Question

An unknown amount of calcium carbonate was neutralized with 35.60 ml of 0.275 M aqueous hydrochloric acid. the resulting solution was titrated with 0.205 M aqueous sodium hydroxide, with the endpoint being reached after the addition of 20.76 ml of base. How many grams of calcium carbonate were neutralized by the aqueous hydrochloric acid?

Answer 1

The first reaction occurs between hydrochloric acid and calcim carbonate:

In this reaction an excess of HCl solution is used in order to neutralize the calcium carbonate.

Afterwards, the excess of hydrochloric acid solution is neutralized with NaOH according to the following reaction:

In order to calculate the mass of calcium carbonate neutralized we need to find the moles of HCl in excess neutralized by the addition of NaOH, this quantity is equal to:

Where Vb and Cb represent the volume and the concentration of the base (NaOH) used to neutralize the acid. Replacing the values we get:

The number of moles of HCl that reacted with calcium carbonate can be calculated as follows:

Replacing the values we obtain:

According to the first equation 2 moles of HCl reacts with 1 mol of CaCO₃, hence the moles of CaCO₃ reacting are:

The mass in grams of calcium carbonate neutralized is:

Remember that the molar mass of CaCO₃ is 100 g/mol approximately.