Question

$1.0g$ carbonate of a metal was dissolved in $50mLN/2HCl$ solution. The resulting liquid required $25mL$ of $N/5NaOH$ solution to neutralise it completely. Calculate the equivalent mass of the metal carbonate.

Answer 1

For $25$ ml of $N/5NaOH,HCl$ is required is;

$N_1{N}_1=N_2{V}_2$

$\frac{1}{5}\ast 25=\frac{1}{2}\ast V_2$

$V_2=2\ast 5=10ml$

So volume of $HCl$ left $=50-10=40ml$

Concentration of $HCl=0.5N$

$0.04$ mol of $HCl$ will be used in neutralisation of carbonate

$M_{2}C{O}_3+2HCl\to 2MCl+H_{2}C{O}_3$

$1$ mol of $M_{2}C{O}_3$ is neutralised by $2$ moles of $HCl$.

If $0.04$ mol of $HCl$ is used, then $\frac{0.04}{2}=0.02$

Mol. of $M_{2}C{O}_3$ is neutralised.

Mass of $M_{2}C{O}_3=1$ g

Moles $=0.02$

Molar mass $=\frac{1}{0.02}=50g/mol$

Equivalent weight $=\frac{50}{2}=25$