Asked by Anonymous
                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.
            
        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.
Answers
                    Answered by
            DrBob222
            
    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?
    
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?
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.