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In a titration with NaOH, CaCO3, and HCl, I found the moles of both the NaOH,...

In a titration with NaOH, CaCO3, and HCl, I found the moles of both the NaOH, HCl, as well as the moles of acid neutralized. How can I find the moles of the base, which I'm assuming is CaCO3?

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Answer #1

HCl is neutralized by Calcium carbonate as follows

CaCO3 (s) + H+ (aq) ------> Ca2+ + HCO3- (aq)

HCO3- (aq) + H+ (aq) -----> H2CO3 (aq) ---> CO2 (g) + H2O (l)

We first dissolve the CaCO3 in the acid of known concentration, some of the HCl will be neutralized by the carbonate, but there will be something remaining. To find this residue acid, we perform a titration with  NaOH, which gives the value of the acid left. This method is called Reverse Titration.

The ratio of CaCO3 to HCl is 1:2 and the molar mass of CaCO3 is 100 g/mol.

The no of moles of NaCl is equal to no of moles of HCl as the ratio is 1:1

No of moles of CaCO3 = mass of CaCO3 used X No of moles of HCl X 2 / ( Molar Mass of CaCO3 X No of moles of HCl neutralized)

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