[Sasha]

Is there equations that i can use to calculate the permanent water hardness in mmol and the mmol of magnesium ions per liter of the water?

[Borek]

Please elaborate, question as posted doesn't make sense.

[DevaDevil]

You beat me to it, Borek;
yeah, if you want to calculate the total amount of Ca²⁺ and Mg²⁺ (most common ions of hard water), you need to know where to start from. What measurement for example.

Also "permanent" hard water is nothing more than hard water after boiling with those doubly charged cations still present. But what is there to calculate? You (We) need more information for a calculation to make sense

[Sasha]

"27.13 ml of 0.01 M EDTA solution was required to reach the end point for titration of  25 ml of the water sample. After boiling and filtering 100 ml of water, the student diluted the filtrate to 100 ml with distilled water. To titrate 25 ml of the diluted filtrate, 26.4 ml of 0.01 M EDTA solution was required. The student precipiated the calcium ion as CaC2CO4 from 75 ml of the water sample and removed the precipiate by filtration. The student then adjusted the water sample volume to 100 ml. to titrate 25 ml of the diluted filtrate , 7.43 ml of 0.01M EDTA solution was required."


calculate the permanent water hardness in terms of total mmol of calcium and magnesium ions.
calculate the number of mmol of magnesium ions per liter of the water.

[chiralic]

Hi Sasha:

Here you can find a Link to get your information about Conversion of different Water Hardness Degree (Permanent o Temporal). It's a table.

http://www.aqua-correct.com/DK1SKW/uk-waterhardness-tabel.html

My best regards,

Chiralic

[Sasha]

thank you chiralic but i m still a bit confused....

[chiralic]

Hi Sasha:

I don't understand what is your confusion. I understood that you are looking for a equation to CONVERT mg of CaCO3/L or ppm of CaCO3 to mmol/L, I "supposed" that these Conversion Table help you and you know HOW TO calculate the Water Hardness.

Anyway...try this:
http://www.cerlabs.com/experiments/10875404367.pdf

Regards,

Chiralic

[DevaDevil]

ok, for the calculation; 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.