Revised Fall 2010 Chemistry 161Lab-K. Marr Lab 7 Prelab Questions Analysis of Ha

Question

Revised Fall 2010 Chemistry 161Lab-K. Marr Lab 7 Prelab Questions Analysis of Hard Water Instructions: Complete the following questions and hand in at the start of your lab period or when instructed by your instructor Name Team No Date Section Show your work with units and correct significant figures for all questions that involve a calculation. Circle numerical answers l. A student prepared a solution of EDTA by placing 3.000g Na,EDTA dihydrate in a 500.0 mL volumetric flask, dissolved the salt in D.I. water and then filled to the mark with D.I. water. a. Calculate the approximate molarity of the EDTA solution. b. Knowing that EDTA is an efflorescent substance, would you expect the actual molarity to be higher or lower than that calculated in part a? Explain. A titration was performed to standardize an EDTA solution. Use the following data to calculate the molarity of the ED solution endpoint. 2· a. 20.00 Lofa standardsolution containing 0.01000 M Ca2+ required 2433mL EDTA to reach adistinct blu

Explanation / Answer

Ans. #a. Moles of Na2EDTA.2H2O = Mass / Molar mass

                                                            = 3.000 g / (372.24 g/ mol)

                                                            = 0.00806 mol

Final volume made upto = 500.0 mL = 0.500 L

Now,

Molarity = Moles of Na2EDTA.2H2O / Final volume in liters

                                    = 0.00806 mol / 0.500 L

                                    = 0.01612 M

#b. In current perspective, “Efflorescence” is the “escaping” of a molecule from the aqueous phase onto the walls of the flask. That is, some EDTA powder will not dissolve properly.

Efflorescence decreases the actual amount of solute is solvated aqueous phase. That is there is less than 3.000 g Na2EDTA.2H2O actually dissolved in the solution –because some will stick to the walls or remains at the surface. Therefore, reduction in the net amount of solute dissolved in aqueous phase, the actual molarity would be lower than the theoretical value.

#c. EDTA forms a 1:1 complex with Ca2+ as follow-

            Ca2+(aq) + EDTA4-(aq) -------> Ca(EDTA)2-

Due to formation of 1:1 complex, the moles of Ca2+ must be equal to that of EDTA4- at the endpoint.

So, using-

            C1V1 (Ca2+ standard) = C2V2 (EDTA solution)

            Or, 0.010 M x 20.0 mL = C2 x 24.33 mL

            Or, C2 = (0.010 M x 20.0 mL) / 24.33 mL

            Hence, C2 = 0.008220 M

Therefore, standardized [EDTA] = 0.008220 M

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