Wessex Water - report as of November 07

Discussion in 'Water Chemistry' started by Matt Holbrook-Bull, 19 Nov 2007.

  1. Matt Holbrook-Bull

    Matt Holbrook-Bull Founder

    Messages:
    963
    Location:
    Dorset, UK
    pH- 7.3
    Ammonium - 0.01 mg NH4/l
    Nitrite - 0.01 mg NO2/l
    Nitrate – 39 mg NO3/l <-------------------------- Im dubious as to this one
    Hardness - 265 mg CaCO3/l
    Free Chlorine – 0.25 mg Cl/l
    Total Chlorine – 0.28 mg Cl/l
    Magnesium – 1.9 mg Mg/l

    Fresh from the horses mouth, as it were
     
  2. JamesC

    JamesC Member

    Messages:
    1,276
    Location:
    Bexley, Kent
    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
     
  3. Matt Holbrook-Bull

    Matt Holbrook-Bull Founder

    Messages:
    963
    Location:
    Dorset, UK
    that was my assumption as well.. i shall carry on dosing it
     
  4. beeky

    beeky Member

    Messages:
    879
    Location:
    Chippenham, Wiltshire
    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!)
     
  5. JamesC

    JamesC Member

    Messages:
    1,276
    Location:
    Bexley, Kent
    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
     
  6. beeky

    beeky Member

    Messages:
    879
    Location:
    Chippenham, Wiltshire
    Thankyou!!
     

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