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crimsonbadger829Lv1

28 Sep 2019

A 50.0 mL sample containing Cd2 and Mn2 was treated with 49.6 mL of0.0600 M EDTA. Titration of the excess unreacted EDTA required 13.7mL of 0.0210 M Ca2 . The Cd2 was displaced from EDTA by theaddition of an excess of CN%u2013. Titration of the newly freedEDTA required 18.0 mL of 0.0210 M Ca2 . What were the molarities ofCd2 and Mn2 in the original solution?

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Beverley SmithLv2

28 Sep 2019

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A student wishes to analyze a seawater sample to determine the amounts of Na+ , Ca2+ and Mg2+ present. She begins with 50.00 mL sample of seawater that she dilutes to 500.00 mL. She removes a 50.00 mL aliquot of the diluted solution, which she analyzes for total cations present by exchanging the cations for H+ on an ion exchange column and titrating the liberated acid with 0.5000 M NaOH. The amount of NaOH required to reach the endpoint was 22.48 mL.

The student then titrates another 50.00 mL aliquot of her diluted seawater solution with 15.31 mL of 0.0100 M EDTA solution at a pH of 10 to the Erio T endpoint.

The third titration, performed on a 50.00 mL aliquot of the diluted seawater solution involves precipitating Mg2+ as the hydroxide and titrating the Ca2+ remaining in solution with 0.010 M EDTA solution. Only 5.15 mL of EDTA were required to reach the end point.

Summary of titration data:

Original seawater = 50.00 mL; diluted seawater = 500.00 mL

Vol. of titration aliquot = 50.00 mL

First titration: [NaOH] = 0.5000 M; V NaOH = 22.48 mL

Second titration: [EDTA] = 0.0100 M; VEDTA = 15.31 mL

Third titration: [EDTA] = 0.0100 M; VEDTA = 5.15 mL

From the titration data, calculate the following (show work):

A). mmol of Ca2+ present in the third 50.00 mL titration aliquot - ans.__________

B). mmol of Ca2+ present in the original 50.00 mL seawater sample - ans.__________

C). mg of Ca2+ present in the original 50.00 mL seawater sample - ans.__________

D). mmol of Ca2+ and Mg2+ present in the second 50.00 mL titration aliquot - ans.__________

E). mmol of Mg2+ present in the 50.00 mL titration aliquot - ans.__________

F). mmol of Mg2+ present in in the original 50.00 mL seawater sample - ans.__________

G). mg of Mg2+ present in in the original 50.00 mL seawater sample - ans.__________

H). meq of Ca2+ and Mg2+ present in the 50.00 mL titration aliquot - ans.__________

I). total meq of Na+ , Ca2+ and Mg2+ present in the 50.00 mL titration aliquot - ans.__________

J). meq of Na+ in the 50.00 mL titration aliquot - ans.__________

K). meq of Na+ present in the original 50.00 mL seawater sample - ans.__________

L). mg of Na+ present in in the original 50.00 mL seawater sample - ans.__________

A 1.000-mL sample of unknown containing Co2+ and Ni2+ was treated with 25.00 mL of 0.03872 M EDTA. Back titration with 0.02127 M Zn2+ at pH 5 required 23.54 mL to reach the xylenol orange endpoint. A 2.000-mL sample of unknown was passed through an ion exchange column that retards Co2+ more than Ni2+. The Ni2+ that passed through the column was treated with 25.00 mL of 0.03872 M EDTA and required 25.63 mL of 0.02127 M Zn2+ for back titration. The Co2+ emerged from the column later. It, too, was treated with 25.00 mL of 0.03872 M EDTA. How many milliliters of 0.02127 M Zn2+ will be required for back titration?

A mixture of AlCl3, MnCl2, and CuCl2 is analyzed to determine its composition. First a 15.00 mL sample of the solution is titrated by 0.01153 F Na2H2Y*2H2O, requiring 43.28 mL to reach the EBT end point. Then excess NH4F is added to this sample to displace Al3+ from its EDTA complex. The liberated EDTA required 10.92 mL of a 0.01037 M Ca2+ solution for complete titration. Then a fresh 25.00 mL sample of the original solution was treated with triethanolamine to mask the aluminum and manganese ions. The remaining metal required 24.31 mL of the 0.01153 F Na2H2Y*2H2O solution to reach equivalence. What is the concentration of the three metal ions in the original solution? â

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