Stoichiometry Problem

Question

One impure sample of CaCO3 is treated with 100.00mL of 1.00mol/L HCl. The resulting solution is diluted to 250.00 ML and 25.00 ML of this solution required 20.00mL of 0.10mol/L NaOH for neutralization. What is the mass of CaCO3 assuming impurities are not reactive substances?

Answer 1

Take a look at what is going on. The first step is to treat the CaCO3 with HCl. That reaction is

CaCO₃ + 2HCl ==> CaCl₂ + CO₂ + H₂O

The next step is to back-titrate the excess HCl with NaOH. That is, in the first reaction, an excess amount of HCl was used to be sure ALL of the CaCO3 reacted. So, some HCl will be left over.

HCl + NaOH ==> NaCl + H2O

If you know the moles of HCl added in the first reaction, and the moles of HCl left over after the reaction, then you can determine the moles of CaCO3 that were present. The rest of the problem deals with all of the dilutions that have taken place.

First reaction: 100.00 ml of 1.00 mol/L HCl = 0.100000 L x 1.00 mol/L = 0.100 moles HCl used

Before second reaction, this 100 ml of reaction is diluted to 250 ml and 25 ml is used for the back-titration. We'll come back to this later.

Second reaction: 20.00 ml of 0.10 mol/L NaOH = 0.02 L x 0.1 mol/L = 0.002 moles NaOH

This must be equal to the moles of HCl present in excess. That is there are 0.002 moles HCl in excess in this 25 mls of the diluted sample. But we want to know moles HCl in excess in the original 100.00 ml sample.

Now to incorporate the dilutions that have been made:

0.002 moles HCl in 25 ml of the sample that was diluted 100 ml to 250 ml. Since the original sample was diluted 2.5 fold (100 ml to 250 ml), this would be 0.002 moles x 2.5 = 0.005 moles excess HCl in 25 ml of the original sample. Since there were 100 mls of original sample, you would have 0.005 x 4 = 0.02 moles excess HCl in the original 100 ml sample. Another way to look at this dilution is that there are 0.002 moles HCl in 25 ml which came from 250 ml, so there would be 0.02 moles HCl in the entire 250 mls. This 250 mls came from the 100 mls so there were 0.02 moles HCl excess in the 100 mls.

In a previous calculation, we found that the total HCl used in the original sample was 0.100 moles. Since 0.0200 moles is left over after reaction with CaCO₃, this means that 0.08 mole HCl was used to react with all of the CaCO₃. From the stoichiometry of the reaction, you see that it takes 2 moles HCl for each mole of CaCO₃. Therefore, moles CaCO₃ present = 0.0800 moles HCl x 1 mole CaCO₃/2 moles HCl = 0.0400 moles CaCO₃.

Mass of CaCO₃ = 0.0400 moles x 100.09 g/mole = 4.00 g (to 3 significant figures)