Question

The Determination of Magnesium by Direct Titration

Preparation of Solutions:

Buffer solution, pH 10: Diluted 57mL of NH₃ and 7g NH₄Cl totaling 100mL solution

Eriochrome Black T: Dissolved 102.4mg in 15mL ethanolamine and 5mL of absolute ethanol

EDTA: 3.8008g of purified dihydrate Na₂H₂Y 2H₂O, mixed with distilled water to total volume of 1L

I received a Magnesium Sulfate unknown, and diluted it to 500mL with distilled water then transferred 3 50mL aliquots to separate conical flasks and added 2mL of pH10 buffer and 4 drops of Eriochrome Black T to each flask and titrated each with 0,01M EDTA until I reached the endpoint. My 1st resulted in 19.23mL of EDTA used, 2nd 19.26mL EDTA and 3rd was 19.25mL EDTA used.

During this, I was not given the weight of the unknown Magnesium Sulfate. I am confused on how to determine the unknown Magnesium Sulfate, and express the results as ppm of Mg²⁺ in the unknown sample.

Answer 1

Disodium salt of ethylene diamine tetra acetic acid is used to determine the total hardness of the given hard water. The hardness causing metal ions (i.e. calcium and magnesium) form a wine-red colored weak complex with Eriochrome Black – T indicator in the presence of a buffer solution. Upon addition of EDTA, which replaces the indicator, a stable complex is obtained. Due to the liberation of Eriochrome Black – T indicator, wine red color changes to steel blue. This is the end point for the titration between EDTA and hard water.

The entire reaction between Mg ions, EBT and EDTA is represented as follows.

Mg²⁺ + EBT → [Mg-EBT] Wine red complex

[Mg-EBT] + EDTA → Mg-EDTA + EBT(steel blue)

1ml of 0.01M EDTA = 1mg/mL of CaCO₃ (standard hard water) = 1ppm

I aliquot:

19.23 mL of 0.01M EDTA = 19.23 ppm

50 mL of made up MgSO₄ solution contains 19.23 ppm

500 mL of the solution contains = 19.23 x 500/50 = 192.3 ppm

II aliquot:

19.26 mL of 0.01M EDTA = 19.26 ppm

50 mL of made up MgSO₄ solution contains 19.26 ppm

500 mL of the solution contains = 19.26 x 500/50 = 192.6 ppm

III aliquot:

19.25 mL of 0.01M EDTA = 19.25 ppm

50 mL of made up MgSO₄ solution contains 19.25 ppm

500 mL of the solution contains = 19.25 x 500/50 = 192.5 ppm

Only Mg²⁺ from Magnesium sulphate reacts with EDTA forming complex. So the result can be taken as the average of the three values, 192.47 ppm in the unknown sample.