Analytical Chemistry Question 79
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
[MP PMT 1997]
Options:
A) 16 ml
B) 32 ml
C) 35 ml
D) 70 ml
Show Answer
Answer:
Correct Answer: A
Solution:
In the neutralization of acid and base N × V of both must be equivalent N × V of HCl = 0.1 × 100 = 10 N × V of NaOH = 0.2 × 30 = 6 as to obtain 10 N × V of base 4 N × V of base is required N × V of KOH = 0.25 × 16 = 4 N1V1 = $ \underset{NaOH}{\mathop{N\times V\ }}+\ \underset{KOH}{\mathop{N\times V}} $ 0.1 × 100 = 0.2 × 30 + 0.25 × V 10 = 6 + 0.25 V $ V=\frac{400}{0.25} $
Þ V = 16 ml
BETA
