Question

0.252 g of acid needed 40 ml, 0.1 N NaOH for complete neutralisation. What is the equivalent weight of the acid?

Answer 1

Dear Student,

Given data:
(V) Volume of base in Liter = 40 ml = 0.04 L
(N) Normality of base (NaOH) (N)= 0.1 N
(W) Weight of acid (gm) = 0.252 gm

The computation for the equivalent weight of acid is expressed below,

$Eq.wtofacid=\frac{W}{V\times N}$
by plugging the given values we get,

$Eq.wtofacid=\frac{0.252}{0.04\times 0.1}=63g$

Hence the equivalent weight of acid = 63 gm

Regards.