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 (b) 32 mL (c) 35 mL (d) 16 mL

In the neutralization of acid and base N x V of both must be equivalent

N x V of HCl = 0.1 x 100 = 10

N x V of NaOH = 0.2 x 30 = 6

To obtain

10N x V of base 4

4N x V of base is required

N x V of KOH = 0.25 x 16 = 4

\(N_{1}V_{1}= \underset{NaOH}{N\times V}+\underset{KOH}{N\times V}\)

0.1 x 100 = 0.2 x 30 + 0.25 x V

10 = 6 + 0.25V

0.25V = 10 – 6 = 4

V = 4/0.25

We get,

V = 16 ml

Hence, the correct option is (d)

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