question archive The reaction is assumed to be CaCO3(aq) + EDTA4-(aq) → CaEDTA2-(aq) + CO32-(aq) Disodium EDTA dihydrate is almost a primary standard1
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The reaction is assumed to be
CaCO3(aq) + EDTA4-(aq) → CaEDTA2-(aq) + CO32-(aq)
Disodium EDTA dihydrate is almost a primary standard1. When pure, its molar mass is 372.2 g/mol but at normal heat and humidity it has a 0.3% excess number of waters of hydration, giving it a molar mass of 372.3 g/mol. For all but the most exacting work, this is reproducible enough to be used as a primary standard. We'll use it.
Calculate the mass of disodium EDTA dihydrate needed to prepare 250 mL of 0.005 M solution.
Show your work completely and clearly.
Use your mass of EDTA to calculate the true molarity of your EDTA solution.
g EDTA and g/mol EDTA → mol EDTA
mol EDTA and volume of EDTA solution → mol/L EDTA
Do this for all of your titrations: use the true concentration of EDTA (the concentration you just calculated), your experimental mL of EDTA and your mL of sample solution to calculate the sample's hardness in mg/L CaCO3. Show your work only for the first titration. Do the other titrations on scrap paper and just show the results. If you discard a titration then write something to justify your decision (ex: obviously overshot endpoint, failed Grubbs test, spilled, bug landed in beaker, exploded).
mL EDTA and mol/L EDTA → mmol EDTA → mmol CaCO3 → mg CaCO3
mg CaCO3 and volume of sample → mg/L of CaCO3 in sample
Calculate the mean and 95% confidence interval of the sample's hardness, showing your work.
Finally, summarize your final results by reporting your sample's ID, hardness and confidence interval using the correct number of significant figures like this:
Sample #9 was found to be 1.23(±0.04) mg/L CaCO3 at the 95% level of confidence
25 mL of unknown sample was used
The volumes of titrations are
26.39
26.33
26.38
mass disodium EDTA dihydrate = 0.4780