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# 0.2 gm of solution of mixture of NaOH and Na2CO3 and inert impurities was first titrated with phenolphthalein and N/10 HCl 17.5ml of HCl was required at the endpoint. After this methyl orange was added and 2.5ml of same HCl was again required for next endpoint. Find out percentage of NaOH and Na2CO3 in the mixture.

A
% NaOH=13.5 , % Na2CO3=35
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B
% NaOH=30 , % Na2CO3=25
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C
% NaOH=18 , % Na2CO3=40
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D
% NaOH=35 , % Na2CO3=13.5
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Solution

## The correct option is D % NaOH=35 , % Na2CO3=13.5 1. In the first titration when we use phenopthalein, NaOH will be neutralized by HClEquivalent of HCl= Equivalent of NaOH→ Equivalent of NaOH=0.1×17.51000=1.75×10−3Weight of NaOH= Equivalent × Equivalent Weight =1.75×10−3×40=0.07 gm%NaOH=0.070.2×100=35%Equivalent of HCl= Equivalent of Na2CO3→ Equivalent of Na2CO3=2.5×0.11000=0.25×10−3Weight of Na2CO3 in the sample = Equivalent × Equivalent Weight =0.251000×106=0.0265 gm%Na2CO3=0.02652×100=13.5%  Suggest Corrections  1      Similar questions  Related Videos   Counting Principle
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