1. Ch. 17– Additional Aspects of Aqueous Equilibria
The Common Ion Effect:
The dissociation, (or solubility), of a weak electrolyte is decreased by the addition of a strong electrolyte that has an ion in common with the weak electrolyte.
Consider the equilibrium established when acetic acid, HC2H3O2, is added to water…
HC2H3O2(aq) H+(aq) + C2H3O2−(aq)
What do you think would happen to the equilibrium if we were to dissolve NaC2H3O2 , (a strong electrolyte), into the solution?
(Remember Le Châtelier’s principle!)

2. Common Ion Effect
As expected, the equilibrium would shift to the left.
The [H+] would decrease by reacting with the extra acetate ions to produce HC2H3O2 .
The [HC2H3O2] increases as the equilibrium is re-established.
Since NaC2H3O2 is a strong electrolyte and HC2H3O2 is a weak electrolyte, the [C2H3O2−] is based on the amount of salt added to the solution not on the weak acid’s dissociation.
Let’s do a practice problem
What is the pH of a M HNO2 solution, (Ka = 4.5 x 10−4) when M of KNO2 is added? ( HNO2  H NO2− )
[HNO2]
[H+]
[NO2−]
Initial
Change
Equilibrium
0.1 M
0.085 M
– x
+ x
+ x
0.085 – x
+ x x

3. Common Ion Effect 4.5 x 10−4 = x[0.10]/[0.085]
[HNO2]
[H+]
[NO2−]
Initial
Change
Equilibrium
0.10
0.085 M
– x
+ x
+ x
0.085 – x
+ x x
Ka = 4.5 x 10−4 = [H+][NO2]/[HNO2] = [x][ x]/[0.085 – x]
Assuming “x” is going to be a very small answer…
4.5 x 10−4 = x[0.10]/[0.085]
Solving for “x”…x = M
This is equal to [H+], so…
pH = −(log[ ]) = 3.42
F.Y.I.—Without the salt, pH ≈ 2.2

4. Common Ion Effect
The previous example used the dissociation of a weak acid. But the ionization of a weak base can also be decreased by the addition of a strong electrolyte…
H2O + NH3  NH OH−
If we were to add NH4Cl to the solution, the equilibrium would shift to the left, and the pH would decrease as OH− reacts with the additional NH4+ in the solution.

5. Buffered Solutions (a.k.a.— “Buffers”)
The solutions we have been describing can resist drastic changes in pH as a strong acid, (H+), or a strong base, (OH−), is added.
These types of solutions are called buffers.
A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X– )… HX(aq)  H+(aq) + X–(aq)
So a buffer contains both:
An acidic species (to neutralize OH–) and
A basic species (to neutralize H+).
Buffers have many important applications in the lab and in medicine…your blood is a buffered solution that functions in the pH range of 7.35 – 7.45

6. Buffered Solutions (a.k.a.— “Buffers”)
When a small amount of OH– is added to the buffer, the OH– reacts with HX to produce X– and water…
HX + OH–  X– + H2O
But the [HX]/[ X–] ratio remains more or less constant, so the pH is not significantly changed.
When a small amount of H+ is added to the buffer, X– is consumed to produce HX…
X– + H+  HX
Once again, the [HX]/[ X– ] ratio is more or less constant, so the pH does not change significantly.

7. Buffers

8. Buffer Capacity
Buffer capacity is the amount of acid or base that can be neutralized by the buffer before there is a significant change in pH.
Buffer capacity depends on the concentrations of the components of the buffer.
The greater the concentrations of the conjugate acid-base pair, the greater the buffer capacity.
The pH of the buffer is related to Ka and to the relative concentrations of the acid and base…
This equation is known as the Henderson-Hasselbalch equation.
When [base] = [acid], the pH = pKa.

9. Let’s tabulate this information…
Buffer Calculations
When a small amount of strong acid or strong base is added to a buffer, we assume that it is completely consumed by the reaction with the buffer.
Buffer problems consist of 2 parts: stoichiometry & equilibrium
Practice Problem:
A buffer is made by adding moles of HOAc and moles of NaOAc to enough water to make 1.00 L of solution. The pH of the buffer equals What is the pH after moles of NaOH is added?
First, we use stoichiometry to determine the effect of the addition of OH− to the buffer…
OH− + HOAc  H2O + OAc−
We see that [HOAc] will decrease and [OAc−] will increase by the concentration of OH− added to the buffer.
Let’s tabulate this information…

10. Buffer Calculations A couple of things to note…
OH− + HOAc  H2O + OAc−
Before reaction: mol mol mol
After reaction: mol mol
Now we can use the Henderson-Hasselbalch equation to find pH…
pH = pKa + log ( [base]/[acid] )
pH = – (log 1.8 x 10–5) + log ( [0.320]/[0.280] )
pH = = 4.80
A couple of things to note…
1) You can use mole amounts in place of concentrations in the equation.
2) If we added moles of a strong acid, then the pH would have decreased by 0.06 instead.

11. Buffer Calculations

12. Acid-Base Titrations In an acid-base titration:
A solution of base of known concentration is added to an acid.
Acid-base indicators or a pH meter are used to signal the equivalence point…the point at which the moles of H+ = moles of OH−.
The end point in a titration is the point where the indicator changes color.
The plot of pH versus volume during a titration is called a pH titration curve.
Let’s look at the titration curve for a strong acid-strong base…
( HCl + NaOH  H2O + NaCl )

13. Strong Acid-Strong Base Titration Curve
Things of interest:
The initial pH is based on the [HCl] at the start.
The pH rises slowly at first then jumps dramatically after the equivalence point.
The equivalence point is at a pH of 7.0
The final pH is based upon the [NaOH] in excess.
We should choose an indicator that changes color in the portion of the graph where the pH rises rapidly.

14. Strong Acid-Strong Base Titration Curve
Here’s what the graph would look like if the strong acid were titrated into a strong
base…the equivalence point is still at a pH of 7.0

15. Weak Acid-Strong Base Titrations
Consider the titration of acetic acid, HC2H3O2 with NaOH…
Again, we divide the titration into four general regions:
(1) Before any base is added… the solution contains only weak acid. Therefore, pH is given by the equilibrium calculation.
(2) Between the initial pH and the equivalence point… the strong base consumes a stoichiometric quantity of weak acid…
HC2H3O2(aq) + OH–(aq)  C2H3O2–(aq) + H2O(l)
However, there is an excess of acetic acid. Therefore, we have a mixture of weak acid and its conjugate base.
Thus the composition of the mixture is that of a buffer. The pH is given by the buffer calculation…(Henderson-Hasselbalch equation.)
moles base added = moles of acid consumed = moles of acetate ion formed

16. Weak Acid-Strong Base Titrations
(3) At the equivalence point… all the acetic acid has been consumed and all the NaOH has been consumed.
- However, C2H3O2– has been generated.
- Therefore, the pH depends on the C2H3O2– concentration.
- The pH > 7 at the equivalence point since C2H3O2− is a weak base…(All weak acid-strong base titrations have equivalence points at pH’s greater than 7!)
- The equivalence point is determined by the Ka of the acid.
(4) After the equivalence point… the pH is based on the excess strong base.

17. Weak Acid-Strong Base Titration Curves
More on choosing indicators…
* Choose the indicator with a pKa 1 less than the pH at equivalence point if you are titrating with base.
* Choose the indicator with a pKa 1 greater than the pH at equivalence point if you are titrating with acid.

18. Weak Acid-Strong Base Titration Curves
Things to notice…
For a weak acid-strong base titration, the initial pH rise is more steep than the strong acid-strong base case.
Then there is a leveling off due to buffer effects.
The equivalence point is higher for the weaker acids.
The shape of the curves after equivalence point is the same because pH is determined by the strong base in excess.

19. Polyprotic Acid Titration Curves
Things to notice…
In polyprotic acids, each ionizable proton dissociates in steps.
Therefore, in a titration there are “n” equivalence points corresponding to each ionizable proton.
In the titration of H3PO3 with NaOH, the first proton dissociates to form H2PO3− then the second proton dissociates to form HPO3−2.

20. Solubility Equilibria
All dissolving is an equilibrium… Solid  Dissolved
Example: BaSO4 (s) Ba+2(aq) SO4−2(aq)
If there is not much solid, it will all dissolve.
As more solid is added the solution will become saturated.
At equilibrium, the solid will precipitate as fast as it dissolves.
Ksp = [Ba+2] [SO4−2]
Ksp is the solubility product. It is the molar concentration of ions raised to their stoichiometric powers…(BaSO4 is ignored because it is a pure solid, so its concentration is constant.)
Solubility product is an equilibrium constant. It doesn’t change except with temperature.

21. Solubility Definition
Solubility is not the same as solubility product.
Solubility is the amount (grams) of substance that dissolves to form a saturated solution. Solubility is an equilibrium position for how much can dissolve.
Molar solubility is the number of moles of solute dissolving to form a liter of saturated solution.
A common ion can change the solubility of a substance.
Relative Solubilities
Ksp will only allow us to compare the solubility of solids that fall apart into the same number of ions.
The bigger the Ksp, the more soluble the substance.
If they fall apart into different number of pieces you have to do an equilibrium calculation to see which is more soluble.

22. Converting Solubility into Ksp
Practice Problem: The molar solubility of CaF2 is 1.24 x 10−3 M at 35º C. What is the solubility product of CaF2 at 35º C?
CaF2(s)  Ca+2(aq) F−(aq)
Ksp = [Ca+2] [F−]2
[Ca+2] = 1.24 x 10−3 M
[F−] = 2 x (1.24 x 10−3 M) = 2.48 x 10−3 M
Ksp = [1.24 x 10−3] [2.48 x 10−3]2 = x 10−9
Note: In order to calculate Ksp , all concentrations must be in units of moles/L (or molarity, M).
Reminder: grams/FormulaWeight = moles

23. Solubility and the Common Ion Effect
This is an application of Le Châtelier’s principle!
For example:
If more F− is added, (let’s say from the addition of a strong electrolyte such as NaF), the equilibrium shifts away from the increase.
Therefore, CaF2(s) is formed and precipitation occurs.
As NaF is added to the system, the solubility of CaF2 decreases. If the F− is removed, then the equilibrium shifts to the right, and more CaF2 dissolves.
F− can be removed by adding a strong acid…
This means that the pH affects the solubility of CaF2!

24. Mg(OH)2(s)  Mg+2(aq) + 2OH−(aq)
Solubility and pH
Another example of pH changing the solubility…
Mg(OH)2(s)  Mg+2(aq) + 2OH−(aq)
What happens to the equilibrium if we were to add a strong acid?
Again we apply Le Châtelier’s principle:
If OH– is removed, then the equilibrium shifts toward the right and Mg(OH)2 dissolves… OH–(aq) + H+(aq)  H2O(l)
As pH decreases, [H+] increases and the solubility of Mg(OH)2 increases.
The effect is most significant if one or both ions involved are at least somewhat acidic or basic.
In general:
The solubility of slightly soluble salts containing basic ions increases as pH decreases.
The more basic the anion, the greater the effect.

25. Formation of Complex Ions
Some metal ions, (cations), can form soluble complex ions when in solutions.
Consider the formation of Ag(NH3)2+:
The Ag(NH3)2+ is called a complex ion.
NH3 (the attached Lewis base) is called a ligand.
The equilibrium constant for the reaction is called the formation constant, Kf… Kf = [Ag(NH3)2+]/[Ag+] [NH3]2

26. Formation of Complex Ions
Consider the addition of ammonia to AgCl (white precipitate):
The overall reaction is…
Effectively, the Ag+(aq) has been removed from solution.
By Le Châtelier’s principle, the forward reaction (the dissolving of AgCl) is favored.

27. Formation of Complex Ions
Finally, amphoteric oxides will dissolve in either a strong acid or a strong base by forming complex ions with several hydroxide ligands attached to the metal.
For example:
We won’t be dealing with these unique solubility situations much, but I wanted you to be aware that they exist.

28. Solubility Guidelines for Common Ionic Compounds in Water
Our dividing line between soluble and insoluble will be 0.01 M at 25 °C. (Some people use 0.1 M as a dividing line, so you may see some discrepancies between solubility guidelines.)
Any substance that can form a solution with a concentration of 0.01 M or more is soluble. Any substance that fails to reach 0.01 M is defined to be insoluble.
This value was picked with a purpose. VERY FEW substances have their maximum solubility near to 0.01 M. Almost every substance of any importance in chemistry is either much MORE soluble or much LESS soluble.
In the past, some teachers would have a third category: slightly soluble. For the most part, that category has been cast aside.

29. Solubility Guidelines for Common Ionic Compounds in Water
All alkali metal…(Group 1A =lithium, sodium, potassium, rubidium, and cesium) and ammonium compounds are soluble.
(2) All acetate, perchlorate, chlorate, and nitrate compounds are soluble.
(3) Silver, lead, and mercury (I) compounds are insoluble.
(4) Chlorides, bromides, and iodides are soluble.
(5) Carbonates, oxides, phosphates, chromates, fluorides, and silicates are insoluble.
(6) Hydroxides and sulfides are insoluble (EXCEPT for Ca+2, Ba+2, and Sr+2 .)
(7) Sulfates are soluble (EXCEPT for Ca+2, Ba+2, and Sr+2).
- These rules are to be applied in the order given. For example, PbSO4 is insoluble because Rule 3 comes before Rule 7. In like manner, AgCl is insoluble because Rule 3 takes precedence over Rule 4.
- Please be aware that these rules are guidelines only. For example, there are some alkali metal compounds that are insoluble. However, these are rather exotic compounds and can be safely ignored at an introductory level.

30. Here are some of these guidelines shown in a table format…

31. (We’ll look at a practice problem in a little bit.)
Precipitation of Ions
We can use our knowledge of Ksp and equilibrium to determine if a precipitate will form when solutions are mixed.
For example…
At any instant in time, Q = [Ba2+][SO42-].
If Q > Ksp, precipitation occurs until Q = Ksp.
If Q = Ksp, equilibrium exists.
If Q < Ksp, solid dissolves until Q = Ksp.
(We’ll look at a practice problem in a little bit.)
Based on our solubility guidelines, ions can be selectively removed from solutions.
Removal of one metal ion from a solution is called selective precipitation.
The textbook goes into detail about the procedure, but we won’t.

32. Qualitative analysis is designed to detect the presence of metal ions.
Quantitative analysis is designed to determine how much metal ion is present.

33. Forming Precipitates
Practice Problem: Will Ag2SO4 precipitate when 100 mL of M AgNO3 is mixed with a 10 mL solution of M Na2SO4?
Ag2SO4(s)  2 Ag+ + SO4–2
Ksp = 1.5 x 10–5 = [Ag+]2 [SO4–2]
Before we can find the concentrations of the ions in the mixture, we need to determine the moles of each ion in the solutions before they were mixed together…(Reminder: moles = Liters x Molarity)
moles Ag+ = 100 mL x (1 L /1000 mL) x ( moles/L) = moles
moles SO4−2 = 10 mL x (1 L /1000 mL) x ( moles/L) = moles
After the solutions are mixed, the total volume is 110 mL or 0.11 L, so:
[Ag+] = moles/0.11 L = M
[SO4−2] = moles/0.11 L = M
Q = [0.045]2 [0.0045] = 9.1 x 10−6 < Ksp , so NO PRECIPITATION will occur.