d University of Min xO FreeEDTA as Y4- Free EDTA as Y4 Welcome to Myu /mod/ibis/

Question

d University of Min xO FreeEDTA as Y4- Free EDTA as Y4 Welcome to Myu /mod/ibis/view.php?id-4540342 ota, Twin Cities -CHEM 2101- Fall17-BANTZ Activities and Due Dates Ch12 10/31/2017 05:00 PM 4 35/70 Calculator Grade Print -d Periodic Table stion 1 of 7 Hardness in groundwater is due to the presence of metal ions, primarily Mg2 and Ca2'. Hardness is generally reported as ppm CaCO3. To measure water hardness Map a chelating agent, in the presence of the indicator eriochrome black T, symbolized here as In Eriochrome black T, a weaker chelating agent than EDTA, is red in the presence of Ca2 and turns blue when Ca2' is removed red Calln"++ EDTA-> Ca (EDTA)», In A 50.00-mL sample of groundwater is titrated with 0.0900 M EDTA. Assume that Ca accounts for all of the hardness in the groundwater. If 12.50 mL of EDTA is required to titrate the 50.00-mL sample, what is the hardness of the groundwater in molarity and in parts per million of CaCOs by mass? Number 0.0225 M CaCo, ppm Caco, Tools . Previous Give Up & View Solution e Check Answer Next Ext Hint about us careers privacy polilcy terms of use contact us he F5 F6 F3 3 4 5 6

Explanation / Answer

we know that one mole of Ca+2 will react with one mole of EDTA.

Moles of EDTA used = molarity of EDTA X volume of EDTA in L = 0.09 X 12.50 / 1000 = 0.001125 moles

so moles of Ca+2 present = 0.001125 moles

Concentration of Ca+2 = Moles / Volume in litres = 0.001125 X 1000 / 50 = 0.0225 Molar

now

1ppm = mg of solute / Litre of solution

Mass of CaCO3 present (g)/ Litre = molarity X molecular weight = 0.0225 X 100 = 2.25 grams / L

Mass of CaCO3 in mg = 2.25 X 1000 mg / L = 2250ppm or 2.25 X 10^3 ppm

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