Example
Calculate the volume of 14.8 M NH3 and the
weight of NH4Cl (FW = 53.5) you would have to
take to prepare 100 mL of a buffer at pH 10.00 if
the final salt concentration is to be 0.200 M. kb =
1.75x10-5
Solution
The key to solving any problem is the equilibrium
of substances in solution.
1
Here, we have ammonia and ammonium which are
combined in the equation:
NH3 + H2O D NH4+ + OHKb = [NH4+][OH-]/[NH3]
We are aware of the pH which means that we can find [OH] and we are given the concentration of NH4+ as 0.200 M.
Therefore:
pOH = 14 – 10 = 4
[OH-] = 10-4 M
Now we can solve the equilibrium relation to find [NH3]
1.75x10-5 = (0.200x 10-4)/[NH3]
[NH3] = 1.14 M
2
We need 100 mL of 1.14 M to be prepared from 14.8 M so
we have
mmol ammonia needed = 1.14x100 = 114 mmol
mL ammonia = mmol/molarity = 114/14.8 = 7.7 mL
Or simply, MiVi = MfVf
14.8* VmL = 1.14 * 100
VmL = 7.7 mL
The weight of NH4Cl can also be found as the volume
and molarity are given in the problem
Mmol NH4Cl = 0.200 x 100 = 20.0 mmol
Mg NH4Cl = 20.0 x 53.5 = 1070 mg
3
Example
How many g ammonium chloride (FW = 53.5) and
how many mL of 3.0 M NaOH should be added
to 200 mL water and diluted to 500 mL to
prepare a buffer at pH 9.5 and a salt
concentration of 0.10 M.
Solution
Once again, the key to solving this problem is the
equilibrium of substances in solution.
4
Here, we have ammonia and ammonium which are
combined in the equation
NH3 + H2O D NH4+ + OHKb = [NH4+][OH-]/[NH3]
We are aware of the pH which means that we can
find [OH-] and we are given the concentration of
NH4+ as 0.10 M. Therefore:
pOH = 14 – 9.5 = 4.5
[OH-] = 10-4.5 = 3.2x10-5 M
5
Now we can solve the equilibrium relation to find
[NH3]
1.75x10-5 = 0.10 x 3.2x10-5/[NH3]
[NH3] = 0.18 M
mmol NH3 = 0.18 x 500 = 90 mmol
mmol NaOH = mmol ammonia
VmL NaOH = mmol/molarity = 90/3.0 = 30 mL
mmol NH4+ = 0.10 x 500 = 50 mmol
Total mmol salt = mmol ammonia + mmol
ammonium = 90 + 50 = 140 mmol
mg NH4Cl = 140 x 53.5 = 7.49x103 mg = 7.49 g
6
Solutions of Polyprotic Acids
Polyprotic acids are weak acids, except sulfuric
acid where the dissociation of the first proton is
complete, which partially dissociate in water in a
multi step equilibria where hydrogen ions are
produced in each step. Examples include
carbonic, oxalic maleic, phosphoric, etc. A
general simplification in the calculation of pH of
such acids is to compare ka1 and ka2 where
,usually ka1/ka2 is a large value (>104) and thus
equilibria other than the first dissociation step
can be ignored.
7
Let us look at the following example:
H3PO4 D H+ + H2PO4ka1 = 1.1 x 10-2
H2PO4- D H+ + HPO42ka2 = 7.5 x 10-8
HPO42- D H+ + PO43ka3 = 4.8 x 10-13
[H+] = [H+]H3PO4 + [H+]H2PO4- + [H+]HPO4Looking at the values of the acid dissociation
constants for the three steps, it is obvious that
the first step occurs about 106 times greater than
the second and thus the amount of protons in
the second step is negligible ( [H+]H2PO4- )
compared with the first.
8
In addition, the third step comes from the second
step and since the second step contributes a
negligible amount of H+ we can also neglect the
third step ([H+]HPO4-) or any other consecutive
steps. Therefore, only the first equilibrium
contributes to the H+ concentration.
9
Example
Find the pH of a 0.10 M H2SO4 (ka2 = 1.7*10-2)
Solution
The first dissociation is 100% complete, therefore,
we have:
[H+] = [HSO4-] = 0.10 M from first dissociation.
The second dissociation is as follows:
HSO4- D H+ + SO42-
10
1.7*10-2 = x(0.10 + x)/(0.10 – x)
assume that 0.10 >> x
x = 1.7*10-2
Relative Error = (1.7x10-2/0.10) * 100% = 17%
Therefore, the assumption is invalid and the equation must
be solved by the quadratic equation.
11
Example
Find the pH of a 0.10 M H3PO4 solution.
Solution
H3PO4 D H+ + H2PO4H2PO4- D H+ + HPO42HPO42- D H+ + PO43-
ka1 = 1.1 x 10-2
ka2 = 7.5 x 10-8
ka3 = 4.8 x 10-13
Since ka1 >> ka2 (ka1/ka2 > 104) the amount of H+ from the
second and consecutive equilibria is negligible if
compared to that coming from the first equilibrium.
12
Therefore, we can say that we only have:
H3PO4 D H+ + H2PO4ka1 = 1.1 x 10-2
13
Ka1 = x * x/(0.10 – x)
Assume 0.10>>x since ka1 is small (!!!)
1.1*10-2 = x2/0.10
x = 0.033
Relative error = (0.033/0.10) x 100 = 33%
The assumption is invalid and thus we have to use
the quadratic equation. If we solve the quadratic
equation we get:
X = 0.028
Therefore, [H+] = 0.028 M
pH = 1.55
14
Example
Find the pH of a 0.10 M H2CO3 solution. Ka1 =
4.3x10-7, ka2 = 4.8x10-11
Solution
We have the following equilibria
H2CO3 D H+ + HCO3ka1 = 4.3 x 10-7
HCO3- D H+ + CO32ka2 = 4.8 x 10-11
Since ka1 is much greater than ka2, we can neglect
the H+ from the second step
15
H2CO3 D H+ + HCO3-
ka1 = 4.3 x 10-7
Ka1 = x * x/(0.10 – x)
Assume 0.10>>x since ka1 is small
4.3*10-7= x2/0.10
x = 2.1x10-4
Relative error = (2.1x10-4/0.10) x 100 = 0.21%
The assumption is valid and [H+] = 2.1x10-4 M
pH = 3.68
16
If we would like to calculate the amount of H+
coming from the second equilibrium
([H+]second step = [CO32-]) we substitute 2.1x10-4
for [H+] as follows:
Ka2 = [H+][CO32-]/[HCO3-]
But from the first step we have
[H+] = [HCO3-]
ka2 = [CO32-] = 4.8x10-11 = [H+]second step
17
Therefore we are justified to omit second
dissociation
where
the
hydrogen
ion
concentration obtained from the first step
(2.1x10-4 M) is much greater than the [H+]
obtained from the second dissociation (4.8x10-11
M).
18