I found this in a textbook; it seems that the question is poorly worded. It goes as follows:

$1.5~\mathrm{g}$ of chalk were treated with $10~\mathrm{ml}$ of $4~\mathrm{N}$ $\ce{HCl}$. The chalk was dissolved and the solution made to $100~\mathrm{ml}$. $25~\mathrm{ml}$ of this solution required $18.75~\mathrm{ml}$ of $0.2~\mathrm{N}$ $\ce{NaOH}$ solution for complete neutralization. Calculate the percentage of pure $\ce{CaCO3}$ in the sample of chalk.

I don't get it when the question says the chalk was dissolved and the solution made to $100~\mathrm{ml}$. Does this refer to the chalk which already reacted with $\ce{HCl}$, or was another $1.5~\mathrm{g}$ sample of $\ce{CaCO3}$ taken?


The question is a bit poorly written. But it also gives a sense to the procedure of dissolution of chalk upon treatment with $\ce{HCl}$ and dilution of the solution.

The reaction that takes place after the treatment with $\ce{HCl}$ is;

$\ce{CaCO3}$+ 2$\ce{HCl}$ → $\ce{CaCl2}$ + $\ce{H2CO3}$

$\ce{H2CO3}$ decomposes into $\ce{H2O}$ and $\ce{CO2}$. $\ce{CO2}$ is a gas. It bubbles up out of the solution.

$\ce{CaCl2}$ is soluble in water. So it should be dissolved by adding water.Further, it should then be diluted to the required volume.

"The chalk was dissolved" might signify the addition of water to dissolve the water soluble $\ce{CaCl2}$ produced during chemical reation with $\ce{HCl}$.

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  • $\begingroup$ Water is already present inside the HCl solution. $\endgroup$ – orthocresol Jan 10 '16 at 9:38
  • $\begingroup$ Obviously it will be present in HCl solution. But it might be needed in case of conc ones. $\endgroup$ – CCR Jan 10 '16 at 10:26

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