1. Additional Aspects of Aqueous Equilibria
Chapter 17
Additional Aspects of Aqueous Equilibria

2. The Common Ion Effect
The presence of a “common ion” will affect the ionization of a weak acid or base and the solubility of a slightly soluble salt.
For example:
Calculate the pH of M acetic acid. (Ka = 1.8 x 10-5)

3. The Common Ion Effect
Now calculate the pH of acetic acid in a 0.500M soution of sodium acetate:

4. Buffers Buffers are solutions of a weak conjugate acid-base pair.
They are particularly resistant to pH changes, even when strong acid or base is added.
Buffers contain a component that will react with H+ ion and a component that will react with OH- ion
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5. Buffers
If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.
OH- + HF ↔ H2O + F-
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6. Buffers
Similarly, if acid is added, the F− reacts with it to form HF and water.
H3O F- ↔ HF + H2O
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7. Buffer Calculations HAc + H2O H3O+ + Ac− [H3O+] [Ac−] [HAc] Ka =
Find the pH of a buffer which is M in acetic acid and M in sodium acetate.
HAc + H2O
H3O+ + Ac−
[H3O+] [Ac−]
[HAc]
Ka =
NOTE that this is essentially a common ion calculation!
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8. Another Way…… If we rearrange the Ka expression:
[H3O+] [Ac−]
[HAc]
Ka =
[A−]
[HA]
Ka =
[H3O+]
And then take the negative log of both side, we get
base
[A−]
[HA]
−log Ka =
−log [H3O+] + −log
pKa pH
acid
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9. Another way…. So pKa = pH − log [base] [acid]
Rearranging, this becomes
pH = pKa + log
[base]
[acid]
This is the Henderson–Hasselbalch equation.
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10. Henderson–Hasselbalch Equation
What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4  10−4.
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11. pH Range and buffer capacity
The pH range is the range of pH values over which a buffer system works effectively.
When preparing a buffer, it is best to choose an acid with a pKa close to the desired pH.
The buffer capacity is the amount of acid or base that can be neutralized before there is a significant change in pH. The buffer capacity depends on the concentrations of the component of the buffer.
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12. When Strong Acids or Bases Are Added to a Buffer…
…it is safe to assume that all of the strong acid or base is consumed in the reaction.
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13. Addition of Strong Acid or Base to a Buffer
Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. (think ICE table!!)
Find the new [ H3O+ ] to determine the new pH of the solution.
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14. Calculating pH Changes in Buffers
A buffer is made by adding mol HC2H3O2 and mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is Calculate the pH of this solution after mol of NaOH is added.
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15. Titration
In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base).
A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.
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16. Titration of a Strong Acid with a Strong Base
Before starting the titration, the pH of the acid is determined by [acid] = [H+].
Before equivalence, the pH is determined by calculating the moles of base added,, determining the moles of H+ remaining AND accounting for the NEW total volume. From the start of the titration to near the equivalence point, the pH slowly increases.
Just before (and after) the equivalence point, the pH increases rapidly.
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17. Titration of a Strong Acid with a Strong Base
At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.
As more base is added, the increase in pH again levels off.
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18. Titration of a Weak Acid with a Strong Base
Before titration, the pH is determined by using the Ka to find the [H+].
Before equivalence, the pH is determined by calculating the moles of base added, determining the moles of weak acid remaining and determining the number of moles of conjugate base that has formed AND accounting for the NEW total volume. Use the Ka expression to find the [H+].
OH- + HW ↔ W- + H2O

19. Titration of a Weak Acid with a Strong Base
At equivalence, all of the acid has reacted and the conjugate base has formed. The base now reacts with water:
W- + H2O ↔ HW + OH-
The equilibrium expression for this reaction is a Kb!!!
Find Kb = Kw/Ka and solve for the [OH-] and [HW] by assuming that the original moles of acid = moles of W- and accounting for the NEW volume.

20. Titration of a Weak Acid with a Strong Base
Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.
At the equivalence point the pH is >7.
After equivalence, the pH depends on the excess OH- ion added…..exactly like a strong acid titration!!
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21. Titration of a Weak Acid with a Strong Base
With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.
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22. Titration of a Weak Base with a Strong Acid
The pH at the equivalence point in these titrations is < 7.
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23. Titrations of Polyprotic Acids
When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.
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24. Solubility Products BaSO4(s) Ba2+(aq) + SO42−(aq)
Consider the equilibrium that exists in a saturated solution of BaSO4 in water:
BaSO4(s)
Ba2+(aq) + SO42−(aq)
The equilibrium constant expression for this equilibrium is
Ksp = [Ba2+] [SO42−]
where the equilibrium constant, Ksp, is called the solubility product.
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25. Solubility Products Ksp is not the same as solubility.
Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).
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26. Factors Affecting Solubility
The Common-Ion Effect
If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.
BaSO4(s)
Ba2+(aq) + SO42−(aq)
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27. Factors Affecting SolubilitypH
If a substance has a basic anion, it will be more soluble in an acidic solution.
Substances with acidic cations are more soluble in basic solutions.
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28. Factors Affecting Solubility
Complex Ions
Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent
The formation of these complex ions increases the solubility of these salts.
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29. Will a Precipitate Form?
In a solution,
If Q = Ksp, the system is at equilibrium and the solution is saturated.
If Q < Ksp, more solid can dissolve until Q = Ksp.
If Q > Ksp, the salt will precipitate until Q = Ksp.
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30. Selective Precipitation of Ions
One can use differences in solubilities of salts to separate ions in a mixture.
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