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Brown's Advanced A Level Chemistry Revision Notes
TheoreticalPhysical
Advanced Level
Chemistry Equilibria Chemical Equilibrium Revision Notes PART 6.4
Equilibria Part 6.4 Buffer solution pH calculations
How do you calculate the pH of a buffer
solution?
How do you calculate the quantities required to make up a
buffer solution of a desired pH?
Chemical Equilibrium Notes Parts 5 & 6 Index
6.4
Buffer pH calculations
theoretical calculation of
a buffer
solution
-
6.4.1 Calculations
involving a buffer made from a weak acid and its salt with a strong
base.
-
Consider the
mixture is made from a monobasic weak acid HA and an alkali metal
salt M+A
-
it is reasonable
to assume for simple approximate calculations that ..
-
[A(aq)]
= [salt(aq)] since salt fully ionised and M+
is a spectator ion, and
-
[HA(aq)]equilib.,
=
[HA(aq)]initial since little of the weak acid
is ionised.
-
Therefore the
weak acid Ka expression is ...
-
(i) Ka =
|
[H+(aq)] [A(aq)] |
mol dm3 |
[HA(aq)]
|
-
becomes
-
(ii)
Ka =
|
[H+(aq)] [salt(aq)] |
mol dm3 |
[acid(aq)]
|
-
therefore: [H+(aq)]
= Ka [acid(aq)] /
[salt(aq)] mol dm3, and taking log10
of both sides gives
-
(iii) pHbuffer
= log10(Ka x [acid(aq)] /
[salt(aq)])
-
or (iv) pHbuffer
= pKa + log10([acid(aq)] /
[salt(aq)])
-
or (v)
pHbuffer
=
pKa +
log10([salt(aq)]
/ [acid(aq)])
-
which is how the
equation is usually quoted, sometimes called the Henderson Equation.
-
Note: For a
given conjugate pair (HA and A), the pH of
the buffer is determined by the acid/salt ratio, though the more
concentrated the buffer, the greater its capacity to neutralise
larger amounts of added/formed in a reaction medium.
-
Another point is
how
to choose which weak acid is best for a desired buffer?
-
The useful range of a
buffer is decided by the weak acid's Ka and the ratio of
the salt and weak acid concentrations.
-
The buffer will be most
useful when the ratio [salt]/[acid] is equal to one i.e. when both
active ingredients are at their maximum concentrations at no expense
to the other by the principles of related chemical equilibrium, if
you increase one concentration you would decrease the other.
-
Therefore the maximum
buffer capacity is when [salt] = [acid]
-
Now Ka =
[H+(aq)] when [salt] = [acid] in equation
(i) or (ii) above
-
therefore taking log10
of both sides gives ...
-
pKa = pH
when [salt] = [acid], because log10(1) = 0 in equations
(iv) or (v)
-
and this simple
mathematical argument gives the necessary guidance ...
-
So for example,
supposing you wanted a buffer to cover a pH range of 4.5 to 6.0,
-
and your choice of weak
acid pKa's was 2.8, 4.2, 5.5 and 6.5,
-
you would choose the
weak acid with a pKa of 5.5 because its pKa
is well in the desired pH range.
-
You would then formulate
it with its sodium (or potassium) salt note that sodium ion and
potassium ion salts are usually used because they have virtually no
acidic or basic character to complicate matters e.g. the hydrated
ions Na+(aq) and K+(aq)
do not donate protons in the way, for example, the hexaaqua ions of
aluminium can ...
-
[Al(H2O)6]2+(aq)
+ H2O(l)
[Al(H2O)5(OH)]+(aq)
+ H3O+(aq)
-
because the polarising power of
the central metal ions, Na+ or K+ is too small
to effect this process.
-
6.4.2 Buffer
calculations
-
Calculating
the pH of a buffer calculating amounts of salt and acid needed
-
Buffer
calculation example 6.4.2b
-
In what ratio
should a 0.30
mol dm3 of ethanoic acid be mixed with a 0.30
mol dm3 solution of sodium ethanoate to give a buffer
solution of pH 5.6?
-
[H+(aq)]
= 10pH = 105.6 = 2.51 x 106
mol dm3
-
Ka
= [H+(aq)] [salt(aq)]/[acid(aq)]
-
[salt]/[acid] = Ka/[H+(aq)] = 1.74 x
105/2.51 x
106 = 6.93
-
Therefore volume
ratio is 6.93 : 1 for salt : acid,
e.g. 6.93 cm3 of 0.30M sodium ethanoate is mixed with 1.0 cm3
of 0.30 M ethanoic acid to give a buffer solution of pH 5.6.
-
Note that the pH
is determined by the ratio of concentrations, but the buffering
capacity of the solution can be increased by increasing the
concentrations of both components in the same molar concentration
ratio.
-
Buffer
calculation example 6.4.2c
-
Buffer
calculation example 6.4.2d
-
Buffer calculation example
6.4.2e
-
Using the Henderson
equation
-
pHbuffer
= pKa + log10([salt(aq)]
/ [acid(aq)])
-
Calculate the pH of
buffer solution made by mixing together 100 cm3 of 0.100M
ethanoic acid and 50 cm3 of 0.400M sodium ethanoate,
-
Now because the volumes
are not equal, the real concentrations in the mixture must be worked
out.
-
The total volume is 150
cm3, therefore the dilutions are given by
-
[acid] = 0.1 x 100/150 =
0.06667
-
[salt] = 0.4 x 50/150 =
0.1333
-
Substituting in the
Henderson Equation gives
-
pHbuffer
= log10(1.74 x 105) + log10(0.1333/0.0667)
-
pHbuffer
= log10(1.74 x 105) + log10(2)
-
pHbuffer
= 4.76 + 0.3010
-
pHbuffer
= 5.06
-
-
Chemical Equilibrium Notes Parts 5 & 6 Index
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Chemical Equilibrium Notes Parts 5 & 6 Index |