A standard EDTA solution was prepared by weighing 1.9345+/-0.001 grams of primar

Question

A standard EDTA solution was prepared by weighing 1.9345+/-0.001 grams of primary standard (Na2H2Y-2H2O, M.W. 372.24+/-0.03), quantitatively transferring the primary standard into a 500.0+/-0.2 mL volumetric flask and bringing the flask to volume.

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A standard EDTA solution was prepared by weighing 1.934510.0001 grams of primary standard (Na Hy 2H20, M.W 372.24±0.03), quantitatively transferring the primary standard into a 500.010.2 mL volumetric flask and bringing the flask to volume. What is the Molarity of the standard EDTA solution Calculate the absolute deviation of this standard EDTA solution and report the Molarity to the corr significant figures. (report Molarity ± absolute deviation) g 1.9345±0.0001 grams of primary standard (NaMY- ect When a flask of 50+1 ml containing 0.1245:0.0001 of an unknown sample was titrated for Mg", with the standard EDTA solution, 26.45 mL was required to reach the titration end point, what is the % Mg (M.W. 24.305tO·003) in the sample? Calculate the absolute deviation for the analysis and report the % Mg to the correct significant figures. (report % Mg ± absolute deviation)

Explanation / Answer

1)Moalrity = mass / molecular weight X volume

Moalrity = 1.9345 X 1000 / 372.24 X 500 = 0.0103 molar

Mass = 1.9345+/-0.001 grams

relative error in the mass is

0.001 / 1.9345 = 5.16 X 10^-4

Molecular weight = 372.24+/-0.03

relative error in the MW is

0.03 / 372.24 = 8.05 X 10^-5

Volume = 500.0+/-0.2 mL

relative error in the volume = 0.2 / 500 = 4 X 10^-4

RSEM = (RSEMW2 + RSEmass2 + RSEvol.2)1/2

RSE molarity = [(0.805 X 10^-4)2 + ( 5.16 X 10^-4)2 + ( 4 X 10^-4)2 ] 1/2

RSE molarity = 10^-4 [0.6480 + 13.73 + 16 ]2 = 10-4 X (30.378)1/2

RSE molarity = 5.511 X 10^-4

So standard error = RSE X molarity = 0.0103 X 5.511 X 10^-4 = 0.0567 X 10^-4

So molarity = 0.0103 +- 0.0567 X 10^-4

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