0.4 g mixture of NaOH, and some inert impurities was first titrated with HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added a

Question

0.4 g mixture of NaOH,  and some inert impurities was first titrated with  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  in the mixture is (Rounded-off to the nearest integer)

Canvas Classes · Accepted Answer

Step 1 — Understand the two-indicator method:

With phenolphthalein (end point ~pH 8.3):
- NaOH is completely neutralised: 
-  is half-neutralised:

With methyl orange (end point ~pH 4):
-  is neutralised:

Step 2 — meq calculations:

HCl normality = N/10 = 0.1 N

meq with phenolphthalein =  meq
meq with methyl orange =  meq

Step 3 — Moles of :

The methyl orange step neutralises  formed from :
meq of  = 0.15 meq → moles of  = 0.15/1000 =  mol

(Since 1 mol  gives 1 mol  in first step, and 1

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