Volume 23, Number 2, November 2013
Answers
Revision quiz
Anne Hodgson
1 pH goes viral (pp. 2–5)
(a)
+
pH = –log10[H (aq)]or more strictly
+
pH = –log10[H3O (aq)]
i.e. it is the negative log of the concentration of hydrogen ions (or more accurately hydroxonium ions)
in solution. The p in pH stands for potens (the Latin for power). So pH is a measure of the power of
hydroxonium ions in a solution.
(b)
There are four possible dipeptides that can arise from condensation reactions in a mixture of
alanine and serine:
(c)
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2 Reclaiming plastic waste (pp. 6–10)
(a)
(b)
(c)
Other factors that would help to make the process more economical include:
•
Keeping the distances over which the raw materials and products need to be transported
short, in order to minimise transport costs and environmental impact.
•
Finding a cheap (local) source of energy, e.g. combustion or anaerobic digestion of other
forms of (non-recyclable) waste at the same plant.
3 Performing the perfect titration (pp. 11–15)
(a)
KOH(aq) + HNO3(aq) à KNO3(aq) + H2O(l)
(b)
19.5 cm of potassium hydroxide = 19.5/1000 dm (= 0.0195 dm )
3
3
3
To calculate the number of moles of KOH in this volume:
3
19.5/1000 dm × 0.050 mol dm
–3
= 0.000975 mol (or 0.975 × 10
–3
mol)
We can see from the equation that 1 mole of KOH reacts with 1 mole of HNO3, so there were 0.975 ×
–3
3
3
10 mol of HNO3 in the 25.0 cm (0.025 dm ) of solution. So the concentration of the nitric acid is:
–3
3
0.975 × 10 mol/0.025 dm = 0.039 mol dm
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–3
2
4 Atropine (pp. 18–22)
(a)
After the first 4.8 hours there will be half the original mass of the compound remaining (i.e. 5
mg × ½ = 2.5 mg). This process is repeated over every period of 4.8 hours. There are 5 periods of 4.8
hours in 24 hours. So the sequence will go 5 mg, 2.5 mg, 1.25 mg, 0.625 mg, 0.3125 mg until at the
end of the 24-hour period there will only be 0.15625 mg remaining in the body.
(b)
From the molecular formula of quinine, C20H24O2N2, we can calculate the relative molecular
mass:
(20 × 12) + (24 × 1) + (2 × 16) + (2 × 14) = 324
1 mole of quinine would have a mass of 324 g.
81 mg (or 81 × 10
–3
3
–3
g dm
–3
or 0.25 mmol dm
g) in 1 litre (i.e. 1 dm ) = 81 × 10
–3
So the concentration of quinine in the tonic water is:
–3
81 × 10–3 g dm /324 g mol
–1
–3
= 0.25 × 10 mol dm
–3
5 Making alkenes (pp. 30–33)
(a)
(i) (Z)-hex-3-ene-1-ol
(ii) (E)-hex-2- ene-1-ol
(b)
(i)
(ii)
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(c)
(i)
(ii) Butanoic acid (CH3CH2CH2COOH) and hydrogen chloride (HCl, which in solution is
hydrochloric acid).
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