Asked by Johnathan
                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.
            
        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.
Answers
                    Answered by
            DrBob222
            
    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?
    
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?
                    Answered by
            Johnathan
            
    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. 
    
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.