Question

what is the molarity of naoh when 22.14 ml% 0.105 m oxalic acid is needed to neutralize 25.0 ml % of the base

Answer 1

Sol :-

From equation of neutralization :

Gram equivalents of oxalic acid = Gram equivalents of NaOH

Because, Normality = Gram equivalents/Volume in L

So,

Normailty of oxalic acid x Volume of oxalic acid = Normailty of NaOH x Volume of NaOH

Because, Normality = Molarity x valency factor

So,

Molarity of oxalic acid x Volume of oxalic acid x Valency facor of oxalic acid = Molarity of NaOH x Volume of NaOH x Valency facor of NaOH

0.105 M x 22.14 ml% x 2 = Molarity of NaOH x 25.0 ml% x 1

Molarity of NaOH = (0.105 M x 22.14 ml% x 2) / (25.0 ml% x 1 )

= 0.186 M

Hence, Molarity of NaOH = 0.186 M