EDTA: Dry Na2H2EDTA • 2H2O (FW 372.25) at 80 oC for 1 h and cool in the desiccat
ID: 917169 • Letter: E
Question
EDTA: Dry Na2H2EDTA • 2H2O (FW 372.25) at 80 oC for 1 h and cool in the desiccator. Accurately weigh out 0.6045g and dissolve it with heating in 400 ml of DI in a 500-ml volumetric flask. Cool to room temperature, dilute to the mark, and mix well.
Pipet a 25.00 ml sample of the unknown into a 250-ml flask. Dilute with an additional 25 ml of DI H2O (you can use a graduated cylinder for the water addition, it is added to help clarify the color change). To each sample add 3 ml of pH 10 buffer and a 2 drops of Eriochrome black T indicator. Titrate with EDTA from your 50-ml buret and note when the color changes from wine red to blue.
Volume
sample [1]
sample [2]
sample [3]
initial
1.290 mL
2.400 mL
3.150 mL
final
12.810 mL
13.850 mL
14.420 mL
total
11.520 mL
11.450 mL
11.270 mL
Using these results,
[1] calculate the molarity of EDTA.
[2] calculate [Ca2+] ion and [Mg2+] ion for each?
[3] calculate the average concentration of Ca2+ and Mg2+ for each.
[4] calculate the water hardness in terms of CaCO3
Please show work!
Volume
sample [1]
sample [2]
sample [3]
initial
1.290 mL
2.400 mL
3.150 mL
final
12.810 mL
13.850 mL
14.420 mL
total
11.520 mL
11.450 mL
11.270 mL
Explanation / Answer
1) Drying process will remove two water molecules from each EDTA.2H2O molecule.
So molecular weight of dry EDTA = 372.25-36 = 336.25
Molarity of EDTA = (weight in g / molecular weight) / volume in liters = (0.6045/336.25) / 0.5 = 0.00359 M
2) The EDTA molecule can form a complex with Ca2+ and Mg2+ in 1:1 (metal:EDTA) molar ratios.When all of the Ca2+ has reacted with EDTA, the Mg2+ in the indicator will react with the EDTA. The indicator then returns to its acidic form which is a sky-blue and signals the end of the process. So, given end point is for the sum of [Ca2+] and [Mg2+].
Sample 1. MunknownVunknown = MEDTAVEDTA
[Ca2+] + [Mg2+]= M unkwon = (0.00359 M X 11.520 mL)/25 mL = 0.00165 M
Sample 2. M unkwon = (0.00359 X 11.450)/25 = 0.00164 M
Sample 3. M unkwon = (0.00359 X 11.270)/25 = 0.00162 M
[3] Average [Ca2+] + [Mg2+] = (0.00165 + 0.00164 + 0.00162)/3 = 0.00163 M
[4] Hardness of water is expressed in ppm (mg/L)
Assuming only CaCO3 is present in unknown,
0.00163 moles of CaCO3 is present in 1 liter
Converting moles into g by mutiplying with molecular weight (100.09)
0.1631 g in of CaCO3 is present in 1 liter
163.1 mg in CaCo3 is present in 1 liter
The hardness of water in terms of CaCO3 = 163.1 ppm
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