Question

if 0.4N, 500ml HCL neutralise 300ml of KOH completely find N of KOH solution. N = Normality​

Answer 1

Answer:

2/3 N or 0.67 N

Explanation:

By the law of equivalence;

Equivalents of acid = Equivalents of base

Here HCl is acid and KOH is base so we equate their equivalents.

Eq. = N × V

Therefore we can equate those of hcl and koh and get the answer. Refer the pic attached.

Hope this solves your doubt!

Answer 2

Answer: N of KOH solution is 0.67

Explanation:

Given:N of HCL=0

         V of HCL=500ml

         V of KOH=300m

Since, the resulting solution is neutral

That means, NaVa=NbVb

                0.4×500=Nb×300

                 Nb=0.4×500÷300

                 Nb=4×500÷10×300

                 N=4÷6

                 Nb=0.66

According to significant rules,

             Nb=0.67