[tobias89]

So i'm generally good with Molarity problems but on a few they threw mg/ml into the mix and its throwing me off. I think i did these right but could someone check them?

The first problem is, you weigh out exactly 0.250g of CaCo3 (form. weight= 100.09g/mol) and disolve it in exactly 250 ml of water. Calculate the Ca concentration in mg/ml AND M.

So for mg/ml I did. 0.250g * 1000mg / 250ml = 1.00 mg/ml

For M I did 0.250 g CaCo3 / 100.09 g/mol / 0.250 L H20 = 9.91x10^-3 M

The second problem is you are titrating 25.00 ml of a 1.00 mg/ml CaCo3 solution, how many ml of 0.010 M EDTA solution will be needed to reach the endpoint.

So I did: 25.00 ml * 1mg/ml = 25mg CaCo3
* 1g/100mg = 0.025g CaCo3
/100.09 g/mol= 2.49x10^-4 mol of CaCo3
/0.010 M EDTA = 0.02497 mL
Rounded to two sig figs = 0.025 mL

Am I anywhere close to being right? Thanks so much! It's been two semesters since I had Intro to Chem lecture and it's making taking the labs now much more difficult

[AWK]

In the first part some problem exists. In 250 you can dissolve just ~1.5 mg CaCO₃, but  not 250 mg. Assuming that you used other soluble salt - you calculate concentration CaCO³ in mg/ml instead of Ca²⁺ in the same unit.
In the second part - always write down though symbolic equation showing molar ratio of reagents. Remark concerning solubility od CaCo₃ is still valid.

[tobias89]

Thanks for the reply! But I'm still kinda confused. In the lab procedure it says to add  a few drops HCl to the CaCo3 slurry until it dissolves so would this make it all dissolve? And how would I figure out the concentration of just calcium and not CaCo3? Use the molecular weigh of calcium over the total weight?

[AWK]

This is a simple scaling problem - mass of 1 mole of stoichiometric units of CaCO₃=100. and  1 mole of Ca²⁺=40.