# Wessex Water - report as of November 07



## Matt Holbrook-Bull (19 Nov 2007)

pH- 7.3
Ammonium - 0.01 mg NH4/l
Nitrite - 0.01 mg NO2/l
Nitrate â€“ 39 mg NO3/l             <-------------------------- Im dubious as to this one
Hardness - 265 mg CaCO3/l
Free Chlorine â€“ 0.25 mg Cl/l
Total Chlorine â€“ 0.28 mg Cl/l
Magnesium â€“ 1.9 mg Mg/l

Fresh from the horses mouth, as it were


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## JamesC (19 Nov 2007)

1.9mg/l Mg = 7.8 mg/l MgCO3 as CaCO3

Hardness - MgCO3 gives CaCO3
265 - 7.8 = 257.2 mg/l CaCO3 = 103 mg/l Ca

If you follow that you have a Ca : Mg ratio of 54 to 1, or in other words not much Mg.

James


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## Matt Holbrook-Bull (19 Nov 2007)

JamesC said:
			
		

> 1.9mg/l Mg = 7.8 mg/l MgCO3
> 
> Hardness - MgCO3 gives CaCO3
> 265 - 7.8 = 257.2 mg/l CaCO3 = 103 mg/l Ca
> ...



that was my assumption as well.. i shall carry on dosing it


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## beeky (19 Nov 2007)

JamesC said:
			
		

> 1.9mg/l Mg = 7.8 mg/l MgCO3



How is this worked out? Are atomic weights used to get the proportions? I'd be interested in the maths for this (my school chemistry was along time ago!)


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## JamesC (19 Nov 2007)

Yes atomic weights are used.

This is how I arrived at the answer:
Atomic weights: Mg=24.3, C=12 and O=16
So MgCO3 we have 24.3 + 12 + (16x3) = 84.3

To get mass MgCO3 from Mg:
1.9 x (84.3/24.3) = 6.6g

But as we are talking in equivalents of CaCO3 and not MgCO3 we have to convert.
So (18/15)x6.6 = 7.9g

Not quite the 7.8g as above but I just use the quick calculation of multiplying by 4.1
1.9 x 4.1 = 7.8

James


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## beeky (19 Nov 2007)

Thankyou!!


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