Question

0.968 g of an acid-are present in 300 ml of a solution. 10 ml of this solution required exactly 20 ml of 0.05 N KOH solution. calculate no. of neutralisable protons and equivalent weight of the acid (Given mol. wt. of acid is 98)

Answer 1

Let the equivalent weight be $W$

Normality of Acid $=\frac{0.968}{W\times 0.3}$ $\left(\frac{Equivalent}{V\left(Litres\right)}\right)$

Milli Equivalent of acid $=$ Milli Equivalent of $KOH$

$⟹10\times \frac{0.968}{W\times 0.3}=20\times 0.05$

$W=32.2667$

$Equivalentweight=\frac{Molecularweight}{No.ofNeutralizableprotons\left(y\right)}$

$⟹\frac{98}{y}=32.2667$

$⟹y=3$