Question

$0.5g$ of a sample of limestone was dissolved in $30mL$ of $N/10$ $HCl$ and the solution diluted to $100mL$. $25mL$ of this solution required $12.5mL$ of $N/25$ $NaOH$ for complete neutralization. Calculate percentage of $Ca{CO}_3$ in the example.

Answer 1

Reactions involved:

$CaC{O}_3+2HCl\to CaC{l}_2+H_{2}C{O}_3$

$H_{2}C{O}_3+2NaOH\to N{a}_{2}C{O}_3+2H_{2}O$

To find the $\%$ of $CaC{O}_3$ present in the limestone sample

Let it be $a\%$

So, in a $0.5$g of sample, $CaC{O}_3$ present$=0.5\times a\%$

$CaC{O}_3=\frac{0.5a}{100}g$

Moles of $CaC{O}_3=\frac{g.wt}{m.wt}=\frac{0.5a}{100\times 100}$ (m.wt of $CaC{O}_3$ is $100$g/mol)

$n=0.5a\times {10}^{-4}$mol

$\therefore$ moles of $H_{2}C{O}_3=$mole of $CaC{O}_3$

So if $25$ml is taken out of $100$ml

$\therefore$ moles of $H_{2}C{O}_3=\frac{molesofCaC{O}_2}{4}=\frac{0.5a\times {10}^{-4}}{4}$

Also, no. of moles of NaOH used$=$vol. $\times$ normality

$=\frac{12.5ml}{1000ml/l}\times \frac{1}{25}$

$=\frac{12.5}{25\times {10}^3}$mol

Also, from the reaction

No. of moles of $H_{2}C{O}_3=\frac{1}{2}\times$(no. of moles of NaOH)

$\frac{0.5a\times {10}^{-4}}{4}=\frac{1}{2}\times \frac{12.5}{25\times {10}^3}$

$a=\frac{4\times {10}^5}{4\times {10}^3\times 5}$

$a=20\%$

$\therefore \%$ of $CaC{O}_3=20\%$.