Some Basic Concepts Of Chemistry Question 281
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 remaining titration was completed by adding 0.25 N KOH solution. The volume of KOH required for completing the titration is [DCE 1999]
Options:
A) 70 ml
B) 32 ml
C) 35 ml
D) 16 ml
Show Answer
Answer:
Correct Answer: D
Solution:
Volume m of HCl neutralised by NaOH = (Caustic soda) = $ V _1 $
$ N _1V _1=N _2V _2 $ ; $ 0.1\times V _1=0.2\times 30 $ ; $ V _1=60ml $
V total (HCl) = 100ml $ V _1 $ = 60ml _________________ 40ml 40ml 0.1N HCl is now neutralised by KOH (0.25N) $ \to $
(HCl) $ N _1V _1=N _2V _2 $ (KOH) $ 0.1\times 40=0.25\times V _2 $ ; $ V _2=16ml $ .
