Posted by darren on .
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