A 0.4515 g sample of pure CaCO3 was transferred to a 500 mL volumetric flask and dissolved using 6M HCl then diluted to volume. A 50.00 mL aliquot was transferred to a 250mL Erlenmeyer flask and the pH adjusted. Calmagite was used as an indicator the solution was titrated using EDTA and required 45.05 mL to reach the endpoint. Calculate the EDTA molarity. [Answer in 4 SF].
(EDTA reacts in a 1:1 stoichiometry with Ca2+)
How many moles CaCO3 did you titrate? That is 0.4515g CaCO3/molar mass CaCO3 = ?
You placed that in 500 mL flask and took a 50 mL aliquot; therefore, that 50 mL is 0.1 of the sample (50/500) so 0.1 of the moles you weighed = amount titrated.
Since the reaction is 1:1, then that is moles of the EDTA used.
moles EDTA (which you now have) = M x L.
You know moles and L, solve for M
EDTA.
To calculate the EDTA molarity, we need to use the information about the titration and the stoichiometry of the reaction between EDTA and Ca2+.
First, let's calculate the number of moles of CaCO3 in the 0.4515 g sample.
We can use the molar mass of CaCO3 to convert the grams to moles:
Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 16.00 g/mol (O) x 3 = 100.09 g/mol
Number of moles of CaCO3 = mass / molar mass
= 0.4515 g / 100.09 g/mol
= 0.004512 mol
Since CaCO3 reacts in a 1:1 stoichiometry with Ca2+, the number of moles of Ca2+ is also 0.004512 mol.
Now, let's calculate the molarity of Ca2+ in the 50.00 mL aliquot.
Molarity (M) = moles / volume (in liters)
= 0.004512 mol / 0.05000 L
= 0.09024 M
We know that EDTA reacts in a 1:1 stoichiometry with Ca2+, so the moles of EDTA used in the titration are the same as the moles of Ca2+.
Now, let's calculate the molarity of EDTA.
Molarity (M) = moles / volume (in liters)
= 0.004512 mol / 0.04505 L
= 0.09997 M
Rounding to 4 significant figures, the EDTA molarity is 0.1000 M.