Maths and Chemistry for Biologists

1
Maths and Chemistry for Biologists
2
Chemistry Examples 2
This section of the course gives worked examples
of the material covered in Chemistry 3, 4, 5 and
6
3
Acids and bases
Where necessary take the pKa of acetic acid as
4.76
Calculate the hydrogen ion concentrations H
of solutions of pH a) 2.5 b) 3.5 pH is
defined as -log H so H is obtained from
antilog (-pH). We obtain a) 3.16 mM (3.16 x
10-3 M) and b) 0.316 mM (3.16 x 10-4 M). Note
that the pH values vary by unity so the H
values differ by 10-fold and that the higher the
pH the lower the H
4
Calculate the pH of a 1.5 mM solution of HCl
assuming that the acid is completely
dissociated HCl is completely dissociated in
water so H 1.5 mM (1.5 x 10-3 M). So pH
-log (1.5 x 10-3) 2.82
5
Calculate the pH of a 1.5 mM solution of acetic
acid using the relationship pH ½pKa - ½log
c where c is the molar concentration pH
(4.76/2) - (log 1.5 x 10-3)/2 2.38 - (-2.82)/2
2.38 1.41 3.79 (Compare this value with
that obtained in the previous example for the
same concentration of a strong acid)
6
Prove the relationship in the previous
example Suppose that the concentration of acid
is c M and that dissociation occurs to yield x M
of H and x M of A-. Then Ka But x is
small compared with c (since the acid is weak)
and hence c - x ? c and the equation becomes Ka
or Ka Taking logs of both
sides gives log Ka 2log H - log c or
-2log H -log Ka - log c and pH ½ pKa -
½ log c
7
Buffers
In the following take the pKa value of acetic
acid as 4.76 and the second pKa value (pKa2) of
phosphoric acid as 7.20
Central to all problems is the Henderson-
Hasselbalch equation i.e.
8
Calculate the pH of a buffer solution made by
dissolving 0.1 mol of acetic acid and 0.025 mol
of NaOH in a volume of 1 L In this case we have
made the salt by partial neutralization of acetic
acid with the strong base NaOH. The salt
concentration salt will be equal to that of the
NaOH added i.e. 0.025 M (0.025 mol in 1 L) and
the free acid concentration acid will be 0.075
M (of the 0.1 mol originally added, 0.025 mol has
been neutralised by the NaOH leaving 0.075 mol
in 1 L, i.e. 0.075 M) contd
9
Hence pH 4.76 log 4.76
(-0.48) 4.28 Note that in cases like this a
useful alternative form of the Henderson-Hasselbal
ch equation is where acid is the
original concentration of acid and base is the
concentration of base added
10
Calculate how many mol of NaOH would need to be
added to 500 ml of 0.2 M NaH2PO4 to make a
buffer of pH 7.00 The pKa2 is 7.20 and the
initial acid concentration is 0.2 M. Therefore
using the eqn on the previous slide
hence
contd
11
and
Multiply top and bottom by (0.2 base) to
give base 0.63(0.2 base) or base
0.126 0.63 x base Rearrange to get 1.63 x
base 0.126 Hence base 0.077 M Hence we
need 0.077 mol of NaOH/L or 0.0385 mol in 500 ml
12
What would be the change in pH if 0.1 ml of 0.1 M
NaOH were added to 9.9 ml of the buffer made in
the previous example? What would be the change
in pH if this amount of NaOH were added to 9.9
ml of water? The addition increases the base
concentration by 1 mM (0.1 ml added to 9.9 ml
i.e. dilution of 100-fold. Hence concentration
of added NaOH is 0.001 M or 1 mM). The total NaOH
concentration is now 0.078 M. So That is,
the pH increases by 0.01 units
contd
13
If 0.1 ml of 0.1 M NaOH were added to 9.9 ml of
water then HO- NaOH 10-3 M . We know
that the product HOH- 10-14 M2 Hence H
10-14/10-3 10-11 M Hence pH -log 10-11
11 The pH increases by 4 units Comparison
with the first part shows how good buffers are at
resisting pH changes
14
Rates of reactions
A first order reaction in which a reactant A is
converted into a product B is found to have a
rate constant k1 0.012 min-1. Starting with a
concentration of A of 1 M, calculate the
concentrations of A remaining and of B formed at
15 min intervals over a period of 2 hours.
Present the results graphically. From the graph
estimate the time taken for the concentration of
A to decrease to 50 of its initial value (the
half life of the reaction). contd
15
The concentration of A at any time t (At ) will
be given by At A0e-kt where A0 1 M
and k 0.012 min-1. Values of 0, 15, 30 etc
were entered into an EXCEL work sheet for the
time values and a function was set up to
calculate A0e-kt. Values of Bt were
calculated from Bt 1 - At. EXCEL was then used
to construct the plot shown from which it can be
seen that t½ is about 60 min
16
Equilibria
Calculate the ?G0 value for a set of reactions
the overall result of which is C6H12O6 6O2
30ADP 30Pi ? 6CO2 36H2O (note that Pi is an
abbreviation for phosphate) We have that
C6H12O6 6O2 ? 6CO2 6H2O ?G0 -3,000
kJ/mol 30ADP 30Pi ? 30ATP 30H2O ?G0
30x31kJ/mol (Note that since the ?G0 value for
the hydrolysis of ATP is -31 kJ/mol then that for
its synthesis is 31 kJ/mol). contd
17
The overall process is the sum of these two and
hence the ?G0 value is also the sum of these two
i.e. ?G0 -3,000 930 -2,070 kJ/mol This
is very large and negative which in turn tells us
that it must be possible to devise a set of
reactions the overall result of which is the
spontaneous oxidation of a mol of glucose coupled
to the synthesis of 30 moles of ATP. What it
does not tell us is what those reactions are