Chapter 15
Applications of
Aqueous Equilibria
Aqueous
Equilibria
THE COMMON-ION EFFECT
• Consider a solution of acetic acid:
HC2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2−(aq)
• If acetate ion is added to the solution,
Le Châtelier says the equilibrium will
shift to the left.
Aqueous
Equilibria
THE COMMON-ION EFFECT
“The extent of ionization of a weak
electrolyte is decreased by adding to
the solution a strong electrolyte that has
an ion in common with the weak
electrolyte.”
Add more of the same ion and there will
be less ions of the weak one.
Aqueous
Equilibria
THE COMMON-ION EFFECT
• The addition of concentrated HCl to a
saturated solution of NaCl will cause
some solid NaCl to precipitate out of
solution. The NaCl has become less
soluble because the addition of
additional chloride ion.
• NaCl + HCl → more NaCl due to
increased Cl-
Aqueous
Equilibria
THE COMMON-ION EFFECT
pH
 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.
Aqueous
Equilibria
THE COMMON-ION EFFECT
• The addition of a common ion to a weak acid
solution makes the solution LESS acidic.
HC2H3O2(aq) ↔ H+(aq) + C2H3O2-(aq)
• If NaC2H3O2 is added to the system, the
equilibrium shifts to undissociated HC2H3O2
raising the pH. The new pH can be calculated by
putting the concentration of the anion into the Ka
equation and solving for the new [H+].
Aqueous
Equilibria
THE COMMON-ION EFFECT
• Adding NaF to a solution of HF causes more
HF to be produced. Major species are HF,
Na+, F-, and H2O. Common ion is F-.
• In a 1.0M NaF and 1.0M HF solution, there is
more HF in the presence of NaF.
HF(aq) ↔ H+(aq) + F-(aq)
• Le Chatelier’s indicates that additional F- due
to the NaF causes a shift to the left and thus
generates more HF.
Aqueous
Equilibria
THE COMMON-ION EFFECT
Finding the pH
1. Always determine the major species.
2. Write the equilibrium equation and
expression.
3. Determine the initial concentrations.
4. Do ICE chart and solve for x.
5. Once [H+] has been found, find pH.
Aqueous
Equilibria
PRACTICE ONE
The equilibrium concentration of H+ in a 1.0M HF solution
is 2.7 x 10-2M and the percent dissociation of HF is 2.7%.
Calculate [H+] and the percent dissociation of HF in a
solution containing 1.0M HF (Ka = 7.2 x 10-4) and 1.0M
NaF.
Aqueous
Equilibria
THE COMMON-ION EFFECT
Notice:
1.0M HF
% dissociation 2.7%
versus
1.0M HF and 1.0M NaF
% dissociation 0.072%
Aqueous
Equilibria
The Common-Ion Effect
H3O+(aq) + F−(aq)
HF(aq) + H2O(l)
Because HCl, a strong acid, is also present,
the initial [H3O+] is not 0, but rather 0.10 M.
Initially
Change
At Equilibrium
[HF], M
[H3O+], M
[F−], M
0.20
0.10
0
−x
0.20 − x  0.20
+x
0.10 + x  0.10
+x
x
Aqueous
Equilibria
EQUATIONS QUIZ
• For each of the following reactions,
write an equation for the reaction. Write
the net ionic equation for each. Omit
formulas for spectator ions or molecules
in the reaction. Put a box around your
final answer.
Aqueous
Equilibria
DO NOW
• Pick handout due tomorrow.
• Turn in lab – make sure the you have:
One Title page, one Prelab, one Data
Table, Calculations Table for everyone, and
Calculations 1-8 for everyone in that order!
• Get out notes.
Aqueous
Equilibria
BUFFERS
• Solutions of a weak
conjugate acid-base
pair.
• They are particularly
resistant to pH
changes, even when
strong acid or base is
added.
• Just a case of the
common ion effect.Aqueous
Equilibria
BUFFERS
• You are always adding a strong acid or
strong base to a buffer solution.
• Buffer system (conjugate acid-base pair)
acts as a net – “catches” acid or base
Aqueous
Equilibria
BUFFERS
If a small amount of hydroxide is added to an equimolar
solution of HF in NaF, for example, the HF reacts with the
Aqueous
OH− to make F− and water.
Equilibria
BUFFERS
If acid is added, the F− reacts to form HF and water.
Aqueous
Equilibria
BUFFERS
Example:
HC2H3O2 / C2H3O2- buffer system (the “net”)
• Add a strong acid:
H+ + C2H3O2- → HC2H3O2 forms a weak acid
• Add a strong base:
OH- + HC2H3O2 → C2H3O2- + H2O
forms a weak base
Aqueous
Equilibria
BUFFERS
Example:
NH3 / NH4+ buffer system (the “net”)
• Add a strong acid: H+ + NH3 → NH4+
forms a weak acid
• Add a strong base:
OH- + NH4+ → NH3
+ H2O
forms a weak base
Aqueous
Equilibria
PRACTICE TWO
• A buffered solution contains 0.050M acetic acid
(HC2H3O2, Ka = 1.8 x 10-5) and 0.50M sodium acetate
(NaC2H3O2). Calculate the pH of the solution.
Aqueous
Equilibria
HELPFUL TIPS
• Buffered solutions are simply solutions of weak
acids and bases containing a common ion.
• The pH calculations for buffered solutions
require exactly the same procedures as
determining the pH of weak acid or weak base
solutions learned previously.
• When a strong acid or base is added to a
buffered solution, it is best to deal with the
stoichiometry of the resulting reaction first. After
the stoichiometric calculations are completed,
Aqueous
then consider the equilibrium calculations. Equilibria
BUFFERS
• Adding a strong acid or base to a
buffered solution
Aqueous
Equilibria
Requires moles
PRACTICE THREE
Calculate the change in pH that occurs when 0.010mol
solid NaOH is added to 1.0L of the buffered solution
contains 0.050M acetic acid (HC2H3O2, Ka = 1.8 x 10-5)
and 0.50M sodium acetate (NaC2H3O2). Compare this pH
change with that which occurs when 0.010mol solid
NaOH is added to 1.0L of water.
Aqueous
Equilibria
HOW BUFFERING WORKS
• When hydroxide ions are added to the
solution, the weak acid provides the source of
protons. The OH- ions are not allowed to
accumulate, but are replaced by A-.
OH- + HA → A- + H2O
• pH can be understood by looking at the EQ
expression.
Ka =
𝐻+ [𝐴−]
[𝐻𝐴]
or
[H+]
=
[𝐻𝐴]
Ka
[𝐴−]
Aqueous
Equilibria
HOW BUFFERING WORKS
Aqueous
Equilibria
HOW BUFFERING WORKS
• So [H+] (and thus pH) is determined by the ratio
of [HA]/[A-]. When OH- ions are added, HA
converts to A- and the ratio decreases. However,
is the amounts of [HA] and [A-] are LARGE, then
the change in the ratio will be small.
• If [HA]/[A-] = 0.50M / 0.50M = 1.0M initially. After
adding 0.010M OH-, it becomes [[HA]/[A-] =
0.49M / 0.51M = 0.96M. Not much of a change.
[H+] and pH are essentially constant.
Aqueous
Equilibria
HOW BUFFERING WORKS
Aqueous
Equilibria
BUFFER CALCULATIONS
Consider the equilibrium constant
expression for the dissociation of a
generic acid, HA:
HA + H2O
H 3O + + A −
[H3O+] [A−]
Ka =
[HA]
Aqueous
Equilibria
BUFFER CALCULATIONS
Rearranging slightly, this becomes
−]
[A
Ka = [H3O+]
[HA]
Taking the negative log of both sides, we get
−]
[A
−log Ka = −log [H3O+] + −log
[HA]
pKa
pH
base
acid
Aqueous
Equilibria
BUFFER CALCULATIONS
• So
[base]
pKa = pH − log
[acid]
• Rearranging, this becomes
[base]
pH = pKa + log
[acid]
• This is the Henderson–Hasselbalch equation.
Aqueous
Equilibria
BUFFER CALCULATIONS
• For a particular buffering system, all
solutions that have the same ratio of [A-]
/[HA] have the same pH.
• Optimum buffering occurs when [HA] =
[A-] and the pKa of the weak acid used
should be as close to possible to the
desired pH of the buffer system.
Aqueous
Equilibria
HENDERSON-HASSELBACH
• The equation needs to be used cautiously.
• It is sometimes used as a quick, easy
equation in which to plug in numbers.
• A Ka or Kb problem requires a greater
understanding of the factors involved and can
ALWAYS be used instead of the HH equation.
• However, at the halfway point (as in a
titration), the HH is very useful.
Aqueous
Equilibria
PRACTICE FOUR
What is the pH of a buffer that is 0.75 M lactic
acid, HC3H5O3, and 0.25 M in sodium lactate?
Ka for lactic acid is 1.4  10−4.
Aqueous
Equilibria
HENDERSON–HASSELBALCH
EQUATION
[base]
pH = pKa + log
[acid]
pH = −log (1.4 
10−4)
(0.25)
+ log (0.75)
pH = 3.85 + (−.477)
pH = 3.37
Aqueous
Equilibria
HINTS
1. Determine the major species involved.
2. If a chemical reaction occurs, write the
equation and solve stoichiometry.
3. Write the EQ equation.
4. Set up the equilibrium expression (Ka
or Kb) of the HH equation.
5. Solve.
Aqueous
6. Check the logic of the answer.
Equilibria
PRACTICE FIVE
A buffered solution contains 0.25M NH3 (Kb =
1.8 x 10-5) and 0.40M NH4Cl. Calculate the pH
of this solution.
Aqueous
Equilibria
PRACTICE SIX
Calculate the pH of the solution that results
when 0.10mol gaseous HCl is added to 1.0Lof
the buffered solution of contains 0.25M NH3 (Kb
= 1.8 x 10-5) and 0.40M NH4Cl.
Aqueous
Equilibria
BUFFERING CAPACITY
• This is the amount of acid or base that
can be absorbed by a buffer system
without a significant change in pH.
• In order to have a large buffer capacity, a
solution should have large concentrations
of both buffer components.
Aqueous
Equilibria
PRACTICE SEVEN
Calculate the change in pH that occurs when 0.010mol
gaseous HCl is added to 1.0L of each of the following
solutions (Ka for acetic acid = 1.8 x 10-5):
• Solution A: 5.00M HC2H3O2 and 5.00M NaC2H3O2
• Solution B: 0.050M HC2H3O2 and 0.050M NaC2H3O2
Aqueous
Equilibria
HINT
• We see that the pH of a buffered
solution depends on the ratio of the
[base] to [acid] (or [acid] to [base]).
• Big concentration difference = large pH
change
Aqueous
Equilibria
PRACTICE EIGHT
A chemist needs a solution buffered at pH 4.30 and can
choose from the following list of acids and their soluble salt:
a. chloroacetic acid Ka = 1.35 × 10-3
b. propanoic acid Ka = 1.3 × 10-5
c. benzoic acid Ka = 6.4 × 10-5
d. hypochlorus acid Ka = 3.5 × 10-8
Calculate the ratio of [HA] / [A-] required for each system to
yield a pH of 4.30. Which system works best?
Aqueous
Equilibria
TITRATIONS and pH CURVES
• Only when the acid AND base are both strong
is the pH at the equivalence point 7.
• Any other conditions and you get to do an
equilibrium problem. It is really a
stoichiometry problem with a limiting reactant.
The “excess” is responsible for the pH
• Weak acid + strong base equivalence pt. >
pH 7
• Strong acid + weak base equivalence pt. <
pH 7
Aqueous
Equilibria
END PT VS EQUIVALENCE PT
• There is a distinction between the
equivalence point and the end point.
• The end point is when the indicator
changes color.
• If you’ve made a careful choice of
indicators, the equivalence point, when
the number of moles of acid = number
of moles of base, will be achieved at the
same time.
Aqueous
Equilibria
VOCABULARY
• Titrant – solution of know concentration
(usually in the buret). The titrant is added to
a solution of unknown concentration until the
substance being analyzed is just consumes
(stoichiometric point or equivalence point).
• Titration or pH Curve – plot of pH as a
function of the amount of titrant added.
Aqueous
Equilibria
pH RANGE
• The pH range is the range of pH values
over which a buffer system works
effectively.
• It is best to choose an acid with a pKa
close to the desired pH.
Aqueous
Equilibria
Titration
A known concentration
of base (or acid) is
slowly added to a
solution of acid (or
base).
Aqueous
Equilibria
Titration
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.
Aqueous
Equilibria
STRONG ACID-STRONG BASE
H+(aq) + OH-(aq) → H2O(l)
• To compute H+, we have to know how much H+
remains at that point in the titration.
• New unit: millimole, mmol – titrations usually
involve small quantities.
• This means Molarity =
•
𝑚𝑚𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑒
.
𝑚𝑖𝑙𝑙𝑖𝑙𝑖𝑡𝑒𝑟𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1.0𝑚𝑜𝑙
So a 1.0M =
1.0𝐿
=
𝑚𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑒
𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1.0𝑚𝑚𝑜𝑙
1.0𝑚𝐿
=
Aqueous
Equilibria
STRONG ACID-STRONG BASE
Example - For the titration of 50.0 mL of
0.200M HNO3 with 0.100M NaOH,
calculate the pH of the solution at the
following selected points of the titration:
A.
B.
C.
D.
0.0 mL of 0.100M NaOH has been added.
10.0 mL of 0.100M NaOH has been added.
20.0 mL of 0.100M NaOH has been added.
50.0 mL of 0.100M NaOH has been added.
Aqueous
Equilibria
STRONG ACID-STRONG BASE
a. 0.0 mL of 0.100M NaOH has been
added to 0.200M HNO3.
• Major Species: H+, NO3-, H2O
HNO3 = strong acid
• pH = -log[H+] = -log (0.200) = 0.699
Aqueous
Equilibria
STRONG ACID-STRONG BASE
b. 10.0 mL of 0.100M NaOH has been added.
• Major Species: H+, NO3-, H2O, Na+, OH• 1.00mmol (10.0mL x 0.100M) OH- reacts with 1.00mmol H+.
H+ + OH- → H2O
• Before: H+ = 10.0mmol; OH- = 1.00mmol
• After: H+ = 10.0mmol – 1.00mmol = 9.0mmol;
OH- = 1.00mmol – 1.00mmol = 0.00mmol
• So after, Major Species: H+, NO3-, H2O, Na+. Determine pH
with H+ remaining.
• pH =
-log[H+]
= -log
9.0𝑚𝑚𝑜𝑙
(
)
50.0𝑚𝐿+10.0𝑚𝐿
= -log[0.15] = 0.82
Aqueous
Equilibria
STRONG ACID-STRONG BASE
c. 20.0 mL (total as opposed to additional) of
0.100M NaOH has been added.
• Major Species: H+, NO3-, H2O, Na+, OH• 2.00mmol (20.0mL x 0.100M) OH- reacts with
2.00mmol H+. H+ + OH- → H2O
• Before: H+ = 10.0mmol; OH- = 2.00mmol
• After: H+ = 10.0mmol – 2.00mmol = 8.0mmol;
OH- = 2.00mmol – 2.00mmol = 0.00mmol
Aqueous
Equilibria
STRONG ACID-STRONG BASE
• So after, Major Species: H+, NO3-, H2O,
Na+. Determine pH with H+ remaining.
• pH =
= -log
-log[0.11] = 0.94
-log[H+]
8.0𝑚𝑚𝑜𝑙
(
)
50.0𝑚𝐿+20.0𝑚𝐿
=
Aqueous
Equilibria
STRONG ACID-STRONG BASE
d. 50.0 mL (total) of 0.100M NaOH has been
added.
• Major Species: H+, NO3-, H2O, Na+, OH• 5.00mmol (50.0mL x 0.100M) OH- reacts with
5.00mmol H+. H+ + OH- → H2O
• Before: H+ = 10.0mmol; OH- = 5.00mmol
• After: H+ = 10.0mmol – 5.00mmol = 5.0mmol;
OH- = 5.00mmol – 5.00mmol = 0.00mmol
Aqueous
Equilibria
STRONG ACID-STRONG BASE
• So after, Major Species: H+, NO3-, H2O,
Na+. Determine pH with H+ remaining.
• pH =
= -log
-log[0.050] = 1.30
-log[H+]
5.0𝑚𝑚𝑜𝑙
(
)
50.0𝑚𝐿+50.0𝑚𝐿
=
Aqueous
Equilibria
STRONG ACID-STRONG BASE
e. 100.0 mL (total) of 0.100M NaOH has been
added.
• Major Species: H+, NO3-, H2O, Na+, OH• 2.00mmol (20.0mL x 0.100M) OH- reacts with
10.0mmol H+. H+ + OH- → H2O
• Before: H+ = 10.0mmol; OH- = 10.0mmol
• After: H+ = 10.0mmol – 10.0mmol = 0.0mmol;
OH- = 10.0mmol – 10.0mmol = 0.0mmol
Aqueous
Equilibria
STRONG ACID-STRONG BASE
• So after, Major Species: NO3-, H2O,
Na+.
• This is the EQUIVALENCE POINT (or
stoichiometric point).
• pH = 7.00, neutral
Aqueous
Equilibria
STRONG ACID-STRONG BASE
f. 150.0 mL (total) of 0.100M NaOH has been
added.
• Major Species: H+, NO3-, H2O, Na+, OH• 15.0mmol (50.0mL x 0.100M) OH- reacts with
10.0mmol H+. H+ + OH- → H2O
• Before: H+ = 10.0mmol; OH- = 15.0mmol
• After: H+ = 10.0mmol – 10.0mmol = 0.0mmol;
OH- = 15.0mmol – 10.0mmol = 5.00mmol
Aqueous
Equilibria
STRONG ACID-STRONG BASE
• So after, Major Species: H+, NO3-, H2O,
Na+. Determine pH with H+ remaining.
5.0𝑚𝑚𝑜𝑙
(
)
50.0𝑚𝐿+150.0𝑚𝐿
• pOH =
= -log
-log[0.025] = 1.60, so pH = 12.40
-log[OH-]
=
Aqueous
Equilibria
STRONG ACID-STRONG BASE
g. 200.0 mL (total) of 0.100M NaOH has been
added.
• Major Species: H+, NO3-, H2O, Na+, OH• 20.0mmol (50.0mL x 0.100M) OH- reacts with
10.0mmol H+. H+ + OH- → H2O
• Before: H+ = 10.0mmol; OH- = 20.0mmol
• After: H+ = 10.0mmol – 10.0mmol = 0.0mmol;
OH- = 20.0mmol – 10.0mmol = 10.0mmol
Aqueous
Equilibria
STRONG ACID-STRONG BASE
• So after, Major Species: H+, NO3-, H2O,
Na+. Determine pH with H+ remaining.
10.0𝑚𝑚𝑜𝑙
(
)
50.0𝑚𝐿+200.0𝑚𝐿
• pOH =
= -log
-log[0.040] = 1.40, so pH = 12.60.
-log[OH-]
=
Aqueous
Equilibria
STRONG ACID-STRONG BASE
The results of a-g are
plotted. The pH
changes gradually
until the titration is
close to the
equivalence point
when there is a
dramatic change.
Why is this?
Aqueous
Equilibria
STRONG ACID-STRONG BASE
Characteristics:
• At equivalence point, pH = 7.
• Before the equivalence point, [H+] and
thus pH can be calculated by dividing
mmol H+ by total volume of solution.
• After the equivalence point, [OH-] and
thus pOH and then pH can be
calculated by dividing mmol OH- by total
Aqueous
volume of solution.
Equilibria
Titration of a Strong Acid with
a Strong Base
From the start of the
titration to near the
equivalence point,
the pH goes up
slowly.
Aqueous
Equilibria
Titration of a Strong Acid with
a Strong Base
Just before and after
the equivalence point,
the pH increases
rapidly.
Aqueous
Equilibria
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.
Aqueous
Equilibria
Titration of a Strong Acid with
a Strong Base
As more base is
added, the increase
in pH again levels
off.
Aqueous
Equilibria
WEAK ACID – STRONG BASE
• We have to do a series of buffer problems
like we did earlier.
• Remember that although the acid is weak,
it reacts to completion with the OH- ion, a
very strong base.
Aqueous
Equilibria
WEAK ACID – STRONG BASE
• A two-step procedure:
A stoichiometry problem: The rxn of OH_
with the weak acid is assumed to run to
completion, and concentrations of the acid
remaining and conjugate base formed are
determined.
An equilibrium problem: The position of
the weak acid equilibrium is determined
and pH is calculated.
Aqueous
Equilibria
WEAK ACID – STRONG BASE
• At halfway to the equivalence point, pH
= pKa.
• At the equivalence point, a basic salt is
present and pH > 7
• After the equivalence point, the strong
base will be the dominant species and a
simple pH calculation can be done after
stoichiometry.
Aqueous
Equilibria
WEAK ACID – STRONG BASE
Example - For the titration of 50.0 mL of 0.10M
HC2H3O2 (Ka = 1.8 x 10-5) with 0.10M NaOH,
calculate the pH of the solution at the following
selected points of the titration:
a. 0.0 mL of 0.10M NaOH has been added.
b. 10.0 mL of 0.10M NaOH has been added.
c. 25.0 mL (total as opposed to additional) of
0.100M NaOH has been added.
d. 40.0 mL (total) of 0.10M NaOH has been added.
Aqueous
e. 50.0 mL (total) of 0.10M NaOH has been added.
Equilibria
WEAK ACID – STRONG BASE
a. 0.0 mL of 0.10M NaOH has been added.
• Major Species: HC2H3O2, H2O
• HC2H3O2 → C2H3O2- + H+
• Ka = [C2H3O2-][H+] / [HC2H3O2] = Ka = 1.8 x 10-5
I
C
E
HC2H3O2 → C2H3O2- + H+
0.10
0
0
-x
+x
+x
0.10-x
x
x
Aqueous
Equilibria
WEAK ACID – STRONG BASE
Ka = 1.8 x
10-5
=
(𝑥)(𝑥)
0.10−𝑥
= x2 / 0.10
x2 = 1.8 x 10-6 x = 0.0013
pH = -log[H+] = -log (0.0013) = 2.89
Aqueous
Equilibria
WEAK ACID – STRONG BASE
b. 10.0 mL of 0.10M NaOH has been added.
• Major Species: HC2H3O2, H2O, Na+, OH• HC2H3O2 + OH- → C2H3O2- + H2O
• Before: OH- = 1.0mmol; HC2H3O2 = 5.0mmol;
C2H3O2- = 0.0mmol
• After: OH- = 1.0mmol – 1.0mmol = 0.0mmol;
HC2H3O2 = 5.0mmol – 1.0mmol = 4.0mmol;
C2H3O2- = 0.0mmol + 1.0mmol = 1.0mmol
formed.
Aqueous
Equilibria
WEAK ACID – STRONG BASE
So after, Major Species: HC2H3O2, C2H3O2-,
H2O, Na+.
Determine pH with HC2H3O2 equilibrium
[Initial]:
[Initial]:
4.0𝑚𝑚𝑜𝑙
HC2H3O2 =
50.0𝑚𝐿+10.0𝑚𝐿
1.0𝑚𝑚𝑜𝑙
C2H3O2 =
50.0𝑚𝐿+10.0𝑚𝐿
[Initial]: H+ ≈ 0
Ka = [C2H3O2-][H+] / [HC2H3O2] = Ka = 1.8 x 10-5
Aqueous
Equilibria
WEAK ACID – STRONG BASE
I
C
E
HC2H3O2 → C2H3O2- + H+
0.067
0.017
0
-x
+x
+x
0.067 - x
0.017 + x
x
(𝑥)(0.017+𝑥)
0.067−𝑥
• Ka = 1.8 x
=
= x(0.017)/0.067
= 0.25x
• x = 7.2 x 10-5
• pH = -log[H+] = -log[7.2 x 10-5] = 4.14
10-5
Aqueous
Equilibria
WEAK ACID – STRONG BASE
Henderson Hasselbach Equation
pH = pKa + log
[𝑏𝑎𝑠𝑒]
[𝑎𝑐𝑖𝑑]
= 4.74 + -0.60
= 4.14
Aqueous
Equilibria
WEAK ACID – STRONG BASE
c. 25.0 mL (total as opposed to additional)
of 0.100M NaOH has been added.
• Major Species: HC2H3O2, H2O, Na+, OH• HC2H3O2 + OH- → C2H3O2- + H2O
• Before: OH- = 2.5mmol;
HC2H3O2 = 5.0mmol; C2H3O2- = 0.0mmol
• After: OH- = 2.5mmol – 2.5mmol = 0.0mmol;
HC2H3O2 = 5.0mmol – 2.5mmol = 2.5mmol;
C2H3O2- = 0.0mmol + 2.5mmol = 2.5mmol
Aqueous
formed.
Equilibria
WEAK ACID – STRONG BASE
So after, Major Species: HC2H3O2, C2H3O2-,
H2O, Na+.
Determine pH with HC2H3O2 equilibrium
[Initial]:
[Initial]:
2.5𝑚𝑚𝑜𝑙
HC2H3O2 =
50.0𝑚𝐿+25.0𝑚𝐿
5.0𝑚𝑚𝑜𝑙
C2H3O2 =
50.0𝑚𝐿+25.0𝑚𝐿
[Initial]: H+ ≈ 0
Ka = [C2H3O2-][H+] / [HC2H3O2] = Ka = 1.8 x 10-5
Aqueous
Equilibria
WEAK ACID – STRONG BASE
I
C
E
HC2H3O2 → C2H3O2- + H+
0.033
0.033
0
-x
+x
+x
0.033 - x
0.033 + x
x
Ka = 1.8 x
10-5
=
(𝑥)(0.033+𝑥)
0.033−𝑥
= x(0.033)/0.033 = x
x = 1.8 x 10-5
pH = -log[H+] = -log[1.8 x 10-5] = 4.74
This is halfway to the equivalence point. Half of the
Aqueous
HC2H3O2 has been converted. [HC2H3O2]0 = [C2H3O2-]0. Equilibria
Ka = [H+] and pH = pKa.
WEAK ACID – STRONG BASE
Henderson Hasselbach Equation
pH = pKa + log
[𝑏𝑎𝑠𝑒]
[𝑎𝑐𝑖𝑑]
= 4.74 + 0.0
= 4.74
Aqueous
Equilibria
WEAK ACID – STRONG BASE
d. 40.0 mL (total) of 0.10M NaOH has been
added.
• Major Species: HC2H3O2, H2O, Na+, OH• HC2H3O2 + OH- → C2H3O2- + H2O
• Before: OH- = 4.0mmol; HC2H3O2 =
5.0mmol; C2H3O2- = 0.0mmol
• After: OH- = 4.0mmol – 4.0mmol = 0.0mmol;
HC2H3O2 = 5.0mmol – 4.0mmol = 1.0mmol;
C2H3O2- = 0.0mmol + 4.0mmol = 4.0mmol
formed.
Aqueous
Equilibria
WEAK ACID – STRONG BASE
So after, Major Species: HC2H3O2, C2H3O2-, H2O,
Na+.
[Initial]:
[Initial]:
1.0𝑚𝑚𝑜𝑙
HC2H3O2 =
50.0𝑚𝐿+40.0𝑚𝐿
4.0𝑚𝑚𝑜𝑙
C2H3O2 =
50.0𝑚𝐿+40.0𝑚𝐿
[Initial]: H+ ≈ 0
Ka = [C2H3O2-][H+] / [HC2H3O2] = Ka = 1.8 x 10-5
Aqueous
Equilibria
WEAK ACID – STRONG BASE
I
C
E
HC2H3O2 → C2H3O2- + H+
0.011
0.044
0
-x
+x
+x
0.011 - x
0.044 + x
x
(𝑥)(0.011+𝑥)
0.044−𝑥
• Ka = 1.8 x
=
= x(0.044)/0.011
= 4.0x
• x = 4.5 x 10-6
Aqueous
+
-6
Equilibria
• pH = -log[H ] = -log[4.5 x 10 ] = 5.35
10-5
WEAK ACID – STRONG BASE
Henderson Hasselbach Equation
pH = pKa + log
[𝑏𝑎𝑠𝑒]
[𝑎𝑐𝑖𝑑]
= 4.74 + 0.60
= 5.34
Aqueous
Equilibria
WEAK ACID – STRONG BASE
e. 50.0 mL (total) of 0.10M NaOH has been
added.
• Major Species: HC2H3O2, H2O, Na+, OH• HC2H3O2 + OH- → C2H3O2- + H2O
• Before: OH- = 5.0mmol; HC2H3O2 =
5.0mmol; C2H3O2- = 0.0mmol
• After: OH- = 5.0mmol – 5.0mmol = 0.0mmol;
HC2H3O2 = 5.0mmol – 5.0mmol = 0.0mmol;
C2H3O2- = 0.0mmol + 5.0mmol = 5.0mmol
formed. EQUIVALENCE POINT!
Aqueous
Equilibria
WEAK ACID – STRONG BASE
So after, Major Species: C2H3O2-, H2O, Na+.
[Initial]: HC2H3O2 = 0
-
[Initial]: C2H3O2 =
5.0𝑚𝑚𝑜𝑙
50.0𝑚𝐿+50.0𝑚𝐿
[Initial]: OH- ≈ 0
Kb = 1.0 x 10-14/1.8 x10-5 = 5.6 x 10-10
Kb = [C2H3O2-][H+] / [HC2H3O2] = 5.6 x 10-10
Aqueous
Equilibria
WEAK ACID – STRONG BASE
I
C
E
C2H3O2- → HC2H3O2 + OH0.050
0
0
-x
+x
+x
0.050 - x
x
x
Kb = 5.6 x
(𝑥)(𝑥)
-10
10 =
0.050−𝑥
= x2 / 0.050
x2 = 2.8 x 10-11 x = 5.3 x 10-6
pOH = -log[OH-] = -log[5.3 x 10-6] = 5.28,
so pH = 8.72
Aqueous
Equilibria
WEAK ACID – STRONG BASE
f. 60.0 mL (total) of 0.100M NaOH has been
added.
• Major Species: HC2H3O2, C2H3O2-, H2O, Na+,
OH• HC2H3O2 + OH- → C2H3O2- + H2O
• Before: OH- = 6.0mmol; HC2H3O2 = 5.0mmol;
C2H3O2- = 0.0mmol
• After: OH- = 6.0mmol – 5.0mmol = 1.0mmol;
HC2H3O2 = 5.0mmol – 5.0mmol = 0.0mmol;
C2H3O2- = 0.0mmol + 5.0mmol = 5.0mmol
Aqueous
Equilibria
formed.
WEAK ACID – STRONG BASE
After, Major Species: C2H3O2-, H2O, Na+, OH-.
[Initial]: HC2H3O2 = 0
[Initial]:
[Initial]:
5.0𝑚𝑚𝑜𝑙
C2H3O2 =
50.0𝑚𝐿+60.0𝑚𝐿
1.0𝑚𝑚𝑜𝑙
OH =
50.0𝑚𝐿+60.0𝑚𝐿
-
this is weak
pOH = -log[OH-] = -log[9.1 x 10-3] = 2.04, so pH
= 11.96
Aqueous
Equilibria
WEAK ACID – STRONG BASE
g. 75.0 mL (total) of 0.100M NaOH has been
added.
• Major Species: HC2H3O2, C2H3O2-, H2O, Na+,
OH• HC2H3O2 + OH- → C2H3O2- + H2O
• Before: OH- = 7.5mmol; HC2H3O2 = 5.0mmol;
C2H3O2- = 0.0mmol
• After: OH- = 7.5mmol – 5.0mmol = 2.5mmol;
HC2H3O2 = 5.0mmol – 5.0mmol = 0.0mmol;
C2H3O2- = 0.0mmol + 5.0mmol = 5.0mmol
Aqueous
Equilibria
formed.
WEAK ACID – STRONG BASE
After, Major Species: C2H3O2-, H2O, Na+, OH-.
[Initial]: HC2H3O2 = 0
[Initial]:
[Initial]:
5.0𝑚𝑚𝑜𝑙
C2H3O2 =
50.0𝑚𝐿+60.0𝑚𝐿
2.5𝑚𝑚𝑜𝑙
OH =
50.0𝑚𝐿+75.0𝑚𝐿
-
this is weak
pOH = -log[OH-] = -log[2.0 x 10-2] = 1.70,
so pH = 12.30
Aqueous
Equilibria
WEAK ACID – STRONG BASE
Note the difference between the two curves on the left and the
curve for a strong acid and strong base titration. The difference is
noted to the right. Why is the shape the same after the
equivalence point?
Aqueous
Equilibria
WEAK ACID – STRONG BASE
• Remember that the equivalence point is
defined by stoichiometry NOT by pH.
• When enough titrant has been added to
react exactly with all the acid or base
being titrated is the equivalence point.
Aqueous
Equilibria
PRACTICE NINE
• HCN is a very weak acid (Ka = 6.2 x 10-10)
when dissolved in water. If a 50.0mL sample
of 0.100M HCn is titrated with 0.100M NaOH,
calculate the pH of the solution
A. After 8.00mL of 0.100M NaOH has been
added.
B. At the halfway point of the titration.
C. At the equivalence point of the titration.
Aqueous
Equilibria
PRACTICE NINE
• HCN is a very weak acid (Ka = 6.2 x 10-10)
when dissolved in water. If a 50.0mL sample
of 0.100M HCN is titrated with 0.100M NaOH,
calculate the pH of the solution
A. After 8.00mL of 0.100M NaOH has been
added. pH = 8.48 HH pH = 8.49
B. At the halfway point of the titration.
pH = 9.21
C. At the equivalence point of the titration.
pH = 10.95
Aqueous
Equilibria
WEAK ACID – STRONG BASE
• It took the same amount of NaOH in both the
example and Practice Nine to reach the
equivalence point. It is the AMT of the acid, not its
strength that determines the equivalence point.
• The pH value at the equivalence point IS affected
by acid strength.
• The strength of a weak acid has a significant
effect on the shape of its pH curve. The weaker
the acid, the greater the pH at the equivalence
point. As the acid becomes weaker, the vertical
Aqueous
region (shaded to the left) becomes shorter. Equilibria
WEAK ACID – STRONG BASE
The pH Curves
for the Titrations
of 50.0-mL
Samples pf 0.10
M Acids with
Various Ka
Values with 0.10
M NaOH
Aqueous
Equilibria
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.
• The pH at the equivalence
point will be >7.
• Phenolphthalein is
commonly used as an
indicator in these
titrations.
Aqueous
Equilibria
Titration of a Weak Acid with a
Strong Base
At each point below the equivalence point, the
pH of the solution during titration is determined
from the amounts of the acid and its conjugate
base present at that particular time.
Aqueous
Equilibria
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.
Aqueous
Equilibria
Titration of a Weak Base with
a Strong Acid
• The pH at the
equivalence point in
these titrations is < 7.
• Methyl red is the
indicator of choice.
Aqueous
Equilibria
TITRATION CURVE
FIVE POINTS OF INTEREST ALONG A
TITRATION CURVE for weak acids/bases:
A. The pH before the titration begins. Treat as
usual, the acid or base in the flask
determines the pH. If weak, an ICE chart is
in order.
Aqueous
Equilibria
TITRATION CURVE
FIVE POINTS OF INTEREST ALONG A
TITRATION CURVE:
B. The pH on the way to the equivalence point.
You are in the “land of buffer” as soon as the
first drop from the buret makes a splash and
reacts to form the salt. Whatever is in the buret
is the “added” part. Use to solve for the
hydrogen ion concentration and subsequently
the pH. Either the acid or the base [whichever is
in the buret] starts at ZERO. Do the
Aqueous
stoichiometry and then an ICE chart.
Equilibria
TITRATION CURVE
FIVE POINTS OF INTEREST ALONG A
TITRATION CURVE:
C. The pH at the midpoint of the titration (½
equivalence point): once the midpoint is
reached, [H+] = Ka since ½ of the acid or
base has been neutralized, AND the
resulting solution in the beaker is composed
of the half that remains AND the salt. That
means that pH = pKa.
Aqueous
Equilibria
TITRATION CURVE
FIVE POINTS OF INTEREST ALONG A
TITRATION CURVE:
D. The pH at the equivalence point.—you are
simply calculating the pH of the salt, all the acid
or base is now neutralized [to salt + water!].
Write the hydrolysis reaction for your ICE chart.
Aqueous
Equilibria
TITRATION CURVE
FIVE POINTS OF INTEREST ALONG A
TITRATION CURVE:
E. The pH beyond the equivalence point—it is
stoichiometry again with a limiting reactant.
Calculate the molarity of the EXCESS and
solve for either pH directly (excess H+) or
pOH (excess OH−) and subtract it from
14.00 to arrive at pH. Be sure to track the
Aqueous
total volume when calculating the molarity! Equilibria
WEAK BASE – STRONG ACID
• We have to do a series of buffer problems like
we did earlier.
• Remember that although the base is weak, it
reacts to completion with the H+ ion, a very
strong acid.
• At the equivalence point, an acidic salt is
present and pH < 7
• After the equivalence point, the strong acid
will be the dominant species and a simple pH
calculation can be done after stoichiometry. Aqueous
Equilibria
PRACTICE TEN
• For the titration of 100.0 mL of 0.050M NH3 (Kb = 1.8 x
10-5) with 0.10M HCl, calculate the pH of the solution at
the following selected points of the titration:
a. 0.0 mL of 0.10M HCl has been added.
b. Before the equivalence point. 10.0mL
c. At the equivalence point 50.0mL
d. Beyond the equivalence point 60.0mL
Aqueous
Equilibria
SUMMARY
Aqueous
Equilibria
SUMMARY
1. Buffered solutions contain relatively large
concentrations of a weak acid and the
corresponding weak base. They can involve
a weak acid, HA, and the conjugate base, A-,
or a weak base, B and the conjugate acid,
BH+.
2. When H+ is added to a buffered solution, it
reacts essentially to completion with the
weak acid present.
H+ + A- → HA
or H+ + B → BH+
Aqueous
Equilibria
SUMMARY
C. When OH- is added to a buffered solution, it
reacts essentially to completion with the weak
acid present.
OH- + HA → A- + H2O or
OH- + BH+ → B + H2O
D. The pH of the buffered solution is determined by
the ratio of the concentrations of the weak acid
and weak base. As long as this ratio remains
virtually constant, the pH will remain virtually
constant. This will be the case as long as the
concentrations of the buffering materials (HA and
A- or B and BH+) are large compared with theAqueous
Equilibria
amounts of H+ and OH- added.
TITRATIONS OF
POLYPROTIC ACIDS
In these cases
there is an
equivalence point
for each
dissociation.
Aqueous
Equilibria
ACID BASE INDICATORS
• Marks the end point of a titration by changing
color.
• The end point is NOT the equivalence point
• With careful selection of an indicator, it can be.
• We want the indicator end point and the titration
equivalence point to be as close as possible –
choose wisely.
• Since strong acid-strong base titrations have a
large vertical area, color changes will be sharp
and a wide range of indicators can be used.
Aqueous
Equilibria
ACID BASE INDICATORS
• With weak acids and bases, we must be more
careful in our choice of indicator.
• Indicators are usually weak acids, HIn. They
have one color in their acidic form, HIn, and
another color in their basic form, In-.
• Common indicator = phenolphthalein
 colorless in its HIn form
 pink in its In- form.
 It changes color in the pH range of 8-10.
Aqueous
Equilibria
ACID BASE INDICATORS
• Usually 1/10 of the initial form of the indicator must
be changed to the other form before a new color is
apparent.
• Equations used to determine the pH at which an
indicator will change color:
 For titration of an ACID: pH = pKa + log1/10 = pKa – 1
 For titration of a BASE: pH = pKa + log10/1 = pKa + 1
• The useful range of an indicator is usually pKa + 1.
• When choosing an indicator, determine the pH at
the equivalence point of the titration and then
Aqueous
choose an indicator with a pKa close to that. Equilibria
Figure 15.8, page 715
Aqueous
Equilibria
INDICATOR
USE
The pH curve
for the titration
of 100.0 mL of
0.10 M of HCl
with 0.10 M
NaOH
Aqueous
Equilibria
INDICATOR
USE
The pH curve for
the titration of 50
mL of 0.1 M
HC2H3O2 with 0.1
M NaOH;
Phenolphthalein
will give an end pt
very close to the
equivalence pt of
the titration
Aqueous
Equilibria
PRACTICE ELEVEN
Bromothymol Blue, an indicator with a Ka of 1.0 x10-7, is
yellow in its HIn form and blue in its In- form. Suppose we
put a few drops of this indicator in a strongly acidic solution.
If the solution is then titrated with NaOH, at what pH will the
indicator color change first be visible?
Aqueous
Equilibria
SOLUBILITY EQ
• Saturated solutions of salts are another
type of chemical equilibria.
• Saturated solution of AgCl:
AgCl(s) ↔ Ag+(aq) + Cl-(aq).
• The SOLUBILITY PRODUCT
expression:
Ksp = [Ag+][Cl-].
The AgCl(s) is left out of the expression
Aqueous
Equilibria
SOLUBILITY EQ
• The Ksp for AgCl = 1.6 x 10-10.
• If the product of [Ag+][Cl-] is < 1.6 x 1010, then the solution is unsaturated and
no solid would be present.
• If the product were 1.6 x 10-10, the
product is exactly saturdated and no
solid would be present.
• If the product > 1.6 x 10-10, the solution
is saturated and a solid (precipitate)
Aqueous
would form.
Equilibria
SOLUBILITY EQ
• 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).
Aqueous
Equilibria
SOLUBILITY EQ
Calculating Concentrations
For Ag2CO3:
Ag2CO3 (s) ↔ 2Ag+(aq) + CO3-2(aq)
Ksp = [Ag+]2[CO3-2], Ksp = 8.1 x 10-12
We can determine the concentrations of
2Ag+(aq) and CO3-2(aq) using the equation
Ksp = [Ag+]2[CO3-2] = 8.1 x 10-12.
Aqueous
Equilibria
SOLUBILITY EQ
Initial [Ag+]2 = 0
Initial [CO3-2] = 0
EQ [Ag+]2 = 2x
EQ [CO3-2] = x
Ksp = [Ag+]2[CO3-2] = 8.1 x 10-12
Ksp = (x)(2x)2 = 4x3 = 8.1 x 10-12
x3 = 2.0 x 10-12
x = 1.3 x 10-4
EQ [Ag+]2 = 1.3 x 10-4M
EQ [CO3-2] = 2.6 x 10-4M
Aqueous
Equilibria
PRACTICE TWELVE
The Ksp value for copper(II) iodate, Cu(IO3)2 is 1.4 x 10-7
at 25°C. Calculate its solubility at 25°C.
Aqueous
Equilibria
PRACTICE THIRTEEN
Copper(I) bromide has a measured solubility of 2.0 x 104M at 25°C. Calculate the K value.
sp
Aqueous
Equilibria
PRACTICE FOURTEEN
Calculate the Ksp value for bismuth sulfide (Bi2S3),
which has a solubility of 1.0 x 10-15M at 25°C.
Aqueous
Equilibria
COMMON ION EFFECT
What happens if we dissolve the
substance into something other than pure
water?
What happens when the solution contains
a common ion?
Aqueous
Equilibria
PRACTICE FIFTEEN
What is the molar solubility of solid calcium fluoride,
CaF2 in a 0.025M solution of sodium fluoride?
Ksp = 4.0 x 10-11
Aqueous
Equilibria
PRECIPITATE AND QUAL
ANALYSIS
• This is now considering the reverse of
the above - forming a solid from
solution.
• The product of the initial concentrations
of the ions (raised to the power of their
coefficients) is called the ION
PRODUCT or Q.
Aqueous
Equilibria
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 will dissolve until Q = Ksp.
 If Q > Ksp, the salt will precipitate until Q = Ksp.
Aqueous
Equilibria
PRACTICE SIXTEEN
A solution is prepared by adding 750.0mL of 4.00 x 10-3M
Ce(NO3)3 to 300.0mL of 2.00 x 10-2M KIO3. Will Ce(IO3)3
with a Ksp = 1.9 x 10-10 precipitate from this solution?
Justify your answer.
Aqueous
Equilibria
SELECTIVE PRECIPITATION
OF IONS
One can use
differences in
solubilities of
salts to separate
ions in a
mixture.
Aqueous
Equilibria
Aqueous
Equilibria