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Question

A solution contains a mixture of Na2CO3 and NaOH. Using phenolphthalein as indicator 35 mL of mixture required 20 mL of 0.80 N HCl for the end point. With methyl orange, 35 mL of solution required 37.5 mL of the same HCl for the end point. What is the weight of Na2CO3 and NaOH respectively in the mixture?

A
1.484 g, 0.08 g
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B
1.5 g, 0.5 g
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C
0.08 g, 1.484 g
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D
0.5 g, 1.5 g
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Solution

The correct option is A 1.484 g, 0.08 g
Suppose milliequivalents of Na2CO3 and NaOH are x and y respectively.

When phenolphthalein is used as an indicator, only 50% of Na2CO3 will be neutralised.
12Milliequivalents of Na2CO3+Milliequivalents of NaOH= Milliequivalents of HCl
x2+y=20×0.80
x2+y=16.....(1)

When methyl orange is used as an indicator, 100% of Na2CO3 will be neutralised.
Milliequivalents of Na2CO3+ Milliequivalents of NaOH= Milliequivalents of HCl
x+y=37.5×0.80
x+y=30.....(2)

From equation (1) and (2),
x=28, y=2

Milliequivalents of Na2CO3=28.0
WNa2CO3MNa2CO3×2×1000=28.0
WNa2CO3=28.0×1062×1000=1.484 g

Milliequivalents of NaOH=2.0
WNaOHMNaOH×1×1000=2.0
WNaOH=2.0×401×1000=0.08 g

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