Analytical Chemistry Question 239
Question: To neutralize 25 ml of 0.25 M $ Na_2CO_3 $ solution how much volume of 0.5 M $ HCl $ is required
[MP PET 1994]
Options:
A) 12.5 ml
B) 25 ml
C) 37.5 ml
D) 50 ml
Show Answer
Answer:
Correct Answer: A
Solution:
$ M_1V_1 $ = $ M_2V_2 $
$ (Na_2CO_3) $ = $ (HCl) $
$ 0.25M\times 25 $ = $ 0.5M\times V_2 $
$ V_2=\frac{0.25M\times 25}{0.5M}=12.5ml $
BETA
