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A 50.0 mL aliquot of solution containing 0.450 g of MgSO4 (MW=120.37 g/mole) in 0.500 L required 37.6 mL of EDTA solution for titration. How many milligrams of CaCO3 (MW=100.09 g/mole) will react with 1.00 mL of this EDTA solution?

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Tod Thiel
Tod ThielLv2
21 May 2020

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Related questions

1 A .5118 g sample of CaCO3 is disolve in concentrated HCl(12M). the solution was diluted to 250 mL in a volumetric flask toobtain a solution for standardizing the EDTA titrant.

a How many moles of CaCO3 are used (MW=100.1)? .0051mol

b What is the molarity of the Ca2+ (calcium ion) in the 250mL ofsolution? .020M
c How many moles of Ca2+ are in a 25 mL aliquot of the solution inpart b? .0005 mol

2. 25.00 mL aliquots of the solution from Part 1 are titratedwith EDTA to the Eriochrome Black T end point. A blank containing asmall measured amount of Mg2+ requires 2.60 mL of the EDTA to reachthe end point. An aliquot to which the same amount of Mg2+ is addedrequires 28.55 mL of the EDTA to reach the end point.

How many milliliters of EDTA are needed to titrate the Ca+2 ion inthe aliquot? 27.56mL

How many moles of EDTA are there in the volume obtained in theprevious question? .0051mol

What is the molarity of the EDTA solution? 0.185M

3. A 100.0 mL sample of hard water is titrated with theEDTA solution from Part 2. The same amount of Mg2+ is added aspreviously, and the volume of EDTA required is 22.44mL

What volume of EDTA is used in titrating the Ca2+ in thehard water?

How many moles of EDTA are there in thatvolume?

How many moles of Ca2+ are there in the 100.0 mL volume ofwater?

If the Ca2+ comes from CaCO3, how many moles of CaCO3 arethere in one liter of the water?

How many grams of CaCO3 per liter?

If 1 ppm (part-per-million) CaCO3 = 1 mg per liter, what isthe water hardness in ppm CaCO3?


A 0.4423 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO2ร‚ยท4H2O and removed by filtration. The resulting filtrate and washings were diluted to a total volume of 200.0 mL. A 15.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 35.87 mL of 0.001481 M EDTA. Thiosulfate was used to mask the copper in a second 20.00 mL aliquot. Titration of the lead and zinc in this aliquot required 33.95 mL of the EDTA solution. Finally, cyanide was used to mask the copper and the zinc in a third 25.00 mL aliquot. Titration of the lead in this aliquot required 25.03 mL of the EDTA solution. Determine the percent composition by mass of each metal in the pewter sample

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