Analytical Chemistry Question 279
Question: The volume of 0.6 M NaOH required to neutralise $ 30cm^{3} $ of 0.4 M HCl is
[Pb. CET 2001]
Options:
A) 40 $ cm^{3} $
B) 30 $ cm^{3} $
C) 20 $ cm^{3} $
D) 10 $ cm^{3} $
Show Answer
Answer:
Correct Answer: C
Solution:
Normality = molarity × basicity or acidity (for HCl) $ N_2=0.4\times 1=0.4N $ Basicity =1 (for NaOH acidity =1) $ N_1=0.6\times 1=0.6N $
$ V_1=-\ V_2=30\ cm^{3} $ From the equation, $ N_1V_1=N_2V_2 $
$ 0.6\times V_1=0.4\times 30 $
$ V_1=\frac{0.4\times 30}{0.6}=20\ cm^{3} $
BETA
