$15.0~\mathrm{mL}$ 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.

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

My attempt:

When the limestone is dissolved,

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

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

When the solution is diluted to $200~\mathrm{mL}$ the molarity of $\ce{HCl}$ can be found by, \begin{align} n_i &= n_f\\ c_i V_i &= c_f V_f\\ c_f &= \frac{c_i V_i}{V_f}\\ \therefore c_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?

  • 2
    $\begingroup$ Hi and welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. Visit the help center for unanswered questions on how it works. Your question is a homework question, but as far as I can see, you have fulfilled the requirements set up in our current homework policy. Unfortunately, you are on the wrong track afaict, though. $\endgroup$ – Jan Jun 16 '16 at 13:29
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    $\begingroup$ Welcome to Chemistry.SE! States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different. Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". We prefer to not use MathJax in the title field, see here for details. $\endgroup$ – Martin - マーチン Jun 16 '16 at 13:42
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    $\begingroup$ Another side note: Your equation is not balanced. $\endgroup$ – Jan Jun 16 '16 at 14:03

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);

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

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

To prove your calculations are correct: I calculate $626~\mathrm{mg}$ of the limestone to be pure $\ce{CaCO3}$.


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