Maths and Chemistry for
Biologists
Chemistry Examples 2
This section of the course gives worked
examples of the material covered in
Chemistry 3, 4, 5 and 6
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+]
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
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)
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-.
2
x
Then Ka =
c-x
But x is small compared with c (since the acid is weak)
and hence c - x  c and the equation becomes
2
x
Ka =
c
[H  ]2
or Ka =
c
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
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 HendersonHasselbalch equation i.e.
[salt]
pH  pK a  log
[acid]
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
Hence
0
.
025
pH = 4.76 + log
= 4.76 + (-0.48) = 4.28
0.075
Note that in cases like this a useful alternative
form of the Henderson-Hasselbalch equation is
[base]
pH  pKa  log
[acid] - [base]
where [acid] is the original concentration of acid
and [base] is the concentration of base added
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
[base]
7.00  7.20  log
0.2 - [base]
hence
[base]
[base]
log
 - 0.20 and
 antilog (-0.20)
0.2 - [base]
0.2 - [base]
contd
and
[base]
 0.63
0.2 - [base]
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
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
0.078
pH  7.20  log
 7.01
0.20 - 0.078
That is, the pH increases by 0.01 units
contd
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 [H+][OH-] = 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
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
Progress curve for reaction
1
0.9
0.8
concentration (M)
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
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
20
40
60
80
tim e (m in)
100
120
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
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