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Calucation of ERRORS - CHEM

  1. Apr 27, 2005 #1
    Calucation of ERRORS - CHEM!!!

    hi im doing one of those determining calcium hardness in swimming pool water here just stuck on a few bits


    1) how do i convert concentration (mol/L) -> ppm ( parts per milliion) example i am given concentration of calcium = 8.595*10^-4 M how do i get parts per milliion?

    2) i am told to calculate the error in calcium hardness found, all i am given are the error values of concentration EDTA in the burette as 0.01022M and the error 0.00003 M and the average titre volume is 8.41ml with error 0.02 cm^3 what am i meant to do with these errors? and how do i calculate the error in calicum hardness? is it something like this:

    eg for first bit 0.00003/0.01022 + 0.02/8.41 dunno please help out
     
  2. jcsd
  3. Apr 27, 2005 #2

    Gokul43201

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    1) The conversion folllows directly from the definitions. Concentration is defined as the number of moles of solute per liter of solvent. How amny particles are there in a mole of the solute ? How many in a liter of water ? And what does ppm mean (explain in words, similarly) ?

    2) See propagation of errors here : http://www.lhup.edu/~dsimanek/errors.htm
    Also, work the examples to make sure you understand. Your text also likely has a chapter right at the beginning, where they talk about error analysis.

    Once you know how errors propagate, the first step towards solving your problem is writing the balanced equation for the titration, and calculating the expression for the end point. Then you may apply the techniques for calculating errors.
     
  4. Apr 27, 2005 #3
    ok i got the part about the errors thats all fine just the conversion from Molarity (mol/l) to parts per million is confusing.

    the concentration of Ca2+ is 8.49*10^04 M how do they go from that to a ppm of Ca2+ 84.900 ???
     
  5. Apr 27, 2005 #4

    xanthym

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    The trick here is to realize that Water Hardness is specified by (ppm CaCO3) and NOT by (ppm Ca+2). However, every Mole Ca+2 in aqueous solution corresponds to 1 original Mole CaCO3. Thus, since "ppm" aqueous solution is equivalent to {mg/(10^6 mg water)}={mg/(10^3 grams water)}={mg/(liter water)}:
    {CaCO3 Concentration (moles/lit)} {Ca+2 Concentration (moles/lit)}
    {ppm CaCO3} = { CaCO3 Concentration (moles/lit)}*{(100 grams)/(Mole CaCO3)}*{1000 mg/gram} =
    = {8.49*10^(-4) M}*{100}*{1000} =
    = (84.9 ppm)


    ~~
     
    Last edited: Apr 28, 2005
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