Example
In the reaction
A+BDC+D
If 0.2 mol of A is mixed with 0.5 mol of B in 1.0 L, find the
concentration of A, B, C, and D. The equilibrium constant of the
reaction is 0.30.
Solution
It is clear that we do not have a clearly small or high equilibrium
constant which suggests that both products and reactants will
coexist appreciably in the reaction mixture at equilibrium.
Therefore, A and B concentration will decrease by x and C and
D will be formed in a concentration equals x for each.
1
K = [C][D]/[A][B]
K = (x)(x)/(0.20 – x)(0.50 – x)
Now from the value of the equilibrium constant we can
not assume that 0.20 >> x. However, let us do that
and find out what we get
0.30 = x2/(0.20x0.50)
x = 0.173 M
2
Now check the validity of the assumption by calculating the
relative error
Relative Error = (0.173/0.20) x 100 = 87%
The assumption is invalid and we must not neglect x in
comparison with 0.20, therefore the equilibrium should be
as follows
K = (x)(x)/(0.20 – x)(0.50 – x)
0.30 = x2 /(0.10 – 0.70x +x2)
Rearrangement of the relation above results in
0.70x2 + 0.21x -0.030 = 0
3
For this type of equation, we can use the quadratic
equation to get a solution for x
The quadratic equation solution of the equation ax2 +
bx + c = 0 is
x = {(- b + (b2 – 4 ac)1/2)/2a}
Substitution gives
x = {(- 0.21 + (0.212 – 4 *0.7*(-0.03))1/2)/2*0.7}
x = 0.11 M
Therefore, we have
[A] = 0.20 – 0.11 = 0.09 M
[B] = 0.50 – 0.11 = 0.39 M
[C] = [D] = 0.11 M
4
The Common Ion Effect
We have discussed earlier that changing the
concentration of a product or reactant has a
significant effect on equilibrium. Addition of a
reactant forces the reaction to proceed forward.
Addition of a product forces the reaction in the
backward direction, a consequence of Le
Chatelier's principle. A common ion is usually a
product ion which means that reactants
concentration will increase on the expense of
products. Let us look at the following example:
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Example
Calculate the equilibrium concentrations of A, B,
and AB in a 0.10 M AB (weak electrolyte) in
presence of 0.20 M B. The equilibrium constant
is 3.0x10-6.
AB D A + B
Solution
Be aware that we have previously solved this
problem but without addition of extra amount of
B. Therefore, compare the answers.
6
The equilibrium constant is very small therefore
only tiny amounts of AB will dissociate to yield A
and B. The initial concentration of B is 0.20,
therefore the equilibrium concentration of B will
be 0.20 + x. The following table summarizes the
situation before and after equilibrium:
7
Writing the equilibrium constant expression we get:
K = x (0.20 + x)/(0.10 – x)
Since K is very small we can assume that 0.1
>> x. This gives:
3.0x10-6 = 0.20x/0.1
x = 1.5x10-6 M
Let us calculate the relative error to check for the
validity of our assumption
8
Relative error = (1.5x10-6/0.10) x 100 = 1.5x10-3 %
The error is less than 5% therefore our
assumption is OK.
The concentrations of the different species
in present of 0.20 M common ion will be:
[A] = 1.5x10-6 M
[B] = 0.20 + 1.5x10-6 ~ 0.20 M
[AB] = 0.10 – 1.5x10-6 ~ 0.10
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Systematic Approach to
Equilibrium Calculations
This Type of approach to equilibrium calculations uses
what is called mass balance and charge balance
concepts. These will be discussed below shortly. By
mass balance we mean that when we dissolve an
amount of dissociating substance, the analytical
concentration of that substance will be equal to the sum
of the concentrations of the different species derived
from it ( mass conservation). Also, solutions are neutral
and thus the concentration of the negative charges is
equal to the concentration of the positive ones (charge
balance, from electroneutrality).
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Example
Write mass and charge balance equations for 0.100 M HOAc
in water.
Solution
First, we write the equilibrium:
HOAc D H+ + OAcH2O D H+ + OHNow we deal with the problem and find the charge balance
by placing the positive ions on one side and the negative
ions on the other side.
[H+] = [OH-] + [OAc-]
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The mass balance equation represents the
concentration of added HOAc where the
analytical concentration added to water was
divided into two parts one represents
undissociated HOAc and the other represents
dissociated HOAc (OAc-). Therefore:
0.100 = CHOAc , the mass balance equation is:
0.100 = [HOAc] + [OAc-]
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Example
Write a mass and charge balance equations for
0.100 M [Ag(NH3)2]Cl.
Solution
First, write all possible equilibria which might take
place in solution. These would be:
[Ag(NH3)2]Cl  Ag(NH3)2+ + Cl13
Ag(NH3)2+ D Ag(NH3)+ +NH3
Ag(NH3)+ D Ag+ + NH3
NH3 + H2O D NH4+ + OHH2O D H+ + OHNow we can write the charge balance equation:
[H+] + [Ag+] + [NH4+] + [Ag(NH3)2+] + [Ag(NH3)+] = [OH-] +
[Cl-]
The mass balance equation can then be written where all
species containing ammonia came from the original
compound [Ag(NH3)2+]Cl.
14
Also, the concentration of ammonia in
Ag(NH3)2+ is twice its concentration since
each mol of Ag(NH3)2+ contains two moles of
ammonia. Therefore, the mass balance
equation for ammonia will be:
0.200 M = 2[Ag(NH3)2+] + [Ag(NH3)+] + [NH3] +
[NH4+]
Another mass balance equation for silver can be
written as
[Ag+] + [Ag(NH3)2+] + [Ag(NH3)+] = [Cl-] = 0.1 M
15
Example
Write mass and charge balance equations for a
solution of H2S.
Solution
First let us set up the equations representing
equilibrium of H2S in water
H2S D HS- + H+
HS- D H+ + S2H2O D H+ + OH16
The charge balance equation can be written as
before but considering that the charge
concentration of S2- is twice as the
concentration of S2-. This means that each
mole of S2- contains two moles of charges. The
charge balance equation is:
[H+] = [OH-] + [HS-] + 2[S2-]
Mass balance equation will be:
CH2S = [H2S] + [HS-] + [S2-]
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Example
Write a mass and charge balance equations for
CdS in solution.
Solution
CdS D Cd2+ + S2S2- + H2O D HS- + OHHS- + H2O D H2S + OHH2O D H+ + OH18
The charge balance equation can be simply
derived from the equilibria above but
taking in consideration that charge
concentration on cadmium is twice the
concentration of cadmium and the charge
concentration on sulfide is also twice the
sulfide concentration. The equation will be:
[H+] + 2[Cd2+] = [OH-] + [HS-] + 2[S2-]
Mass balance equation:
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[Cd2+] = [H2S] + [HS-] + [S2-]