Back titration of excess HCl used for dissolving limestone

Question

$15.0~\mathrm{mL}$ of $1.4~\mathrm{M}\ \ce{HCl}$ was mixed with $1.00~\mathrm{g}$ of limestone (impure $\ce{CaCO3}$) until all the solid had dissolved. The solution was then transferred to a conical flask and made up to $200~\mathrm{mL}$ with water. A $20.0~\mathrm{mL}$ portion was then neutralised by $8.50~\mathrm{mL}$ of a $0.1~\mathrm{M}\ \ce{NaOH}$ solution.
Calculate:

1. Amount of excess $\ce{HCl}$ in the $20.0~\mathrm{mL}$ portion
2. Amount of excess $\ce{HCl}$ in the $200~\mathrm{mL}$ portion
3. Amount of $\ce{HCl}$ which reacted with $\ce{CaCO3}$

My attempt:

When the limestone is dissolved,

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

$$n(\ce{HCl}) = c \times V = (15 \times 10^{-3}~\mathrm{L}) (1.4~\mathrm{M}) = 0.021~\mathrm{mol}$$ $$n(\ce{CaCO3}) = x~\mathrm{mol}$$

When the solution is diluted to $200~\mathrm{mL}$ the molarity of $\ce{HCl}$ can be found by, \begin{align} n_\mathrm i &= n_\mathrm f\\ c_\mathrm i V_\mathrm i &= c_\mathrm f V_\mathrm f\\ c_\mathrm f &= \frac{c_\mathrm i V_\mathrm i}{V_\mathrm f}\\ \therefore c_\mathrm f (\ce{HCl}) &= \frac{(15\times 10^{-3}~\mathrm{L})(1.4~\mathrm{M})}{(200\times 10^{-3}~\mathrm{L})} = 0.105~\mathrm{M} \end{align}

I'm having doubts about my working up until this stage. Am I correct so far?

Answer 1

It is generally a good idea to approach problems in the order that sub-questions are asked in. There is often a paedagogic reasoning in that order that can help you approach the answer.

In this case, the key point is that the limestone is impure $\ce{CaCO3}$, and that thus we do not know how much of the acid was initially neutralised with $\ce{CO2}$ evolution. Approaching the question from this side, as you did, is not very helpful, since we are missing an important value to calculate from.

What we do know, however, is the final end of the experiment: that $8.50~\mathrm{ml}$ of $0.1~\mathrm{M}\ \ce{NaOH}$ were able to neutralise a certain portion of the remaining hydrochloric acid. You should think of this as being a titration.

Once you have realised that, it makes sense why the first question asks for the amount of substance in the $20.0~\mathrm{ml}$ aliquot — it is the one you can actually calculate quickly, and it will lead you to the entire solution.

Thus, your plan of action is:

• find the concentration or amount of $\ce{HCl}$ in the $20.0~\mathrm{ml}$ portion by known formulae used for titration problems (your work so far shows you can do this);

We have $\ce{HCl + NaOH -> NaCl + H2O}$, so $$\begin{align} n_\ce{HCl} = n_\ce{NaOH} &= (\pu{8.50 mL})(\pu{0.1 M}) \\ &= \pu{0.00085 mol} \end{align}$$

• thencefrom calculate the concentration or amount of $\ce{HCl}$ in the entire $200~\mathrm{ml}$ sample from similar formulae;

A $\pu{20 mL}$ portion of the $\ce{HCl}$ simply has one-tenth of the original amount of $\ce{HCl}$, so in the entire $\pu{200 mL}$, we have: $$n_\ce{HCl} = (\pu{0.00085 mol}) \times \frac{\pu{200 mL}}{\pu{20 mL}} = \pu{0.0085 mol}$$

• Next, decide how much hydrochloric acid was lost in the dissolution of the limestone.

At the start, we had a grand total of $$\begin{align} n_{\ce{HCl},\mathrm{tot}} &= (\pu{15.0 mL})(\pu{1.4 M}) \\ &= \pu{0.021 mol} \end{align}$$ which you already calculated. After reacting with the limestone, we were left with only $\pu{0.0085 mol}$ of that. So, the amount that reacted is simply: $$n_{\ce{HCl},\mathrm{reacted}} = \pu{0.021 mol} - \pu{0.0085 mol} = \pu{0.0125 mol}.$$