posted by Anonymous on .
A 25.00 ml sample of a standard solution containing 1 g of CaCO3/L required 25.20 ml of EDTA to fully complex the Ca present. Calculate the volume of EDTA stoichiometrically equivalent to 1.0 mg of CaCO3.
From the volume of EDTA required in titrating the sample, calculate the equivalent mass of CaCO3 in the aliquot of sample titrated.
Volume of sample: 100 ml
Volume of CaCO3 taken: 25.0 mL
Volume EDTA added: 3.225 mL
Volume EDTA needed for 25 ml CaCO3 :25.20 mL
Volume EDTA needed for sample alone: 3.2225 mL
Do you have two problems or one problem here?
It's two problems.
The first one I got 1.01 mL EDTA.
Oh, it's 25.00 ml sample of a standard solution containing 1 mg of CaCO3/L.
The second problem, I'm not sure how to approach it.
I am totally confused about the concn. I don't know if it is 1 g/L as in the original post or if the 1 mg/L is a correction.
I'm sorry, it's a correction.