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Question

20 mL of 0.1 M solution of compound Na2CO3NaHCO32H2O is titrated against 0.05 M HCl, x mL of HCl is used when phenolphthalein is used as an indicator and y mL of HCl is used when methyl orange is the indicator in two separate titrations. Hence (yx) is:

A
40 mL
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B
80 mL
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C
120 mL
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D
None of these
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Solution

The correct option is B 80 mL
When phenolphthalein is used as an indicator:
Only one part of base reacts
By using formula,
meq. of acid used=meq. base
V×M×n=V×M×n
Where,
M=MolarityV=Volumen=nfactor
0.05x×1=20×0.1×1;x=40 mL
When methyl orange is used as an indicator:
All parts of base reacts
By using formula,
meq. of acid used=meq. base
0.05y×1=(20×0.1×1)+(20×0.1×2)
y=120 mL
yx=80 mL

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