# Application of limiting reagents and acid-base stoichiometry

I'm having a bit of trouble with the latter questions of this problem.

$$\pu{15.0 mL}$$ of $$\pu{1.4 M}$$ $$\ce{HCl}$$ was mixed with $$\pu{1.00 g}$$ of $$\ce{CaCO3}$$ until all the solid had dissolved. The solution was then transferred to a conical flask and made up to $$\pu{200 mL}$$ with water. A $$\pu{20.0 mL}$$ portion of this solution was then neutralized by $$\pu{8.50 mL}$$ of a $$\pu{0.100 M}$$ $$\ce{NaOH}$$ solution.
Calculate:

1. Amount of substance in excess of $$\ce{HCl}$$ in the $$\pu{20.0 mL}$$ portion
2. Amount of substance in excess of $$\ce{HCl}$$ in the $$\pu{200.0 mL}$$ portion
3. Amount of substance in excess of $$\ce{HCl}$$ which reacted with $$\ce{CaCO3}$$

My approach

$$\ce{2HCl(aq) + CaCO3(s) -> CaCl2(aq) + H2O(l) + CO2(g)}$$

Finding the amount of substance of each reactant and hence the limiting reactant,

For $$\ce{HCl}$$: \begin{align} M[\ce{HCl}] &= \frac{\mathrm{mol}}{\mathrm{L}}\\ \mathrm{mol}[\ce{HCl}] &= \mathrm{M}\times \mathrm{L}\\ \therefore \mathrm{mol}[\ce{HCl}] &= 1.4\times (15.0\times 10^{-3}) = 0.021 \end{align}

For $$\ce{CaCO_3}$$: \begin{align} \mathrm{mol}[\ce{CaCO_3}] &= \frac{\mathrm{m}}{\mathrm{mm}}\\ \mathrm{mol}[\ce{CaCO_3}] &= \frac{1.00}{100.0869 }\\ \therefore \mathrm{mol}[\ce{CaCO_3}] &= 9.9913\times10^{-3} \end{align}

Therefore (after dividing by stoichiometric coefficients) $$\ce{CaCO_3}$$ is limiting.

From here I'm not sure if what I did next is correct. I found the amount of substance of $$\ce{CaCl_2}$$ produced the remaining $$\ce{HCl}$$ in excess. Which were,

\begin{align} \mathrm{mol}[\ce{CaCl_2}] &= 9.9913\times10^{-3} \text{ and,}\\ \mathrm{mol}[\ce{HCl}] &= 1.1009\times10^{-2} \end{align}

(Meaning that we have $$9.9913\times10^{-3}\mathrm{mol}$$ of $$\ce{CaCl_2}$$ and $$1.1009\times10^{-2}\mathrm{mol}$$ of $$\ce{HCl}$$ in in $$\pu{15 mL}$$ of solution?)

From here on I'm stumped.

I understand the procedure but not the questions. So there are two reactions:

$$\ce{2HCl(aq) + CaCO3(s) -> H2CO3(aq) + 2 Cl-(aq) + Ca2+(aq)}$$ $$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + Cl-(aq) + Na+(aq)}$$

The first reaction has reactants in non-stoichiometric amounts. The second reaction, a titration, is done with stoichiometric amounts, yielding a neutral reaction mixture.

The strange thing is that the problem is over-determined: I can calculate the amount of HCl not neutralized by calcium carbonate either by looking at the amounts of the two (assuming the reaction goes to completion) or by the titration (assuming that neutralizing means to add sufficient NaOH to be at stoichiometry ratio with the remaining HCl).

There is also a technical detail: it would be better to boil off the carbonic acid made by protonating the carbonate. Otherwise, the analyte will contain strong and weak acid, and buffer will form during the titration (with the endpoint of the titration ill-defined: is it when the HCl is depleted, or when all acid is depleted?).

However, it seems like the question wants us to make an argument about amounts starting with the titration. We will assume that the titration and the reaction between $$\ce{CaCO3 and HCl}$$ go to completion, and chalk up (get it?) any inconsistencies to impure $$\ce{CaCO3}$$.

Amount of NaOH

$$c_{\mathrm{\ce{NaOH}}} = 0.100\ \mathrm{M}$$

$$n_{\mathrm{\ce{NaOH}}} = c_{\mathrm{\ce{NaOH}}} \cdot V_{\mathrm{\ce{NaOH}}}$$

$$\ \ \ =0.100\ \mathrm{M} \cdot 8.50\ \mathrm{mL}$$

$$\ \ \ =8.50\times 10^{-4}\ \mathrm{mol}$$

Amount of HCl

The amount of HCl in the 20 mL sample was $$\pu{8.50e-4 mol}$$ (from the titration). So the amount in the entire 200 mL sample after adding carbonate was $$\pu{8.50e-3 mol}$$. The amount used up by reacting with carbonate is the difference between the original amount, $$\pu{0.021 mol}$$ or $$\pu{21e-3 mol}$$, and what is left after the first reaction.

$$n_\mathrm{reacted} = \pu{21e-3 mol} - \pu{8.50e-3 mol} = \pu{12.5e-3 mol}$$

Amount of $$\ce{CaCO3}$$

As the OP writes, there is a 2:1 stoichiometry (HCl vs $$\ce{CaCO3}$$), so

$$n_\ce{CaCO3} = \pu{12.5e-3 mol} / 2 = \pu{6.2e-3 mol}$$

On the other hand, if the sample were 100% and everything has 100% yield and we had boiled off the carbonic acid, we would expect a higher amount in 1.00 g of calcium carbonate, i.e. $$\pu{9.99e-3 mol}$$ as correctly pointed out by the OP.

Why the discrepancy?

This problem looks like a typical job, that I have often given in the lab sessions to my high school classes. The student has to analyze a sample of chalk, which is made of CaCO3, plus some insoluble impurities like alumina $$\ce{Al2O3}$$, silica $$\ce{SiO2}$$ or gypsum $$\ce{CaSO4·2H2O}$$. So the student adds an excess of $$\ce{HCl}$$, waits until no effervescence is observed, and then titrate the excess of $$\ce{HCl}$$ with a standard $$\ce{NaOH}$$ solution. This is what was done here.
Here the titration shows that the sample contains $$0.625$$ g pure $$\ce{CaCO3}$$. So it is a rather impure sample, with only $$62.5$$ % $$\ce{CaCO3}$$. This may happen.