In an experiment for determining water hardness, 75.00 ml of a water sample required 13.03 ml of an EDTA solution that is 0.009242 M. What is the ppm CaCO3 in this sample?
Could someone please tell me if this is correct?
Total hardness, ppm CaCO3 = mg CaCO3 / L
= 13.03 ml x 1.00 mg CaCO3/ml EDTA x 1000 ml/L
75 ml
= 173.733 mg CaCO3/L
= 173.7 ppm
Thanks for your help.
I don't know where you obtained 1 mg CaCO3/mL.
0.01303 x 0.009242M = ? mols CaCO3.
?mols CaCO3 x molar mass CaCO3 = grams CaCO3and that's in 75 mL. Convert to mg/L for ppm. Something like 160 ppm?
To determine the ppm (parts per million) CaCO3 in the water sample, you need to use the following equation:
ppm CaCO3 = (mg CaCO3 / L) x 1000
Let's calculate the ppm CaCO3 in the water sample using the given information.
Step 1: Calculate the mg CaCO3 used in the titration.
Given that 13.03 ml of the 0.009242 M EDTA solution is used in the titration, we can calculate the mg of CaCO3 used as follows:
mg CaCO3 = 13.03 ml x (1.00 mg CaCO3/ml EDTA)
Step 2: Calculate the ppm CaCO3 in the water sample.
Substituting the value calculated in step 1 into the equation:
ppm CaCO3 = (mg CaCO3 / L) x 1000
Given that the sample volume is 75.00 ml, we have:
ppm CaCO3 = (mg CaCO3 / 75 ml) x 1000
Now, substituting the previously calculated value of mg CaCO3:
ppm CaCO3 = (13.03 ml x 1.00 mg CaCO3/ml EDTA) / 75 ml x 1000
Calculating this expression, we get:
ppm CaCO3 = 173.733 mg CaCO3/L
Rounding this value to one decimal place, we have:
ppm CaCO3 ≈ 173.7 ppm
So, based on the calculations, the correct answer is indeed 173.7 ppm CaCO3.
Remember to double-check your calculations and units to ensure accuracy.