- #1
kuahji
- 394
- 2
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.
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.