ANSWERS TO EXERCISES AND REVISION QUESTIONSUnit 23
Exercises
23A
The time interval, 8 h, is double the half-life, so that the
calculation is straightforward since eight hours earlier
the concentration would have been four times greater
i.e. 200 mg dm-3. Alternatively, you could re-arrange
equation 23.2 for [A]o. It is worth doing this to check that
you can use your calculator to calculate [A]t/e-kt: the
answer should be the same.
23B
We want [A]o. Rearranging equation 23.2 gives [A]o= [A]
+kt
Since [A] = 60
[A]o= 100 + (6 x 15) = 190 mg of ethanol per 100 cm^3 of
blood. This is dangerously high (see Table 23.2).
Revision questions
23.1 (i)The data is plotted as follows:From the graph, the time it takes for the initial
concentration of poison to fall by one half is about 15 h.
This is true whatever the starting time (for a first-order
decay, the half-life is independent of concentration).Conclusion: t (^1) / 2 = 15 h.
(ii)k = 0.693/15 = 0.0462 h.
[A] 0 = [A]t/ e –kt= 1.02 / e(–0.046216)
1.02/0.4752, or approximately 2.1 mg dm–3.
23.2We will work out the concentrations in mol per
dm–3. Molar mass of ethanol = 46 u.
One microgram ( g) is one millionth of a gram. The
amount of ethanol in 35 g of ethanol is (35 10 –6/46)
mol. The amount of ethanol per dm^3 ofbreathis (35
10 –6/46) 10, i.e. [ethanol]breath= 7.60 10 –6mol
dm–3. One milligram (mg) is one thousandth of a gram.
The amount of ethanol in 80 mg of ethanol is (80
10 –3/46) mol. The amount of ethanol per dm^3 ofbloodis
(80 10 –3/46) 10, i.e. [ethanol]blood= 1.74 10 –2
mol dm–3.
Concentration ratio = 1.74 10 –2/7.60 10 –6 2300
(to three significant figures). Put another way, there are
about 2300 more molecules of ethanol in blood than in
the same volume of breath.
Comment
If we express Henry’s law in the form:
[ethanol]bloodKH [ethanol]breath
The ratio 2300 is sometimes called the Henry’s law
constant for the ethanol in blood–ethanol in breath
system. It is assumed to be independent of temperature
over the narrow ranges encountered in forensic work.
However, in using molar concentration units for both
ethanol in blood and in breath, KHis unitless. This differs
from the units for the Henry’s law constant used in the
book (Unit 11, p. 184; i.e. mol dm–3atm–1) because,
although we expressed the concentration of the solute in
the liquid in the units of mol dm–3, the dissolving gas was
expressed in terms of partial pressure (in atm). Working
with partial pressures is more useful when the solubility
of gases like O 2 , CO 2 , N 2 is being considered. The
conversion factor is:
KH(in mol m–3atm–1) = KH(unitless)/RT
Where R = 0.08205 dm 3 atm mol–1K–1. Here, KH(mol
m–3atm–1) = 2300/(0.08205 298) = 94 mol dm–3
atm–1.
23.3 (i)Reduction is 15 mg per 100 ml of blood per
hour. Therefore two hours earlier the blood alcohol
concentration is estimated to be 79 + 15 + 15 = 129
mg/100 ml.
(ii)Urine: 1.34 80 = 107 mg alcohol per 100 ml of
urine.
Saliva: (109/100) 80 = 87 mg per 100 ml of saliva
23.4 (i)It is converted to an acid, probably ethanoic
acid.
(ii)(a) Two, one in each of the O-CO units.
(b) C 17 H 21 NO 4
(c) 303 u
(d) C=O
(e) parent molecule ion (M+)
23.5At first sight we could choose any two peaks. There
is, however, a subtle reason why we should not choose
the peaks due to the hydroxyl group, OH. Such peaks are
greatly affected by hydrogen bonding. The extent of
hydrogen bonding turns out to be dependent on the
concentration of ethanol, and this interferes with a
simple analysis of peak heights based on concentration.
For the other peaks, such as those due to the C–H
stretch and C–O bend, we would expect the absorbance
and not the transmittance to be proportional to ethanol
concentration, in accordance with the Beer–Lambert law.
23.6 (a)A generous allowance of 6 mg/100 ml is
subtracted from the measurement. The statement would
be that ‘the sample contained not less than 94 mg of
ethanol per 100 ml of blood’.
(b)100 – (3 15) = 100 – 45 = 55 mg/100 ml.
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