Aqueous Equilibria
1.
What is the common ion effect?
Consider the following equilibrium:
If NaCl were added to the solution how would this effect the equilibrium of the above reaction? Think terms of Le Châtelier’s principle. The reaction would have to shift to the left in order to return concentrations back to equilibrium proportions. This means that the common ion between the precipitate and the solution cause the precipitate to be less soluble.
2.
What is a buffer solution?
A buffer solution is one in which the solution contains both a weak acid or base and it’s conjugate salt.
For example:
If a solution contained both HOCl and NaOCl – it would be a buffer as it meets both criteria – a weak acid and a salt that contains its conjugate base.
3.
When solving problems for a buffer solution you can bypass the ICE chart and use the Henderson-Hasselbalch equation instead. What is the Henderson-
Hasselbalch equation?
Remember that “p” means “- log of …”
4.
What is an equivalence point?
The point in a titration where
5.
For a strong acid and weak base reaction the equivalence point has a pH < 7.
Why?

At the equivalence point the moles of HA = moles of B. Which would make one assume neutralization and thus a pH = 7. However, because a weak base is being titrated – it’s conjugate that is formed (BH + ) has a measurable amount of strength as an acid. Meaning that after the first reaction has occurred – there will be a second reaction…
This formation of H + leads to an acidic pH indicated by a value less than 7.
6.
For a strong base and weak acid reaction the equivalence point has a pH > 7.
Why?
At the equivalence point the moles of HA = moles of B. Which would make one assume neutralization and thus a pH = 7. However, because a weak acid is being titrated – it’s conjugate that is formed (A
–
) has a measurable amount of strength as a base. Meaning that after the first reaction has occurred – there will be a second reaction…
This formation of OH
– leads to a basic pH indicated by a value greater than 7.
7.
Identify the buffer? a.
HCl and NaClO
4
This is not a buffer because HCl is not a weak acid and ClO
4
– is not the conjugate of HCl. b.
HClO and LiClO
2
This is not a buffer because, although, HClO is a weak acid. The salt
LiClO
2
does not contain its conjugate base.
c.
HF and KF
It is a buffer. HF is a weak acid and KF contains the conjugate base, F
–
. d.
CH
3
NH
2
and CH
3
NH
3
Cl
It is a buffer. CH
3
NH
2
is a weak base and the salt contains its conjugate acid, CH
3
NH
3
+
.

8.
Determine the pH of the following solutions a.
1.00 L solution of 1.00M HNO
2
and 1.50M NaNO
2
. (K a
= 4 x 10 -4 ) b.
The solution in part a after 0.200 moles HCl has been added.
Whenever you add a strong acid or base to a solution that contains weak substances, it is helpful to write a list of the major species in solution.
“Major Species” are substances that are found in the solution in the greatest quantity. Strong substances and salt always fully dissociate into its ions. Weak substances primarily stay intact. Also, remember water will always be a major species as we are dealing with aqueous solutions.
In this solution the major species are:
The strongest acid/base in solution takes priority. In this case, H
+
is the strongest substance in the solution. It is going to react with the best base in solution, which in this case is, NO
2
–
.
Remember that strong substances react to completion that means you would use stoichiometry as your calculation methods and stoichiometry is in terms of units of moles.

Thus the first step in the calculation involves the following reaction along with a stoichiometry chart – similar to the ICE chart.
The change row is determined by the limiting reactant. Because there is a
1:1 mole ratio between reactants, you can determine the limiting reactant by simply looking at the moles of each. The reactant with smallest number of moles is the limiting factor. So you subtract that number of moles from everything on the reactant side… and because there is a 1:1 mole ratio between reactants and products, you add that many moles to the product side.
To determine how to proceed with the calculation of pH, you will need to determine what kind of solution is present after the first reaction. This information is found in the final row of the chart above. Looking at the chart you will notice that you do not have any of the strong acid left, but there is a mixture of weak acid and conjugate base in the solution. This means that the resulting solution is a buffer.
Because this solution is a buffer, we can use the Henderson-Hasselbalch equation in step 2 to calculate the pH of the solution.

You may have noticed that I plugged moles into the equation instead of concentration. The reason that I can do this is because the acid and base are in the same solution… which means that they have the same value for volume.
Remember that So if they are in the volume the L cancels out when you divide one molarity by the other… thus plugging the number of moles results in the same answer as plugging in molarity.
If this is not something that you are totally comfortable with… then it is totally fine to convert moles

molarity by diving the moles by the total volume of solution. And then plugging that value into the Henderson equation. c.
25.0g of NH
3
and 40.0g NH
4
NO
3
in 501.0 mL of water. (K a
= 5.6 x 10
-10
)
This is a buffer solution. NH
3
is a weak base and the salt contains its conjugate acid NH
4
+ . This means that we can use the Henderson-
Hasselbalch equation to solve for the pH. We will just need to convert to the correct units first.
Because the NO
3
–
is just a spectator ion, we are only really interested in the amount of conjugate base this salt provides, we switch from moles of
NH
4
NO
3
to moles of NH
4
+
.
As you can see the volumes cancel out – this illustrates the idea described

in question above – since the volumes will ultimately cancel out, it is valid to simply plug in moles. d.
The solution in part c after 0.02 mol of NaOH has been added.
Remember whenever you add a strong acid or base to a solution that contains weak substances, it might be helpful to write out your list of major species.
The major species in solution are:
The strongest acid/base in solution takes priority. In this case OH
–
is the strongest substance in solution. It is going to react with the best acid in solution, which in this case is, NH
4
+
.
Remember that reactions involving strong substances happen to completion. This means that you are going to be using stoichiometry (and units of mols) to calculate.
Looking at the chart you will notice that you do not have any of the strong acid left, but there is a mixture of weak acid and conjugate base in the solution. This means that the resulting solution is a buffer. Because this solution is a buffer, we can use the Henderson-Hasselbalch equation in step 2 to calculate the pH of the solution.

9.
Calculate the pH when 80. mL of 0.25M KOH is added to 40. mL of 0.50M
HClO. (K a
= 3 x 10
-8
)
When you first start these problems determine whether you have a buffer – that’ll let you know whether or not you may use the Henderson-Hasselbalch equation in the initial step.
This solution is not a buffer as the salt des not contain the conjugate base of
HClO. So the first step would be to determine the major species in solution.
Major species in solution:
Because there is a strong base in solution, it takes priority. So we are going to look at the reaction between hydroxide and the best acid in solution, HClO.
Because this is a strong reaction, we are going to complete a stoichiometry calculation which means we need to determine the moles of HClO and OH
–
in solution.
Now that we have the moles of each in solution we can plug into the stoichiometry chart.

If we now take a look at what is left in solution after the strong reaction has taken place, we notice that the only species left are H
2
O and ClO
–
. This is the equivalence point of the titration (where moles of acid equal moles of base) so we are only left with the conjugate in solution after the reaction has run to completion. This means that the next reaction we have to look at is the equilibrium reaction between ClO
–
(a conjugate base) and H
2
O. Because this is an equilibrium reaction we will use an ICE chart to solve.
ICE charts are in terms of molarity so we will need to begin by converting from moles of ClO
– to [ClO
–
]. Remember that we will have to divide the moles of ClO – by the sum of the two volumes of solution added.
Now we can plug into the ICE chart.
Remember that this is K b
because we are looking at a reaction of a base producing OH
–
. So we will need to solve for K b
using the relationship:
Looking up the K a
value of the weak acid HClO, 3.0 x 10
–8
we can plug in and solve.

Because K b
is such a small number we can make the small K, small x approximation and we have the equation:
Solving for x we get
10.
What is buffer capacity?
The amount of acid or base a buffer can absorb without having a significant change in pH.
11.
Consider
0.1 M NaF and 0.1M HF
1.0M NaF and 1.0M HF
0.01M NaF and 0.01M HF a.
Which has the highest pH?
They all have the same pH because they all have the same K a
value and ratio of [base]:[acid]. So in a buffer the pH doesn’t depend on how concentrated the acid and base are; it depends on their ratio to one another.
b.
Which has the highest buffer capacity?
1.0 M NaF and 1.0 M HF has the highest buffer capacity as it has a 1:1 acid to base ratio with the greatest concentration of each.

12.
Consider a solution that contains both NH
3
and NH
4
NO
3
. If a solution has a pH = 8.5, calculate the ratio of
K a
= 5.6 x 10
-10
Looking at the solution you will notice that it contains a weak base and the salt contains its conjugate acid. That means that this solution is a buffer and we can use the Henderson - Hasselbalch equation.
All we have to do is plug the information provided into the equation and solve for the .
13.
What volumes of 0.22M CH
3
COOH and 0.46M NaCH
3
COO must be mixed to prepare a 1.00L solution buffered at pH = 5.00? (K a
= 1.76 x 10
-5
)
Once again we start by determining whether or not this solution is a buffer. As it contains CH
3
COOH ( a weak acid) and a salt that contains its conjugate base (CH
3
COO
–
), this solution would be a buffer. This means that we can use the Henderson-Hasselbalch equation – plugging in all the relevant data.

Remember:
Since the volume of the solution is equal to 1.00L we know that the sum of the volume of base added plus the volume of acid added is equal to 1.00 L.
Thus we can set up the following equation
14.
Draw the titration curves of the following reactions a.
Strong Acid/Strong Base

b.
Strong Base/Weak Acid c.
Strong Acid/Weak Base

15.
Calculate the pH for 100.0 mL of 0.400M CH
3
COOH with 300. mL of
0.200M NaOH.
16.
Do precipitation reactions reach a measurable equilibrium?
Yes they do. K sp
is a measure of this equilibrium.
17.
How does the common ion effect affect the solubility of a precipated solid?
The common ion effect is when 2 solutions have a common ion. The addition of NaCl to a solution that contains an AgCl precipitate demonstrates a case of having a common ion. The Cl
–
is an ion found in both - hence “common ion”.
Now we have to consider how the presence of this common ion would affect the solubility of AgCl. Let’s look at the equilibrium reaction
Now thinking in term so f Le Châtelier’s principle… how would equilibrium shift upon addition of Cl
–
. Equilibrium would shift away from the added substance (left, in this case). Shifting left would mean that the AgCl would dissociate less or become less soluble.
Thus, the common ion effect is to decrease the solubility of a precipitate salt.
18.
Calculate the solubility for the following solutions
You handle these problems as you have for every equilibrium calculation thus far – an ICE chart. In this case you are looking at the equilibrium reaction for the solids dissociation into its component ions.
When the question asks for the “solubility” of the salt, it is asking you to solve for the x value in the ICE chart.
a.
Ag
2
CO
3
. (K sp
= 8.1 x 10 -12 )
Remember that because solids do not affect equilibrium, we do not include any values for them in the ICE chart calculation. They are irrelevant most

of the time.
Now all we have to do is plug our equilibrium values into the K sp expression for this equilibrium reaction. b.
Ag
2
CO
3
in 0.10M AgNO
3
.
The variation in this question, as compared to the previous one, is that we have to take the common ion effect into consideration. Both compounds contain Ag
+
. Remember that this is supposed to limit the solubility of the precipitate in solution. The only difference between is that there will now be an initial concentration of Ag
-+
in the ICE chart calculation. c.
Ag
2
CO
3
in 0.20M Na
2
CO
3