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Homework Help: Total Water Hardness

  1. Oct 16, 2007 #1
    I'm having trouble figuring out the total water hardness in ppm of equivalent CaCO3.

    In the lab manual it shows the example equation as
    (150 mg/1L H2O) x (1 L H2O/1000 mL H2O) x (1.00 mL H2O/1.00 g H2O) x (1 g/1000 mg) = 150 ppm of equivalent CaCO3.

    The professor stated the equation is correct, but unclear. Clearly if the equation is calculated out, it does not give that result, but gives 1.50 x 10^-4 ppm of equiv CaCO3. Is there something missing?

    In class the professor showed another way to calculate the problem.
    First start with millimoles, example
    (.2713 mmol EDTA) x (1 mmol ions/1 mmol EDTA) / (.02500 mL H2O) = 1.08 mmol (L H2O)^-1 ions
    Use that number & then use the following formula
    (1.08 mmol ions / 1 L) x (1 mmol CaCO3 equiv. / 1mmol ions) x 100.0 mg CaCO3 equiv / 1 mmol CaCO3 equiv. = 108 ppm of equivalent CaCO3.

    So do either of these even look remotely right? If not, which steps am I missing in the calculations.
  2. jcsd
  3. Oct 19, 2007 #2


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    Gold Member

    Yes and no. Yes, 1.5 X 10^-4 is 150 parts per million but no, it does not calculate out to 1.5 X 10^-4 ppm. Remember, you took an expression, 150 mg/(1000)(1000)mg, or 150 mg / 1,000,000 mg (I like to say 150 ppm) and converted it to 1.5 / 10,000 or 1.5 X 10^-4. The units cancel! You can't just throw a 'ppm' back in! That would actually be 1.5 parts per ten thousand...

    s'claro, no?
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