[JohnTan]

Hi everyone,
I have encountered a difficult question and I do not know how to begin going about solving it. What information does the titration data give? Any help would be greatly appreciated.
The amount of carbonate present in a sample may be determined by firstly, reacting the carbonate with an excess of strong acid. The excess acid is then determined by titration with a strong base.
In this example a sample of limestone is dissolved by adding 60 mL of 1.025 M HCl and the solution is transferred to a 250 mL volumetric flask and made up to the mark using distilled water. Aliquots of 10mL of this solution are then titrated against a standard solution of 0.0998 M NaOH and the average titre is 23.81 cm3.
Calculate the molarity of CaCO3 in the 250 mL volumetric flask.

[Biopolmonkey]

Write balanced equations for the reactions between HCl and CaCO3 and NaOH.

How many moles of HCl did you add (to the carbonate)?

How many moles of HCl were then available to react with NaOH?

What do think the difference tells you?

[JohnTan]

Still struggling.
2HCl + CaCO3 --> CaCl2 + CO2 +H2O
HCl + NaOH --> NaCl + H2O

I calculated there to be 0.0615 moles of HCl in the solution that reacted with the calcium carbonate. As for the number of moles that reacted with NaOH, I'm unsure.

[AWK]

_{Calculate the molarity of CaCO3 in the 250 mL volumetric flask}
There is no CaCO3 in volumetric flask. Instead you have CaC_l2. But: moles of calcum carbonate = moles of calcium chloride

Now calculate moles of HCl found in titration sample. Then multiply them by 250/10 and you get moles of HCl remained after reaction with CaCO₃.
Subtaction of your result and this one will give you moles of consumed HCl in the reaction.
And finally calculate moles of CaCO₃