If the 50.00mL sample from problem 3 above was titrated with EDTA,what volume of a 0.00420 M EDTA solution would be needed to reach the endpoint? (Your answer should be given in milliliters.) (Problem 3 found that .0000625 moles of calcium carbonate were present in the sample)
EDTA forms 1:1 complexes with metals; therefore, 6.25E-5 mols from the previous problem will be that amount with EDTA.
M = mols/L
0.00420 = 6.25E-5/L. Solve for L and convert to mL if desired.
To determine the volume of a 0.00420 M EDTA (Ethylene diamine tetraacetic acid) solution needed to reach the endpoint, we need to use the stoichiometry of the reaction between EDTA and calcium carbonate.
From problem 3, we know that the sample contained 0.0000625 moles of calcium carbonate.
The balanced chemical equation for the reaction between EDTA and calcium carbonate is:
CaCO3 + EDTA → Ca(EDTA) + CO2
According to the equation, one mole of EDTA reacts with one mole of calcium carbonate.
So, the moles of EDTA required to react with 0.0000625 moles of calcium carbonate is also 0.0000625 moles.
Now, we can use the formula:
Moles = Concentration x Volume
to find the volume of EDTA solution needed.
0.0000625 moles = 0.00420 M x Volume
Solving for Volume:
Volume = 0.0000625 moles / 0.00420 M
Volume = 0.0148809524 L
Finally, convert the volume from liters to milliliters:
Volume = 0.0148809524 L x 1000 mL/L
Volume ≈ 14.88 mL
Therefore, approximately 14.88 mL of the 0.00420 M EDTA solution would be required to reach the endpoint.