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

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

70 ml

32 ml

35 ml

16 ml

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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 = 100 mlV1=60 ml HCl remains unreacted=100-60=40 ml40 ml 0.1 N HCl is now neutralised by KOH 0.25 N N1V1 = N2V2KOH0.1 × 40=0.25 × V2 ; V2=16 ml.

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th