Professor Key

Chemistry 3600
Final Examination
December 13 1999

Directions: There are 5 questions, each worth equal points. Some are more difficult than others. Read over all questions before you start to answer. Answering the harder questions last will insure that you obtain the maximum score. You must show your work and pertinent formula's to get full credit. The final number answer is only ten percent of the total score.

Useful formulas:

1. Gravimetric Analysis

A 0.5664 g ore sample is dissolved in nitric acid and then filtered. The aluminum is present in solution as Al3+. The solution is made basic with ammonium hydroxide, NH4OH, and the aluminum hydroxide, Al(OH)3 (FW 78.004), precipitates. This gel is filtered in a porous glass crucible, rinsed with dilute ammonium hydroxide, ignited, cooled in a desiccator, and weighed. The resulting alumina, Al2O3 (FW 101.94), weighed 0.1605 g.

  1. Answer the following questions;
  1. Why is the solution filtered after acid dissolution?

    2. To remove insoluble matrix material (like sand, etc.)
  2. Why rinse with ammonium hydroxide solution?

    3. To avoid peptization or loss of precipitate. Ammonium hydroxide is a volatile electrolyte.
  3. What chemical transformation takes place during ignition? (Show the balanced chemical equation)

    4. 2Al(OH)3 -> Al2O3 + 3H2O
  4. Why use a desiccator during cooling?

To avoid weight gain by adsorption of atmospheric water
  1. Calculate the weight percent Al (AW 26.9815) in the sample.

2. Acid-Base Titrations/ Advanced Acid-Base Chemistry

20 mL of a diprotic acid, H2A, is titrated with 0.1 F NaOH titrant. The resulting titration curve is shown below.

  1. Label the two points where the solution has the maximum buffer capacity by A and C, and label first and second equivalence points as B and D.
  2. Estimate the formal concentration of H2A from the graph.

  3. There are simple expressions for the pH at points A, B, and C. Give the equations below.

  4. Estimate Ka1 and Ka2 for this acid from your estimate of pH at points A, B, and C.

  5. At what indicator pH would you want your color-change indicator to have in order to titrate to the second equivalence point?

At pH=10

3. Potentiometry and Redox Titrations

  1. A chemist wants to know the molar solubility of nickel sulfide (NiS). An experiment is performed where the potential of a nickel electrode in sulfide is solution is measured relative to a standard hydrogen electrode (SHE). By buffering the pH of solution, sulfide activity was AS2-=0.001 The Ni/NiS electrode potential was measured at -0.899 V. Using the Nernst equation, calculate the molar solubility of NiS.

    2. The molar solubility is moles of Ni2+ or S2- in solution for the salt. Since the complex is 1:1

    The molar solubility is 10-12
  2. 25 mL of a Fe2+ analyte is titrated with 0.02 M Ce4+ titrant. Both are in 1 F HCl. The formal reduction potentials are

1. What is the potential half-way to the equivalence point?

Half way to eq. pt., [Fe2+]=[Fe3+], E=EFe0'=0.70 V

2. What is the potential at the equivalence point?

3. What potential is produced at twice the volume of titrant required to reach the equivalence point?

Here, [Ce3+]=[Ce4+], E=ECe0'=1.23 V

4. Spectroscopy

  1. Define the following terms

    2. 1 - %T Fraction transmitted light power in percent

    2 – molar absorptivity Beer's Law absorption coefficient expressed in per molar per cm.

    3 – luminescence Emitted light

    4 – Beer’s law A=ebc

    5 - absorption spectrum absorbance as a function of wavelength

  2. 0.570 g of a steel sample is dissolved in acid. The manganese is oxidized to permanganate, MnO4- (FW 118.936), using potassium persulfate and is diluted to 100 mL. A few mL of solution is placed in a 1 cm pathlength cell and the transmission is found to be 30% at 525 nm. The molar absorptivity of permanganate is 2.24´ 103 at 525 nm. Calculate weight percent manganese (Mn, AW 54.938) in the steel.

 

5. Analytical Separations

  1. A solute with a partition coefficient of KD=1.0 is extracted from 10 mL of phase 1 into phase 2.

    2. 1 - What volume of phase 2 is needed to extract 75% of the solute in a single extraction?

    2 - What percent of the solute is extracted upon three equal-volume extractions using 10 mL each?

  2. The van Deemter equation describes the "height equivalent to a theoretical plate" in terms of mobile phase flow rate and three phenomenological constants. Give the names for and/or describe the effects thought to be responsible for these three terms.

    3. A: Eddy diffusion: due to several paths through stationary phase

    B: Longitudinal diffusion: due to mass diffusion in mobile phase

    C: Dis-equilibrium: slow kinetics causes non-equilibrium between stationary and mobile phases

  3. Circle the correct response

1 - Is it better to | minimize| maximize | HETP when performing chromatographic separations.

2 - Resolution | increases | decreases | with increasing HETP.

3 - Increasing column length will | increase | decrease | peak width.

4 - Increasing column length will | increase | decrease | elution time.

5 - Increasing column length will | increase | decrease | resolution.

6 - The amount of material is proportional to peak | height | width | area | in chromatography.