    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 mlIn the complete neutralization of acid and base, the Milli-equivalent of acid and base must be equivalent.Milli-equivalent of HCl = ${\mathrm{N}}_{1}×{\mathrm{V}}_{1}=0.1×100=10$Milli-equivalent of NaOH = ${\mathrm{N}}_{2}×{\mathrm{V}}_{2}=0.2×30=6$Let the titration be completed by adding a ‘V3’ volume of 0.25 N KOH solution.${\mathrm{N}}_{1}×{\mathrm{V}}_{1}={\mathrm{N}}_{2}×{\mathrm{V}}_{2}+{\mathrm{N}}_{3}×{\mathrm{V}}_{3}$$0.1×100=0.2×30+0.25×{\mathrm{V}}_{3}$$10=6+0.25×{\mathrm{V}}_{3}$$4=0.25×{\mathrm{V}}_{3}$$\frac{4}{0.25}={\mathrm{V}}_{3}$$16={\mathrm{V}}_{3}$The titration is completed by adding a ‘16ml’ volume of 0.25 N KOH solution.Hence, option (d) is correct.  Suggest Corrections  1      Similar questions  Explore more