How much titrant will be required if your EDTA titrant is exactly 0.01 M and you weighed out exactly 0.4 g calcium carbonate?
mols CaCO3 = grams/molar mass
mols EDTA = mols CaCO3 (it's always 1:1)
Then M EDTA = mols EDTA/L EDTA
Solve for L EDTA and convert to mL if desired.
To calculate the amount of titrant (EDTA) required, we need to use the stoichiometry of the reaction between the titrant and calcium carbonate.
The balanced chemical equation for the reaction between EDTA and calcium carbonate is as follows:
CaCO3 + EDTA → Ca(EDTA) + CO2
From the balanced equation, we can see that the mole ratio between calcium carbonate (CaCO3) and EDTA is 1:1. This means that for every 1 mole of calcium carbonate, we will need 1 mole of EDTA.
Now, let's calculate the number of moles of calcium carbonate (CaCO3) in 0.4 g using its molar mass.
The molar mass of calcium carbonate (CaCO3) is:
molar mass of Ca = 40.08 g/mol
molar mass of C = 12.01 g/mol
molar mass of O = 16.00 g/mol (3 oxygen atoms in carbonate)
molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + (16.00 g/mol × 3) = 100.09 g/mol
Number of moles of calcium carbonate (CaCO3) = mass / molar mass
= 0.4 g / 100.09 g/mol
Now, we know the number of moles of calcium carbonate. Since the stoichiometry between calcium carbonate and EDTA is 1:1, the number of moles of EDTA required is also 0.4 moles.
Finally, we can calculate the amount of titrant (EDTA) required using its molarity:
Amount of titrant (EDTA) required = number of moles of EDTA × molarity of titrant
= 0.4 moles × 0.01 M
Therefore, the amount of titrant required is 0.004 moles (or 4 millimoles).
Note: It's essential to always double-check the stoichiometry of the reaction, the molar mass values, and the units to ensure accurate calculations.