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Question

100 mL solution of 0.1 M HCl was titrated with 0.2 NaOH solution. The titration was discontinued after the addition of 30 mL of NaOH. The remaining titration was completed with the addition of 0.5 M KOH. The volume of KOH required for completing the titration is:

A
8mL
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B
16mL
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C
32mL
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D
64mL
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Solution

The correct option is B 16mL
For HCl,KOH & NaOH normality = molarity
Moles of HCl=0.1L×0.1N=0.01
Moles of OH needed =0.01
Moles of NaOH added =0.03×0.2=0.006
Moles of KOH needed =0.010.006=0.004
V=0.004/0.25=0.016L=16ml

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