Calculate Total Water Hardness (ppm of Equiv. CaCO3)

[kuahji]

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.

 

[chemisttree]

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?

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...

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.

s'claro, no?