I need to determine the moles of EDTA in a 1:1 stoichemetric titration of a 10.0 mL aliquot of .2108 g of CaCO3 dissolved in 100ml of h2o please help
0.2108g CaCO2 x (1 mol/molar mass CaCO3) = moles CaCO3.
You took 10 of 100 mL for the aliquot which is 1/10 of the sample so moles from above x 0.1 = moles titrated. If the ratio is 1:1 then that will be the moles EDTA.
To determine the moles of EDTA in a 1:1 stoichiometric titration, you will need to follow these steps:
1. Calculate the moles of CaCO3 in the 10.0 mL aliquot:
- Convert the mass of CaCO3 (.2108 g) to moles using the molar mass of CaCO3 (100.09 g/mol):
Moles of CaCO3 = Mass of CaCO3 / Molar mass of CaCO3
2. Use the balanced equation of the reaction between CaCO3 and EDTA to determine the moles of EDTA:
- The balanced equation is: CaCO3 + EDTA → Ca-EDTA + CO2
- According to the balanced equation, the stoichiometric ratio between CaCO3 and EDTA is 1:1.
3. Convert the moles of CaCO3 to moles of EDTA:
Since the stoichiometric ratio is 1:1, the moles of EDTA will be the same as the moles of CaCO3.
Here's an example calculation to help you understand the process:
1. Moles of CaCO3:
Moles of CaCO3 = 0.2108 g / 100.09 g/mol
Moles of CaCO3 = 0.002107 mol
2. Moles of EDTA:
Moles of EDTA = Moles of CaCO3 = 0.002107 mol
Therefore, the moles of EDTA in the 10.0 mL aliquot of the solution would be 0.002107 mol.