You want to measure the CaCo3 contents in chalk. To this end, you grind a piece of chalk, dissolve the slurry to make 10ml of solution and drive the carbonate out by adding 10 ml of 2M HCl. You titrate the remaining HCl back with 0.1M potassium hydroxide solution, of which you need 47 ml. Assume all carbonate was CaCO3, how much was it in grams?

Answer 1

moles HCl added initially = M x L = ??
Some was not used (there was an excess of HCl); how much of an excess did you have? That will be moles KOH = M x L = ??

So the difference between moles initially and moles of excess must be the amount CaCO3 in the chalk. Then moles x molar mass CaCO3 = g CaCO3.
Post your work if you get stuck.

Answer 2

How would you calculate the excess of HCl?

Answer 3

Go back and re-read my response. The excess HCl was titrated with 47 mL of 0.1 M KOH. M x L = moles KOH added which is the same as moles HCl in excess.

You need to make a correction, then, for the fact that CaCO3 used two moles HCl for every one mole CaCO3.
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2

So initial moles HCl - moles HCl in excess = moles HCl used to neutralize the CaCO3.
Then 1/2 moles HCl used = moles CaCO3. Then g CaCO3 = moles CaCO3 x molar mass CaCO3

Answer 4

How do you find the initial moles of HCl?