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Q.

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 of KOH required for completing the titration is

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a

70 ml

b

32 ml

c

35 ml

d

16 ml

answer is D.

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Detailed Solution

Volume of HCl neutralised by NaOH (Caustic soda) = V1

 N1V1=N2V2;   0.1  ×  V1=0.2  ×30;  V1=60  ml(here 2 stands for NaOH)Vtotal  HCl=100mlV1=60  ml HCl remains unreacted=100-60=40  ml40  ml  0.1  N  HCl  isnow  neutralised  by  KOH  0.25  NN1V1=N2V2KOH0.1×40=0.25×V2;  V2=16  ml.

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