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!

Answer 1

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. M_unknownV_unknown = M_EDTAV_EDTA

[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