Chemistry 121

QUANTITATIVE ANALYSIS
Problem Set #5: Redox Titrations/Equilibria; Potentiometry; Spectrophotometry

November 15, 1999

DUE: December 8, 1999 (optional)

  1. Calculate Eo and K for each of the following reactions:

    a) Cr2+ + Fe(s) <----> Fe2+ + Cr(s)
    b) 5MnO2(s) + 4H+ <----> 3Mn2+ + 2MnO4- + 2H2O
    c) CuI(s) <----> Cu+ + I-

  2. Calculate the potential of the half-cell:

    ||Pb(NO3)2 (0.0010 M)|Pb

  3. Calculate the potential of the half-cell:

    ||FeCl3 (0.001 M), FeCl2 (0.010 M)|Pt

  4. Calculate the potential of the following cell:

    Cu(s)|Cu(NO3)2 (0.020 M)||Fe(NO3)2 (0.050 M)| Fe(s)

  5. For M2+ + 2e- --> M(s), Eo = 0.0118v.

    The following cell:

    M|MX2(Sat'd), X-(0.400M)||Std Hydrogen Electrode

    was found to have a potential of 0.204v with the M-electrode as anode. Calculate the Ksp for MX2.

  6. Calculate the equivalence point electrode potential of the system for:

    a) 0.0400N Fe3+ (analyte),
    0.100N Sn2+ (titrant)

    b) 0.0400N HNO2 (analyte),
    0.100N KMnO4(titrant)
    [H+]=0.200 N

    HINT: The reduction half-cells involve:

    MnO4- --> Mn2+
    NO3- --> HNO2

  7. Calculate the potential (versus S.C.E. anode) at each point listed for the titration of 25.00 mL of 0.02000M Cr2+ with 0.01000 M Fe3+: 5.00, 25.00, 50.00, and 100.0 mL.

  8. A solution of KMnO4 was standardized against oxalate:

    5C2O42- + 2MnO4- + 16H3O+ <---> 10CO2 + 2Mn2+ + 24H2O

    and found to be 0.0761N. The solution was then used to determine Mn via the Volhard procedure:

    2MnO4- + 3Mn2+ + 4OH- <---> 5MnO2 + 2H2O

    A 0.543-gram sample required 29.4 mL of the KMnO4. Calculate the %-Mn in the sample.

  9. Nitrite (NO2)- can be determined by oxidation with excess Ce4+, followed by back titration of the unreacted Ce4+. A 4.030-g sample of solid containing only NaNO2 and NaNO3 was dissolved in 500.0 mL. A 25.00-mL sample of this solution was treated with 50.00 mL of 0.1186 M Ce4+ in strong acid for five minutes, and the excess Ce4+ was back-titrated with 31.13 mL of 0.04289 M ferrous ammonium sulfate.

    2Ce4+ + NO2- + H2O ---> 2Ce3+ + NO3- + 2H+

    Ce4+ + Fe2+ ---> Ce3+ + Fe3+

    Calculate the weight percent of NaNO2 in the solid.

  10. What value of absorbance corresponds to 45.0% T?

  11. If a 0.0100 M solution exhibits 45.0% T at some wavelength, what will be the %-T for a 0.0200 M solution of the same substance?

  12. An 8.64 ppm solution of FeSCN2+ has a transmittance of 0.295 at 580 nm when measured in a 1.00-cm pathlength cell. Calculate the molar absorptivity coefficient for the complex at this wavelength.

  13. When I was a boy, Uncle Wilbur let me watch as he analyzed the iron content of runoff from his banana ranch. A 25.0-mL sample was acidified with nitric acid and treated with excess KSCN to form a red complex. (KSCN itself is colorless.) The solution was then diluted to 100.0 mL and put in a variable-pathlength cell. For comparison, a 10.0-mL reference sample of 6.80 x 10-4 M Fe3+ was treated with HNO3 and KSCN and diluted to 50.0 mL. The reference was placed in a cell with a 1.00-cm light path. The runoff sample exhibited the same absorbance as the reference when the pathlength of the runoff cell ws 2.48 cm. What was the concentration of iron in Uncle Wilbur's runoff?

  14. EDTA will steal the cation from the Magnesium/Erio T complex at a pH=10:

    MgIn- + HY3- ----> HIn2- + MgY2-
    (Red)         (Blue)

    The Mg/EDTA chelate absorbs at 225 nm. Predict the shape of the curve for the photometric titration of:

    a) Mg2+ with EDTA at 225 nm.
    b) A Mg2+ solution containing a small amount of Erio T, with EDTA at 640 nm (abs. max for indicator).
    c) The solution in (b) at a wavelength corresponding to the Mg/Erio T chelate absorbance maximum.

Created and copyright by Joel M. Goldberg. Last updated: November 15, 1999


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