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?

 

[Blueshift5]

I would suggest checking the units and conversions in both equations to ensure they are correct. It is also important to note that the first equation is using milligrams of water (mg/1L H2O), while the second equation is using milliliters of water (.02500 mL H2O). This could be causing the discrepancy in results.

I would also recommend double checking the values used in the equations to make sure they are accurate. If there are any uncertainties or errors in the values, it could affect the final result.

Additionally, it may be helpful to consult with your professor or a colleague to clarify any confusion or misunderstandings in the equations. They may be able to provide further guidance and insight on the calculations.

Overall, it is important to carefully check all calculations and units to ensure accurate and precise results. In science, attention to detail is crucial in obtaining reliable data.