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?
Please Please help, any equation?
Thanks
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?
Please Please help, any equation?
Thanks
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.
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?
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.
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.
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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