6. A 0.3205 g sample of CaCO, was dissolved in HCl and the resulting solution di

Question

6. A 0.3205 g sample of CaCO, was dissolved in HCl and the resulting solution diluted to 250.0 mL in a volumetric fask A 25.00 mL sample of the solution required 18.75 ml. of an EDTA solution for titration the Eriochrome Black T end point. a. How many moles of CaCO, were used? b. What is the concentration of Ca (in mol/L) in the 250.0 ml. of CaCl, solution? How many moles of Ca? are contained in a 25.00 mL. sample? c. d. How many moles of EDTA are contained in the 18.75 ml used for titration? e. What is the concentration (in mol/L) of the EDTA solution? Copy this answer into postlab question #7 29

Explanation / Answer

a) Molar mass of CaCO3 = 40 + 12 + 3 * 16 = 100 gm/mol

Number of moles of CaCO3 = Mass/molar mass = 0.3205/100 = 3.205 * 10^(-3) moles

b) Concentration of Ca(+2) in mol/L

Molarity = Number of moles of solute/Volume of solution in L = 3.205 * 10^(-3)/0.250 = 1.282 * 10^(-2) M

c) Number of moles of Ca(+2) in 25 mL sample = Molarity * Volume of solution in L

=> 1.282 * 10^(-2) * 25/1000 = 3.205 * 10^(-4)

d) Number of moles of EDTA = Number of moles of Ca(+2)

Number of moles of EDTA = 3.205 * 10^(-4) moles

e) Molarity of EDTA = number of moles of EDTA/volume of solution in L = 3.205 * 10^(-4)/(18.75 * 10^(-3)) = 1.7093 * 10^(-2) M

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