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21 Cards in this Set
- Front
- Back
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According to the DPIC method, what might you do in the second step (the P part) of solving
a problem? (Give some detail here, not just a single word.) |
Plan an approach to solving the problem (this may include the use of a block diagram or an
outline); identify assumptions in the approach. |
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For the numbers 116.0, 97.9, 114.2, 106.8, 108.3 find the mean and the 95% confidence
interval (NOT the 90% CI in the original practice problem). [Values of t are listed in Table 4-2 on p. 58 of your text.] |
Mean = 108.6, s = 7.14, t = 2.776 (4 d.o.f) CI = ± 2.776 × s / 51/2 = ± 8.85 = ± 8.8
Therefore, 108.6 ± 8.8 (or 109 ± 9) @95% CL |
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Could any of the values listed in the previous question be rejected at the 90% confidence
level based on the Q test? [Q values are given in Table 4-6 on p.75.] |
Range = 116.0 − 97.9 = 18.1 largest gap = 108.3 − 97.9 = 10.4
Q = 10.4/18.1 = 0.57 Q-test value for 5 data is 0.64 0.57 < 0.64, so 97.9 cannot be rejected with 90% confidence. |
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Suppose 23 data points were used in calculating the Abest fit@ straight line calibration
(y = mx + b) for a buret by the least squares method. How many degrees of freedom are there for this calculation? a) 20 b) 21 c) 22 d) 23 e) 24 |
Two degrees of freedom are lost in determining m and b from the data, so
N – 2 = 21 |
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If a quadratic fit were made (y = ax2 + bx + c), what would be the number of degrees of
freedom? a) 20 b) 21 c) 22 d) 23 e) 24 |
In this case three constants must be determined, leaving 20 degrees of freedom.
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What is the difference between the titration reaction and the cell reactions in a potentiometric
titration? For which of these reactions do we measure the potential? [You may wish to aid your explanation with examples from Lab 2.] |
The titration reaction is between the two reactants in solution; it rapidly goes to
equilibrium with each addition of titrant. The cell reactions are between the species in solution and the reference electrode; in potentiometry, these reactions do not proceed to any significant extent. We measure the cell potential between the solution and the reference electrode. For example, in Lab 2, the titration reaction Ce(IV) + Fe(II) → Ce(III) + Fe(III) goes to equilibrium with every addition of Ce(IV) to the solution, and therefore its potential is zero at every step in the titration. However, the reaction of the iron half-cell is not at equilibrium with the calomel half-cell, and the potential for the iron-calomel reaction is what we measure. As the ratio of Fe(III) to Fe(II) changes during the titration, so does the potential measured relative to calomel. Alternatively, we can say that we measure the potential of the cerium half-cell relative to calomel during the titration; because the cerium and iron are in equilibrium with each other, they must each give the same potential relative to the calomel half-cell. |
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What is a junction potential?
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This is a voltage that develops across any interface of dissimilar phases. For example, at
the point of contact between two different metals (the basis on which your lab thermometers work) or across the liquid junction between two half cells. In your electrochemistry experiments, a junction potential arises across the interface between your solution and the calomel reference electrode. |
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Ferrous ion in acidic solution is readily oxidized by oxygen in the solution to ferric ion.
Why then can we get away with using a ferrous compound as our standard in Lab 2? Check ALL correct answers below—only one may be correct, all may be correct, or some number in between. (Note, the order of the answers was determined by a random number generator.) a) Calculation of the equilibrium concentrations shows that, if we keep our FEDS solution capped, the partial pressure of oxygen in the air that is trapped in the flask will be too low to cause much oxidation. b.)Calculation of the equilibrium concentrations shows that the virtually no ferrous ion can be present when it has reached equilibrium with the air. Fortunately, the rate at which oxygen dissolves into solution is too slow to cause an appreciable problem over the course of a single lab period. c) Any ferric ion produced is reduced back to ferrous ion by the sulfuric acid. d) Calculation of the equilibrium concentrations shows that oxygen will oxidize only a little of the ferrous ion and then the ferric ion concentration, though small, will be high enough that the reaction will reach equilibrium. |
a.) Keeping the solution capped is a good strategy, but even so, the amount of oxygen
present in the airspace in the flask is sufficient to oxidize a few percent of the FEDS. b.) Correct; see problem S-11. c.) No. Sulfuric acid is an oxidizing agent, not a reducing agent. d.) No. |
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4 - What is EEN (E-naught-prime), and why is it preferred over EE (E-naught) in biochemical
measurements? |
E°′ is the formal reduction potential under conditions of pH = 7 (and otherwise standard
state conditions—i.e., activities of all other species are one), whereas E° is the standard reduction potential under conditions of pH = 0. Biological systems are usually close to pH = 7, so E°′ is more useful for biochemical measurements. |
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What is a selectivity coefficient for an ion selective electrode? Is it best to have a small or a
large selectivity coefficient? |
No electrode is perfectly selective for only the species of interest; there are always a few
other species that will interfere with the measurement by also producing a response by the electrode. The selectivity coefficient gives the relative response of an interfering ion to that of the ion of interest. Therefore, it is best to have a selectivity coefficient for the interfering ion that is small so that the signal it produces is small compared to that from the analyte ion. |
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2. What would be the pH of 0.00100 F HCl that is also 0.010 F in KNO3?
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HCl is a strong acid that completely dissociates in water, so the [H+] = 0.00100 M.
However, pH = -log (aH+); therefore, as ionic strength of the solution increases, activity coefficients decrease, and the activity of the H+ will be less than its concentration, meaning that the pH would be higher (less acidic) in the presence of KNO3. Doing the EDHE calculation gives pH 3.04. |
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3. In the multiple standard addition experiment of Lab 3, students fit four data points (one
unknown and three additions of standard) to a straight line (y = mx + b). How many degrees of freedom are there for the standard deviation about this fit (sy)? |
4 – 2 = 2. Two degrees of freedom are lost in determining m and b from the data.
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4. Which of the choices below is the correct charge balance for CaCl2 in water if the species
from the salt are Ca2+, Cl-, CaCl+, and CaOH+. |
[Cl-] + [OH-] = 2[Ca2+] + [CaCl+] + [CaOH+] + [H+]
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5. Which of the following bases would be the most suitable for preparing a buffer of pH 9.00?
a) ammonia (Kb = 1.8 x 10-5) b) aniline (Kb = 4 x 10-10) c) hydrazine (Kb = 1.05 x 10-6) d) pyridine (Kb = 1.6 x 10-9) |
The correct answer is ammonia.
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1 - A spectrum was collected with an integration time of 0.5 s; the signal-to-noise ratio of the
smallest peak in the spectrum was estimated to be 2. If ensemble (or signal) averaging were employed to improve the S/N to 10, how many spectra would need to be taken? |
S/N ~ n1/2 S/N2 / S/N1 = n2
1/2 / n1 1/2 10 / 2 = n2 1/2 / 11/2 n2 = (10 / 2)2 = 25 |
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2 - If the selectivity coefficient for Na+ with your glass pH electrode is kH, Na = 10-12, at what pH
will the presence of 0.1 M Na+ cause an error of 10% in the measured hydrogen ion concentration? [Neglect activity coefficients for this problem.] |
Assuming activities are equal to concentrations, the response of a pH electrode is
proportional to log([H+] + kH, Na × [Na+]). We want to determine the pH at which kH, Na × [Na+] = 0.1 × [H+] Solving for [H+] gives: [H+] = 10 × kH, Na × [Na+] = 10 × 10−12 × 0.1 = 10-12 pH = 12 |
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4. The acid dissociation constants for citric acid (a triprotic acid) are
K1 = 7.44 x 10-4 K2 = 1.73 H 10-5 K3 = 4.02 10-7 The pH of a lemon is about 2.3. What is (are) the predominant form(s) of citric acid in lemons? |
At a pH of 2.3, [H+] = 10–2.3 = 5.01 × 10–3 M. Using the general form for calculating α
for a polyprotic acid, we can calculate α0, α1, α2, α3. [H+]3 = 1.26 × 10–7 [H+]2 K1 = 1.87 × 10–8 [H+]K1K2 = 6.45 × 10–11 K1K2K3 = 5.17 × 10–15 For H3Cit α0 = 0.87 For H2Cit– α1 = 0.13 For HCit2– α2 = 4.5 × 10–4 For Cit3– α3 = 3.6 × 10–8 H3Cit and H2Cit– are the predominant forms of citric acid at a pH of 2.3 |
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1 - When 1.06 mmol of pentanol and 1.53 mmol of 1-hexanol were separated by gas
chromatography, they gave relative peak areas of 922 and 1570 units. When 0.57 mmol of pentanol was added to an unknown containing hexanol, the relative peak areas were 843:816 (pentanol:hexanol). How much hexanol did the unknown contain? [Assume that the addition of pentanol did not significantly affect the sample volume or mass—unlike in our lab]. |
0.47 mmol
See solution to problem 24-B in the text book. |
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2 - What is temperature programming in gas chromatography, and what advantage does it have?
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Temperature programming is changing the temperature of the column from low to high
during the course of a run. It allows good separation of the most volatile, earliest eluting compounds at the beginning of the run, but decreases the elution time and sharpens the peaks of later eluting compounds. In this way it solves the Ageneral elution problem.@ |
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3 - A mixture of benzene, toluene, and methane was injected into a gas chromatograph.
Methane was unretained and gave a sharp peak at 42 s. Benzene eluted at 251 s, and toluene eluted at 333 s. What are the capacity factors for benzene and toluene? |
k' = (tr – tm)/tm = (251 – 42)/42 = 5.0 for benzene
= (333 – 42)/42 = 6.9 for toluene |
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4 - What is the “dynamic range” of a detector?
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It is the range of concentrations over which the signal output of the detector changes.
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