na2O3+ Meq. of NaOH = Meq. of HCl (W/E)×1000+(W/E)×1000=NV (Suppose Na2O3 = a g, NaOH = b g) (a/106)×1000+(b/40)×1000=300×0.1 .......(1) (II) Methyl orange indicate complete neutralisation HCl HCl N1V1 = N1V1, 25 × 0.2 = 0.1 × V2 so V2 = 50 ml excess ∴ (a/50)×1000+(b/40)×1000=350×0.1 .......(2) From (1) and (2) b=1 gm.

$0.6 g$

$1.0 g$

$1.5 g$

$2.0 g$

Q. A solution containing Na2​CO3​ and NaOH requires 300ml of 0.1NHCl using phenolphthalein as an indicator. Methyl orange is then added to the above titrated solution when a further 25ml of 0.2NHCI is required. The amount of NaOH present in solution is (NaOH=40,Na2​CO3​=10)

0.6g

1.0g

1.5g

2.0g

$0.6 g$

$1.0 g$

$1.5 g$

$2.0 g$