1. Additional Aspects of Aqueous Equilibria

2. Aspects of Aqueous Equilibria:
The Common Ion Effect
Salts like sodium acetate are strong electrolytes
NaC2H3O2(aq)  Na+(aq) + C2H3O2-(aq)
The C2H3O2- ion is a conjugate base of a weak acid
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq)
[H3O+] [C2H3O2-]
Ka =
[HC2H3O2]

3. Now, lets think about the problem from the
The Common Ion Effect
Now, lets think about the problem from the
perspective of LeChatelier’s Principle
What would happen if the concentration of the
acetate ion were increased?
[H3O+] [C2H3O2-]
Ka =
[HC2H3O2]
Q > K and the reaction favors reactant
Addition of C2H3O2- shifts equilibrium, reducing H+

4. HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq)
[H3O+] [C2H3O2-]
The Common Ion Effect
Ka =
[HC2H3O2]
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq)
Since the equilibrium has shifted to favor the reactant, it would appear as if the dissociation of the weak acid(weak electrolyte) had decreased.

5. Aspects of Aqueous Equilibria: The Common Ion Effect
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq)
So where might the additional C2H3O2-(aq) come from? Remember we are not adding H+. So it’s not like we can add more acetic acid.
How about from the
sodium acetate?

6. NaC2H3O2(aq)  Na+(aq) + C2H3O2-(aq)
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq)
In general, the dissociation of a weak electrolyte (acetic acid) is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte
The shift in equilibrium which occurs is called the
COMMON ION EFFECT

7. HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq)
Let’s explore the COMMON ION EFFECT in a little more detail
Suppose that we add 8.20 g or mol sodium acetate, NaC2H3O2, to 1 L of a M solution of acetic acetic acid, HC2H3O2. What is the pH of the resultant solution?
NaC2H3O2(aq)  Na+(aq) + C2H3O2-(aq)
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq)

8. Calculate the pH of a solution containing
0.06 M formic acid (HCH2O, Ka = 1.8 x 10-4)
and 0.03 M potassium formate, KCH2O.
Now you try it!

9. HCl (aq) + H2O  H3O+(aq) + Cl-(aq)
Calculate the fluoride ion concentration and pH of a solution containing 0.10 mol of HCl and 0.20 mol HF in 1.0 L
HCl (aq) + H2O  H3O+(aq) + Cl-(aq)
HF (aq) + H2O  H3O+ (aq) + F-(aq)

10. Addition of NH4+ shifts equilibrium, reducing OH-
Now, lets think about the problem from the perspective of LeChatelier’s Principle
But this time lets deal with a weak base and a salt containing its conjugate acid.
NH3(aq) + H2O  NH4+(aq) + OH- (aq)
[NH4+] [OH-]
Q > K and the reaction favors reactant
Kb =
[NH3]
Addition of NH4+ shifts equilibrium, reducing OH-

11. Calculate the pH of a solution produced by mixing 0
Calculate the pH of a solution produced by mixing 0.10 mol NH4Cl with 0.40 L of 0.10 M NH3(aq), pKb = 4.74?
NH4Cl(aq)  NH4+(aq) + Cl-(aq)
NH3(aq) + H2O  NH4+(aq) + OH-

12. HC2H3O2 (aq) + OH-(aq)  H2O + C2H3O2-(aq)
Common Ions Generated by Acid-Base Reactions
The common ion that affects a weak-acid or weak-base equilibrium may be present because it is added as a salt, or the common ion can be generated by reacting an acid and base directly (no salt would be necessary….which is kind of convenient if you think about it)
Suppose we react 0.20 mol of acetic acid (weak) with mol of sodium hydroxide strong)
HC2H3O2 (aq) + OH-(aq)  H2O + C2H3O2-(aq)
0.20 mol
0.10 mol
-0.10 mol
-0.10 mol
+0.10 mol
0.10 mol
0.10 mol

13. Suppose we react 0.20 mol of acetic acid with
0.10 mol of sodium hydroxide
HC2H3O2(aq) + OH- (aq) H2O + C2H3O2-aq)
0.20 mol
0.10 mol
-0.10 mol
-0.10 mol
0.10 mol
0.10 mol
0.10 mol
Let’s suppose that all this is occurring in 1.0 L of solution
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq)
0.10 M
0.10 M

14. NH4Cl(aq)  NH4+(aq) + Cl- (aq)
Sample problem: Calculate the pH of a solution produced by mixing 0.60 L of 0.10 M NH4Cl with 0.40 L of 0.10 M NaOH
NH4Cl(aq)  NH4+(aq) + Cl- (aq)
NH OH  NH H2O
0.06 mol
0.04 mol
-0.04 mol
-0.04 mol
0.04 mol
0.02 mol
0.04 mol
Don’t forget
to convert to
MOLARITIES
NH H2O  H3O NH3
0.04 M
0.02 M

15. Calculate the pH of a solution formed by mixing 0. 50 L of 0
Calculate the pH of a solution formed by mixing 0.50 L of M NaOH with
0.50 L of 0.30 M benzoic acid
(HC7H5O2, Ka = 6.5 x 10-5)
Now you try it!

16. adding acid or base
calculate the pH of a solution that has .2 mol of NaOH added to a solution that is .25 M HC2H3O2 and .32M NaC2H3O2 HC2H3O2(aq) + OH- (aq) H2O + C2H3O2-aq

17. HC2H3O2(aq)  H C2H3O2-aq
-X X X
X (.52+X) =1.8 x 10-5
.05-X

18. BUFFERED SOLUTIONS Suppose we have a salt: MX  M+(aq) + X-(aq)
A buffered solution is a solution that resists change in pH upon addition of small amounts of acid or base.
Suppose we have a salt: MX  M+(aq) + X-(aq)
And we’ve added the salt to a weak acid containing the same conjugate base as the salt, HX:
HX +H2O  H3O+ + X-
And the equilibrium expression for this reaction is
[H+ ] [ X-]
Ka =
[HX]
Note that the concentration of the H+ is dependent upon the Ka and the ratio between the HX and X- (the conjugate
acid-base pair)
[HX]
[H +] = Ka
[X-]

19. to change to an appreciable degree.
Two important characteristics of a buffer are buffering capacity and pH. Buffering capacity is the amount of acid or base the buffer can neutralize before the pH begins
to change to an appreciable degree.
The pH of the buffer depends upon the Ka
This capacity depends on the amount of acid and base from which the buffer is made
The greater the amounts of the conjugate acid-base pair, the more resistant the ratio of their concentrations, and hence the pH, to change
[HX]
[H +] = Ka
[X-]
Henderson-Hasselbalch Equation
[HX]
-log[H +] =
-log Ka
[X-]
pH =
pKa
log
[X-] +
[HX]

20. NaC2H3O2(aq)  Na+(aq) + C2H3O2-(aq)
0.1 M
0.1 M
0.1 M
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-(aq)
0.1 M
0.1 -x x x
0.1 - x x
0.1 + x
x(0.1 + x )
pH = 4.74
x = 1.8 x 10-5
1.8 x 10-5 =
0.1 - x
Using the Henderson-Hasselbalch Equation
Note that these are initial
concentrations
[.1]
pH =
4.74 +
log
[.1]

21. A liter of solution containing 0. 100 mol of HC2H3O2 and 0
A liter of solution containing mol of HC2H3O2 and mol NaC2H3O2 forms a buffered solution of pH Calculate the pH of this solution (a) after mol NaOH is added, (b) after adding mol HCl is added.
NaC2H3O2(aq)  Na+(aq) + C2H3O2-(aq)
0.1 M
0.1 M
0.1 M
Step 1
HC2H3O2(aq) + OH-  H2O C2H3O2- (aq)
0.1 M
0.02 M
0.1 M
0.02 M
-0.02 M
-0.02 M
0.08 M
0.00 M
0.12 M
Step 2
Henderson-Hasselbalch Equation
Note that these are
initial concentrations
[.12]
4.74 +
log
pH =
[.08]
pH = 4.92

22. initial concentrations pH = 4.74 log [.08] + [.12] pH = 4.56
A liter of solution containing mol of HC2H3O2 and mol NaC2H3O2 forms a buffered solution of pH Calculate the pH of this solution (a) after mol NaOH is added, (b) after adding mol HCl is added.
C2H3O2-(aq) + H+ HC2H3O2
0.10 M
0.02 M
0.10 M
Step 1
-0.02 M
-0.02 M
0.02 M
0.08M
0.00 M
0.12 M
Henderson-Hasselbalch Equation
Note that these are
initial concentrations
Step 2
pH =
4.74
log
[.08] +
[.12]
pH = 4.56

23. HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-aq
Now consider, for a moment, what would have happened if I had added mol of NaOH or 0.02 mol HCl to .1 M HC2H3O2.
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-aq
0.1 M -x x x
0.1 - x x x x2
1.8 x 10-5 =
0.1 - x
x =
pH = 2.9

24. HC2H3O2(aq) + OH-  H2O + C2H3O2-(aq)
0.1 M
0.02 M
0.00
-0.02 M
-0.02 M
0.02 M
0.08 M
0.00 M
0.02 M
Henderson-Hasselbalch Equation
pH =
[.02]
4.74 +
log
[.08]
pH = 4.13

25. pH = -log [0.02] pH = 1.7 HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2-aq
0.10
0.02 M
H+ from HCl Completely dissociates therefore the
pH is calculated without regard for the weak acid
pH = -log [0.02]
pH = 1.7

26. add 2 ml
10 M HCl
to .1M HC2H3O2
+ .1M NaC2H3O2
pH = 4.74
add 2 ml
10 M HCl
to 1.8 x 10-5M HCl
pH = 4.74
to .12M HC2H3O2
+ .06M NaC2H3O2
pH = 4.56
a drop of .18
.02 M HCl
pH= 1.7
a drop of 3.04

27. before any acid or base are added,
Sample exercise: Consider a litter of buffered solution that is M in formic acid (HC2H3O) and M in sodium formate (NaC2H3O ). Calculate the pH of the buffer
before any acid or base are added,
(b) after the addition of mol HNO3,
(c) after the addition of mol KOH

28. Titration Curves End points
Stoichiometrically equivalent quantities of acid and base have reacted
HCl(aq) + NaOH(aq)  H2O + NaCl

29. Titration of a weak acid and a strong base results in pH curves that look similar to those of a strong acid-strong base curve except that the curve (a) begins at a higher pH, (b) the pH rises more rapidly in the early part of the titration, but more slowly near the equivalence point, and (3) the equivalence point pH is not 7.0

31. Calculating pH’s from Titrations
Calculate the pH in the titration of acetic acid by NaOH after 30.0 ml of M NaOH has been added to 50 mL of acetic acid
HC2H3O2(aq) + OH-  H2O(l) + C2H3O2-
0.005 mol
0.003 mol
mol
0.003 mol
mol
mol
0.003 mol
[.0370]
log
pH =
4.74 +
[.0250]
pH = 4.91

32. Determining the Ka From the Titration Curve
pKa = pH = 4.74

33. Solubility Equilibria Ksp
The equilibrium expression for the following reaction is CaF2(s)  Ca2+(aq) + 2F-(aq)
[Ca2+] [F-]2
[CaF2]
Look at the table on page 759 or the appendices A26, where you will find the value of the ksp to be 4.1 x 10-11
= Ksp

34. Calculating Ksp from solubility
The solubility of Bi2S3 is 1.0 x 10-15M what would be the Ksp?
Bi2S3(s)  Bi3+(aq) S2-(aq)
1.0 x (1.0 x 10-15) 3(1.0x 10-15)
Ksp = [Bi3+]2 [S2-]3
Ksp = (2.0 x 10-15)2(3.0 x 10-15)3 = 1 x 10-73

35. Calculating solubility from Ksp
The Ksp of Cu(IO3)2 is 1.4 x 10-7 what would be the solubility?
Cu(IO3)2(s)  Cu2+(aq) IO31-(aq)
X X X
Ksp= 1.4 x 10-7 = (X)(2X)2 = 4X3
X = 3.3 x 10-3 mol/L

36. Drop the 2X as insignificant
Common ion effect
What would be the solubility of Ag2CrO4 in a solution that is .1M AgNO3,
the Ksp = 9.0 x 10-12
Ag2CrO4  2Ag1+ + CrO42-
X X X
9.0 x = (2X+.1)2(X)
Drop the 2X as insignificant
X = 9.0 x mol/L

37. pH and Solubility Mg(OH)2  Mg2+ + 2OH-
If the pH is raised (the OH- is raised) then we have the common ion effect and the solubility is decreased.
If the pH is lowered (the H+ is raised) then OH- is reacted with H+ to make water and the solubility is increased.

38. pH and solubility of salts
Ag3PO4  3Ag1+ + PO43-
If the pH is lowered (the H+ is raised) the PO43- reacts with H+ to make HPO42- , which essentially removes PO43-
from the equation, shifting the reaction to the right and the solubility is increased.

39. Will a precipitate form?
If 750 mL of 4.00E-3M Ce(NO3)3 is mixed with 300 mL of 2.00E-2M KIO3, will a precipitate form?
Ce(NO3)3(aq) + KIO3 (aq) → Ce(IO3)3(s) + KNO3(aq)
(750)(4 x 10-3) = (1050) X; X= 2.86 x 10-3M for Ce3+
(300)(2 x 10-2) = (1050) Y; Y= 5.71 x 10-3 M for IO31-
Ce(IO3)3(s)  Ce3+ (aq) + 3IO31- (aq)
Ksp = 1.9 x 10-10
Q = (2.86 x 10-3)(5.71 x 10-3)3 = 5.32 x 10-10
Q is greater than K so a precipitate will form