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  • A 100 ml solution of 0.1 N HCl was titrated with 0.2 N NaOH solution. The titration was discontinued aft

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 remaining titration was completed by adding 0.25 N KOH solution, the volume (in ml)  of KOH required for completing the titration is

Option: 1

16


Option: 2

32


Option: 3

35


Option: 4

70


Answers (1)

best_answer

For HCl, NaOH and KOH, normality = molarity

mmoles of HCl = 100 x 0.1N = 10

Thus, mmoles of base required for titration = 10

Now, mmoles of NaOH added = 30 x 0.2 = 6

Now, for complete titration, mmoles of KOH needed = 10 - 6

                                                                        = 4

Therefore, the volume of KOH required = 4/0.25

                                                        = 0.016L or 16ml

 

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vinayak

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