A 15.00 ml sample of a standard solution containing 1 g of CaCO3/L required 8.45 ml of EDTA to fully complex the Ca present. Calculate the volume of EDTA stoichiometrically equivalent to 1.0 mg of CaCO3.'
Sorry, forgot to write the entire question.
1 g CaCO3/L x 0.015 L = 0.015 g or 15 mg CaCO3 in the aliquot.
That is equivalent to 8.45 mL EDTA; therefore,
15 mg/8.45 mL = 1.775 mg CaCO3/mL EDTA. I wold round that to 1.78.
DrBob222, how are you sure this is the right answer. could you explain a little bit more?
Relating Density to a 1:1 stoichiometry equation
To calculate the volume of EDTA stoichiometrically equivalent to 1.0 mg of CaCO3, we need to determine the stoichiometry relationship between CaCO3 and EDTA and use the given information.
1. First, let's convert the given concentration of CaCO3 from grams per liter (g/L) to milligrams per milliliter (mg/mL):
- Given concentration of CaCO3 = 1 g/L
- Conversion factor: 1 g = 1000 mg; 1 L = 1000 mL
- Concentration of CaCO3 = 1 g/L * (1000 mg/1 g) * (1 mL/1000 mL) = 1 mg/mL
2. Now, let's calculate the number of moles of CaCO3 in the given 15.00 ml sample using its concentration:
- Number of moles of CaCO3 = Concentration of CaCO3 * Volume = 1 mg/mL * 15.00 mL = 15.00 mg
3. Since the stoichiometry relationship between CaCO3 and EDTA is 1:1, the number of moles of EDTA required to fully complex the Ca present is also 15.00 mg.
4. Finally, let's calculate the volume of EDTA stoichiometrically equivalent to 1.0 mg of CaCO3 using the volume of EDTA used to complex the Ca present in the 15.00 ml sample:
- 15.00 mg of EDTA corresponds to 8.45 ml
- 1.0 mg of EDTA would correspond to (8.45 ml * 1.0 mg) / 15.00 mg = 0.5633 ml (rounded to four decimal places)
Therefore, the volume of EDTA stoichiometrically equivalent to 1.0 mg of CaCO3 is approximately 0.5633 ml.