Name: Professor Key
Chemistry 3600
Final Examination
December 14 1998
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:
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Useful data:
Values of Q @ 90% confidence
Q |
0.94 |
0.74 |
0.64 |
0.56 |
0.51 |
0.47 |
0.44 |
0.41 |
N |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Values of Student's t @ 95% confidence
t |
12.71 |
4.303 |
3.182 |
2.776 |
2.571 |
2.447 |
2.365 |
2.306 |
2.262 |
2.228 |
d.f. |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
Values of F @ 90% confidence
Degrees |
Degrees of freedom s1 |
||||
2 |
3 |
4 |
5 |
6 |
|
2 |
9.00 |
9.16 |
9.24 |
9.29 |
9.33 |
3 |
5.46 |
5.39 |
5.34 |
5.31 |
5.28 |
4 |
4.32 |
4.19 |
4.11 |
4.05 |
4.01 |
5 |
3.78 |
3.62 |
3.52 |
3.45 |
3.90 |
6 |
3.46 |
3.29 |
3.18 |
3.11 |
3.05 |
1. Statistics
3.274, 3.258, 3.265, 3.258, 3.350, 3.483
Qcalc = gap/range
Qcalc = (3.483-3.350)/(3.483-3.258) = 0.591
Qtable = 0.56
Since Qtable< Qcalc datum can be rejected from the data set
B) Based on the above results, calculate the 95% confidence interval and report the statistically probable mean value at 95% confidence. Show your work using appropriate formulas.
The table t value for 5-1=4 degrees of freedom is 2.776 Using the mean
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and standard deviation

the confidence interval is

C) Calculate the variances of the data in part A with and without the 3.483 value. Apply the F-test to determine if there is a significant difference in the two variances (@ 90% confidence). Does your F-test result agree with the Q test in part A?
Without the rejected datum, there are N=6 points, with a mean and standard deviation of 3.315 and s1=0.08958 The calculated F is
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The whole data set, used to calculate s1, has 6-1=5 degrees of freedom. The data set with one point rejected, used to calculate s2, has 5-1=4 degrees of freedom. Ftable = 4.05 Since Ftable < Fcalc the two standard deviations are statistically different. This result agrees with that of the Q test. The later indicated that one datum could have come from a different measurement population. Including this datum results in an overall standard deviation that is statistically different than that obtained with the rejected datum.
D) Gravimetric analysis of nickel by precipitation with dimethylglyoxime gave the linear (y=mx+b) calibration shown below. Regression values for the slope and intercept, and their standard deviations, are given in the inset box. Calculate the weight % nickel, and estimate the error using propagation formulas, for a sample with precipitate weight 5.120± 0.001 g.
From y=mx+b, solve for x (wt. %) as a function of y (wt)
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The error estimate is obtained from the slope and intercept standard deviations (0.020 and 0.35) and the weight error (0.001). Using the above equation and the error propagation rules formulas, for a sample with precipitate weight 5.120± 0.001 g.

2. Acid-Base Titrations/Advanced Acid-Base Chemistry
The amino acid, methionine, is diprotic with a carboxylic acid pKa1=2.20 and an ammonium pKa2=9.05. In this problem, 10 mL of a 0.1 F solution of this amino acid is titrated with 0.1 F NaOH titrant.
(a) Below is
the resulting titration curve. Indicate the two points where the solution has the maximum
buffer capacity by A and C, and indicate the first and second equivalence
points as B and D. Label the Titrant Volume axis markers.
(b) There are simple expressions for the pH at points A, B, and C. What are the pH at these points? Give the equations below and put numbers on the pH axis on the above plot.

(c) At what pH would you want your indicator color change to occur at in order to titrate to the first equivalence point?
Around pH=6
3. Potentiometry
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Start with the Nernst equation for copper reduction
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Next, solve the Ksp equation for Cu+

Substitution into the Nernst equation yields

B. The ideal potential response for the solid-state fluoride ion selective electrode is given by
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But the solid-state conductor also responds to hydroxide when [OH-]» 0.1 M, and hydrogen fluoride is a moderate acid with pKa=3.17. Give reasons for two pH-dependent sources of error associated with fluoride measurements
4. Spectroscopy


1 - %T Percent transmission. The fraction of light passing through the sample
2 excited state A quantum state other than the ground state.
3 luminescence Light emission
4 Beers law A=ebC relates sample absorbance to molar absorption coefficient, pathlength and concentration.
5 - absorption spectrum Wavelength or photon energy dependent absorbance. Plot of absorbance versus wavelength or photon energy.
5. Analytical Separations
1 - What volume of phase 2 is needed to extract 75% of the solute in a single extraction?
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2 - What percent of the solute is extracted upon three equal-volume extractions using 10 mL each?
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Since m1=m2, each extraction removes 50%. The amount remaining after 3 extractions is (0.5)3=0.125. The amount removed is (1-0.125)´ 100%=87.5%
| Solute equilibrates between mobile phase and film of liquid attached to stationary phase | Partition |
| Different-sized solutes penetrate voids in stationary phase to different extents | Size |
| Solute equilibrates between mobile phase and surface of stationary phase | Adsorption |
| Mobile phase ions attach to counter-ions covalently attached to stationary phase | Ion |
| Solute is attracted to specific groups covalently attached to stationary phase | Affinity |
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1 - Give the names for and/or describe the effects thought to be responsible for A, B, and C.
A: Eddy Diffusion: due to different paths through column
B: Longitudinal Diffusion: due to mass diffusion along the length of the column
C: Dis-equilibrium: due to dynamic effects between stationary and mobile phases
2 - Is it better to minimize or maximize H.E.T.P. when performing chromatographic separations?
Minimizing the HETP increases the number of theoretical plates resulting in better separations.