Question

A student collected the following data while using the procedure described in this module: 27.13 mL of 1.00 x 10^-2M EDTA soltuion was required to reach the ErioT end point for titration of 25.00 mL of the sample water. After boiling and filtering 100.00 mL of the water, the student diluted the filtrate to 100.0 mL with distilled water. To titrate 25.00 mL of the diluted filtrate, 26.40 mL of 2.00 x 10^-1 M EDTA solution was required. The studetn precipitated the Ca^2 ion as CaC2O4 from 75.00 mL of the water sample and removed the precipitate by filtration. The student then adjusted the water sample volume to 100.0 mL. To titrate 25.00 mL of the diluted filtrate, 7.43 of 1.00 x 10^-2 M EDTA solution was required Calculate the permanent water hardness in terms of total mmol of Ca^2 and Mg^2 ions.

Answer 1

1 mol EDTA forms a complex with 1 mol of cations. 1 ml 0.01M EDTA contains 0.01 mol/l * 0.001 l = 1 x 10^-5 M EDTA

Remember, the filtrate is used in the second titration, those are the compounds that cannot be removed by boiling/filtration. Hence the filtrate is the "permanent" hard water.

100ml was boiled/filtrated, the filtrate filled back up to 100 ml again (no net. dilution), then 25 ml titrated.

26.4ml EDTA was needed = 2.64 x 10^-4 mol.
Assuming the compounds are mainly Ca²⁺ and Mg²⁺, the combined concentrations of these in "permanent" hard water is 2.64 x 10^-4 mol / 25 ml = 0.264 mmol / 25 ml = 0.0106 M = 10.6 mmol/l (choose whatever unit you need

Now the last experiment is done with the Calcium removed, hence the result of the last experiment gives the Mg²⁺ concentration.
Try to calculate this; keeping in mind that for the last one the student did dilute the sample from 75 to 100 ml