hslove142331 Posts: 71, Reputation: 1 Junior Member #1 Feb 11, 2010, 10:09 PM
how can i calculate total hardness in ppm CaCO3?
35.0mL sample of water is titrated with 0.0100M EDTA. Exactly 9.70mL of EDTA are required to reach the EBT endpoint. How can I calculate total hardness in ppm CaCO3?

Thanks
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #2 Feb 16, 2010, 09:25 AM

Well, I looked at the first site I found concerning the formula for calculating water hardness.

water-hardness

So, you need to find the concentration of Ca^2+ ions in your 35.0 mL of sample water.

I'm not familiar with EDTA... depending on the basicity of that acid, you will need to write a balanced equation, and then obtain the mole ratio from that.

You'll get the number of moles of OH^- in 35.0 mL of sample water. That means the concentration of Ca^2+ is half that value and find the concentration for 1000 mL. Multiply that by 2.5 x 10^-3 to get the concentration in ppm.
 sveegaard Posts: 13, Reputation: 3 New Member #3 Feb 21, 2010, 07:47 AM

EDTA complexbinds calcium and is not used for pH titration but colourimetric titration. Hslove, you need to have a reaction scheme or similar, do you have that?
 madhavan2001c Posts: 1, Reputation: 1 New Member #4 Mar 7, 2010, 11:05 PM
Titre value = 9.7 ml
Normality = 0.0100 EDTA solution
Atomic weight of Ca = 20
Volume of water taken for analysis = 35 ml i.e 0.035 litre

ppm of Hardness as Ca2+ = ( 9.7 * 0.0100 * 20 ) / ( 0.035 )
= 55.43 ppm i.e 55.43 mg / kg of water

Hope answer is correct.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #5 Mar 8, 2010, 06:34 AM

Hmm... not quite.

Number of moles of EDTA = (9.7 * 0.01)/1000 = 9.7 x 10^-5 mol
Therefore, number of moles of Calcium carbonate present = 4.85 x 10^-5 mol

That amount is present in 35 mL sample. In 1 L, there is 0.00139 mol.
Hence [CaCO3] = 0.00139 M (this cannot be used since not all CaCO3 are involved in the hardness of water).

[Ca^2+] = 0.00139 M

Using the formula given from the site I posted, hardness = 2.5(0.00139) = 0.00346 M

In grams, this gives 0.00346 * 40.1 = 0.139 g/L = 139 mg/L or 139 ppm

Looking in the given table in from the site, you'll see this water is rated as a hard water.
 solver3000 Posts: 1, Reputation: 1 New Member #6 Oct 7, 2011, 11:02 AM
informatins are given :
Titre value = 9.7 ml
Normality = 0.01 EDTA solution
Atomic weight of CaCO3 = 100g/mole
Volume of water taken for analysis = 35 ml = 0.035 litre
solution:
1-NO.of moles of EDTA = M*volume=0.01*9.7*10^-3 = 9.7*10^-5 moles
2- where EDTA reacts with caco3 based on 1mole:1mole , No.moles of caco3=9.7*10^-5 moles
3-mass of caco3 = 9.7*10^-5 *100 = 9.7*10^-3 g as caco3
4- the hardness = 9.7/0.035 = 277.1428 ppm

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