Hard A Level Chemistry Question

12.0 g of a mixture of calcium carbonate and sodium chloride was treated with 100 cm3 of 2.00 mol dm-3 hydrochloric acid (only the calciumcarbonate reacts). The resulting solution was made up to 250 cm3 with water and a 25.0 cm3portion of this needed 34.1 cm3 of 0.200 mol dm-3 sodium hydroxide for neutralisation. Find the % by mass of the calcium carbonate in the mixture.

This was a relatively hard question that I got. I know how to do the question but this is because I did an almost identical question and learnt how to do it without really understanding it.

My question is why is the "resulting solution" here HCl.
and if it makes up a solution of 250cm^3 *with H20* then why do we not include the water in the titration equation.

The equation for the CaCO3 reaction is:

HCl + CaCO3 --> CaCl2 + H20 +C02

The equation for the titration is HCl + NaOH --> NaCl + H20

The answer is 55% btw.

The resulting solution here is HCl because after all the calcium carbonate reacted there was still HCl that was left unreacted. This happens because the HCl is in excess and there was not enough calcium carbonate added to completely neutralise the solution. The water that is produced in the reaction does not disappear or anything but simply just dilutes the HCl solution.

You also ask why H20 is not included. Water does not take part in the titration - only the acid and base take part in it. Water is added in a stage beforehand to prepare the solution for the titration. Also, I am sure you are familiar with dilution - which is when water is added to acids and bases to decrease their concentration. This is what is happening in this situation: the water added is diluting the HCl. We do not end up with a water and HCl mixture, we end up with a diluted HCl solution - so H20 is not needed to be included in the equation.


You said you did not really understand how to answer this question. If you have any other questions regarding this problem, feel free to ask.

Okay so I got 55% too
Firstly the equation is CaCO3 + NaCl + 2HCl -> CaCl2 + H2CO3 +NaOH !
NaCl doesn't take part in the equation
The moles of HCl are 0.2 which are to be used later.

The moles of NaCl are 6.82× 10^-3 which when multipled by 10 give ans 0.0682. We have multipled because the mixture has been transferred from 250 to 25cm3

Now we will subtract the moles we calculated of HCl (0.2) with 0.0682.
The ans 0.1318 will be divided by 2 because of the mole ratio giving ans 0.0659.
0.0659 is multipled with Mr of CaCO3(100) giving ans 6.59.
The Q mentions the use of 12 g So 6.59÷12 ×100= 54.92 % As the question asks the % mass of CaCO3


Hope this helps