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Question

A 100 mL solution of 0.1 N HCl was titrated with 0.2 N NaOH solution. The titration was discontinued after adding 30 ml of NaOH solution. The titration was completed by adding 0.25 N KOH solution. The volume of KOH required for completing the titration is:


A

70 ml

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B

32 ml

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C

35 ml

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D

16 ml

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Solution

The correct option is D

16 ml


In the complete neutralization of acid and base, the Milli-equivalent of acid and base must be equivalent.

Milli-equivalent of HCl = N1×V1=0.1×100=10

Milli-equivalent of NaOH = N2×V2=0.2×30=6

Let the titration be completed by adding a ‘V3’ volume of 0.25 N KOH solution.

N1×V1=N2×V2+N3×V3

0.1×100=0.2×30+0.25×V3

10=6+0.25×V3

4=0.25×V3

40.25=V3

16=V3

The titration is completed by adding a ‘16ml’ volume of 0.25 N KOH solution.

Hence, option (d) is correct.


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