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Question

A mixed solution of potassium hydroxide and sodium carbonate required 15 mL of an N20 HCl solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as 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
0.25 g
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Solution

The correct option is A 0.014 g
KOH(x mmol)+Na2CO3(y mmol)
i. (x×1)+(y×2)×12=120×15

ii. (x×1)+(y×2)=120×25y=0.5 and x=0.25

KOHx mmoles =0.25×103×56 g=0.014 g

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