A .600 g sample of CaCO3 is disolve in concentrated HCl (12M). the solution was diluted to 250 mL in a volumetric flask to obtain a solution for standardizing the EDTA titrant.

Question

1 What is the molarity of the Ca2+ (calcium ion) in the flask?
2 How many moles of Ca2+ are in a 25 mL aliquot of the solution in question 1?

Answer 1

So you have 0.600 g in 250 mL. That is 0.600/100 = 0.006 moles CaCO3. (I estimated the molar mass CaCO3; you need to confirm and update it.)
Molarity = moles/L = moles/0.250 L = ?

How many moles do you have in the 250. How many moles in 1/10th that?

Answer 2

yes the calcium carbonate has (MW= 100.09 g)hence the moles for CaCo3 is correct but how do you calculate the mole for calcium ion Ca2+

Answer 3

Don't you have 1 Ca^+2 ion for every mole CaCO3 there?

Answer 4

Let me redo that. Don't you have 1 atom Ca^+2 for every molecule CaCO3; done another way, don't you have 1 mole Ca^+2 for every mole CaCO3?

Answer 5

2HCl+CaCO3=CaCl2 + H2O + CO2... i think so....

Answer 6

I don't know why you need an equation. There is 1 mole of Ca +2 ions in every mole of CaCO3; therefore, the concn of CaCO3 and the concn of Ca +2 ion are the same. For that matter, C ion is the and CO3 ion is the same as CaCO3. The oxygen concn (although not there as a gas) is 3 times that.