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Question

A mixed solution of potassium hydroxide and sodium carbonate required 15 mL of N20HCl solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as an indicator required 25 ml of the same acid. The amount of KOH present in the solution is :

A
0.014 g
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B
0.14 g
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C
0.028 g
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D
1.4 g
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Solution

The correct option is A 0.014 g
Titration of sodium carbonate with HCl by using methyl orange as the indicator requires twice the volume as required by titration with phenolphthalein indicator.
Thus, out of 15 ml, the volume required for titration with sodium carbonate is 2515=10 ml.
The volume required for titration with KOH is 1510=5 ml.
The number of millimoles of KOH present in the solution is 5×120=0.25 millimoles.
The amount of KOH present in the solution is 0.25×56.11000=0.014 g.

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