The details you give from the lab manual are for the calculation of the molarity of the EDTA solution.
M = moles/L.
moles = grams/molar mass (formula weight) and you have all of that information. Then divide moles/0.500 L = molarity.
Regarding the remaining part of the problem, what did you use as an indicator and how was the "double" titration done?
Determine the concentration of Mg2+ (in mmol/L) in the unknown solution (Note: details and additional information to solve the problem are provided in your lab manual):
Mass of EDTA: 0.606 g
Volume of EDTA used in titration (step 1): 10.48 mL
Volume of EDTA used in titration (step 2): 5.64 mL
FW of EDTA: 372.25 g/mol
The lab manual says: Accurately weigh out 0.6 g and dissolve it with gentle heating in 400 mL of water in a 500 mL volumetric flask. Gentle heating is best accomplished by standing the flask in hot water, not placing it directly on a hot plate. Cool to room temperature, dilute to the mark, and mix well.(in regards to EDTA)
I am completely lost I have tried multiple times and have gotten no where near the right answer (I know what it should be)...any help would really be appreciated.
2 answers
The indicator used was Eriochrome Black T, and for the double titration... I don't know. I know that NaOH was added to precipitate Mg(OH)2, and then it was titrated with EDTA. Then it was left to sit for 5 minutes to dissolve any precipitated Ca(OH)2, and titrated again to go back to endpoint color.
1 answer