Question

$1.725g$ of a metal carbonate is mixed with $300mL$ of $\frac{N}{10}HCl.10mL$ of $\frac{N}{2}$ sodium hydroxide were required to neutralise excess of the acid. Calculate the equivalent mass of the metal carbonate.

Answer 1

$10mL$ of $\frac{N}{2}NaOH$ solution
$=10mL$ of $\frac{N}{2}HCl$ solution
$\equiv 50mL$ of $\frac{N}{10}HCl$ solution
Volume of $\frac{N}{10}HCl$ used for neutralisation $=300-50=250mL$
$250mL$ of $\frac{N}{10}HCl\equiv 20mL$ of $\frac{N}{10}$ metal carbonate solution
Let the equivalent mass of metal carbonate be $E$.
Mass of metal carbonate present in solution
$=\frac{N\times E\times V}{1000}=1.725$
$=\frac{1\times E\times 250}{10\times 1000}=1.725$
$=\frac{E}{40}=1.725$
$E=40\times 1.725=69$.