Chapter 16
Applications of Aqueous Equilibria
Chapter 16
Cartoon from: http://www.nearingzero.net/sci_chemistry.html
1
Strong Acid/Base Neutralization
• When a strong acid and a strong base react, the
products are always water and a salt.
– The formula of the salt depends on the anion of the
acid and the cation of the base.
– The overall pH of the solution will be neutral
because the salt does not effect the pH and water is
neutral!
1
1
H+ (aq) + OH- (aq) → H2O (l) K C = +
=
[H ][OH ] K W
Chapter 16
Net Ionic Reaction
2
1
Strong Acid/Base Neutralization
• To write the products of a neutralization
reaction, you MUST know how to put together
an ionic compound (Review Section 2.10)
• Complete and Balance the following reactions:
H3PO4 (aq) + Ca(OH)2 (aq) →
HNO3(aq) + NaOH(aq) →
HClO4 (aq) + Mg(OH)2 (aq) →
Chapter 16
3
Predicting Neutralization Reactions
• We can also identify the strong acid and base that react in
a neutralization reaction to produce a given salt.
• For example, what strong acid and strong base react to
give the salt, calcium sulfate?
Strong Acid + Strong Base → CaSO4(aq) + H2O(l)
– The calcium must be from a strong base, calcium hydroxide.
– The sulfate must be from a strong acid, sulfuric acid
H2SO4 (aq) + Ca(OH)2 (s) → CaSO4(aq) + 2 H2O(l)
Chapter 16
4
2
Neutralization: Weak Acid / Strong Base
• When a weak acid and a strong base react, the
products are also a water and a salt.
– Again, the formula of the salt depends on the anion of
the acid and the cation of the base.
– However, in this case, the anion of salt will effect the
pH of the solution. In other words, the pH is not neutral
HCN (aq) + OH- (aq) → H2O (l) + CN- (aq)
KC =
Chapter 16
[CN - ]
Ka
=
[HCN][OH - ] K W
Net Ionic Reaction
5
Neutralization: Strong Acid / Weak Base
• When a strong acid and a weak base react, the
products are also a water and a salt.
– Again, the formula of the salt depends on the anion of the
acid and the cation of the base.
– However, in this case, the cation of the salt will effect the
pH of the solution. In other words, the pH is not neutral
H3O+ (aq) + NH3 (aq) → H2O (l) + NH4+ (aq)
+
[ NH 4 ]
Kb
KC =
=
[H 3O + ][ NH 3 ] K W
Chapter 16
Net Ionic Reaction
6
3
Neutralization: Weak Acid / Weak Base
• When a weak acid and a weak base react, the
products are only a salt.
– The formula of the salt depends on the anion of the acid and
the cation of the base.
– In this case, the both the cation and anion of the salt will
effect the pH of the solution. In other words, the pH is not
neutral
HCN (aq) + NH3 (aq) ⇆ CN- (aq) + NH4+ (aq)
[CN - ][ NH 4 ] K b × K a
KC =
=
[HCN][ NH 3 ]
KW
+
Chapter 16
7
Predicting Neutralization Reactions
• Write a balanced net ionic equation for the neutralization
of the following acids and bases. Indicate whether the
pH after neutralization is acidic, basic or neutral.
– HNO2 and KOH
– HF and NH3
– HBr and NH3
Chapter 16
8
4
Buffers
• A buffer is a solution that
resists changes in pH when an
acid or a base is added.
• A buffer always contains a
combination of an acid-base
conjugate pair.
• For example, a buffer can be
made up from equal
concentrations of a weak acid
and a salt of the conjugate base
of that acid.
• May also contain a weak base
and a salt with the conjugate
acid.
Chapter 16
9
Identifying Buffer Solutions
• When trying to determine whether a solution is a
buffer, you need to determine: 1) whether a conjugate
acid-base pair is present; and 2) whether that pair
consists of weak acids/bases.
• State whether the following pairs will be a buffer:
NaCl and Na2CO3
NaCl and HCl
NaF and HF
CH3COOH and NaCH3COO
Chapter 16
10
5
Buffer Capacity
• Buffer Capacity is a measure of the amount of acid or base
that a buffer system can neutralize while maintaining its
own pH
• The buffer capacity depends on the number of moles of the
conjugate pair.
• As the concentration of the buffer species increases, the
buffer capacity increases.
Chapter 16
11
Calculating the pH of a Buffer
• We cannot use the pH formula to determine the pH of a
buffer because there is an acid AND a base present.
• We use the Henderson-Hasslebach Equation for this
calculation.
Conjugate Pair
pH = pKa + log
[Base]
[Acid]
• When the [Base]/[Acid] = 1, then…
pH = pKa
Chapter 16
12
6
Calculating the pH of a Buffer
• Calculate the pH of a buffer system containing 1.0 M
CH3COOH and 1.0 M CH3COONa. What is the pH of
the system after the addition of 0.10 mole of gaseous
HCl to 1.0 L of solution?
• Calculate the pH of 0.30 M NH3/0.36 NH4Cl buffer
system. What is the pH after the addition of 20.0 mL of
0.050 M NaOH to 80.0 mL of the buffer solution?
Chapter 16
13
Buffer Preparation
•
To prepare a buffer:
1)
Choose weak acid with pKa close to required pH.
2) Substitute into Henderson–Hasselbalch equation.
3) Solve for the ratio of [conjugate base]/[acid].
•
This will give the mole ratio of conjugate base to acid.
•
You can use this ratio and the volume of your solution to
calculate the amount of each species you will need to prepare the
buffer.
•
How would you prepare 1.00 L of “carbonate buffer” at a pH of
10.10? You are provided with carbonic acid (H2CO3), sodium
hydrogen carbonate (NaHCO3), and sodium carbonate
(Na2CO3).
Chapter 16
14
7
Titrations
Chapter 16
15
Acid-Base Titrations
• An acid-base titration is used to analyze an acid solution using a
solution of a base (with a known concentration) or vice versa.
– A measured volume of base of known concentration is added to an acid
solution of unknown concentration.
– Titrations involving only strong acids and bases are straightforward
– Titrations involving weak acids and/or bases are complicated by the
formation of a salt that effects pH
• When a stoichiometric amount of acid and base have reacted,
you have reached the endpoint or equivalence point.
• We use an indicator or a pH meter to show us when the solution
reaches the endpoint.
Chapter 16
16
8
Strong Acid/Base Titration
There are four stages of a titration:
A) Before the addition of any titrant:
•
Here, the pH is dependent only on
the amount of acid or base being
titrated
B) Before the equivalence point:
•
Here, you must determine how
much of the acid or base being
titrated has been neutralized (by
subtracting added from initial), then
calculating pH.
C) At the equivalence point:
•
D
C
B
A
For a strong acid/base titration, the
pH = 7.0 at the equivalence point.
D) Beyond the equivalence point:
•
Now, the pH is dependent solely on
the excess titrant that has been
Chapter 16
added beyond the
pH at equivalence point
will always be 7.0
17
Strong Acid/Base Titration
• What is the pH of a 25.0 mL 0.10 M HCl sample
after each of the following additions:
– No addition of 0.10 M NaOH.
– 10.0 mL (total) of 0.10 M NaOH.
– 25.0 mL (total) of 0.10 M NaOH.
– 35.0 mL (total) of 0.10 M NaOH.
Chapter 16
18
9
Titration of a Weak Acid with a Strong Base
There are four stages of a titration:
A) Before the addition of any base:
•
Here, the pH is dependent only on the
amount of weak acid.
D
B) Before the equivalence point:
•
Here, both the weak acid and its
conjugate will be present. You must
determine how much of the acid has
been converted to its conjugate, then
use the H-H equation to calculate pH.
C) At the equivalence point:
•
C
A
B
Here, the acid has been completely
converted to its conjugate base. The pH
is dependent on the amount of CB
present.
D) Beyond the equivalence point:
•
Now, there is excess strong base in
solution. The pH is dependent solely on
this excess base. The CB has a
negligible effect and is ignored.
Chapter 16
pH at equivalence point
will always be > 7.0
19
Weak Acid / Strong Base Titration
• The pH of a 25.0 mL 0.10 M CH3COOH sample
can be determined after the addition of:
–
–
–
–
No addition of 0.10 M NaOH.
10.0 mL (total) of 0.10 M NaOH.
25.0 mL (total) of 0.10 M NaOH.
35.0 mL (total) of 0.10 M NaOH.
Chapter 16
20
10
Titration of a Strong Acid with a Weak Base
There are four stages of a titration:
A) Before the addition of any acid:
•
Here, the pH is dependent only on the
amount of weak base.
A
B) Before the equivalence point:
•
Here, both the weak base and its
conjugate will be present. You must
determine how much of the base has
been converted to its conjugate, then
use the H-H equation to calculate pH.
C) At the equivalence point:
•
B
C
D
Here, the base has been completely
converted to its conjugate acid. The pH
is dependent on the amount of CA
present.
D) Beyond the equivalence point:
•
Now, there is excess strong acid in
solution. The pH is dependent solely on
this excess acid. The CA has a
negligible effect and is ignored.
Chapter 16
pH at equivalence point
will always be < 7.0
21
Strong Acid / Weak Base Titration
• The pH of a 25.0 mL 0.10 M NH3 sample can be
determined after the addition of:
–
–
–
–
No addition of 0.10 M HCl.
10.0 mL (total) of 0.10 M HCl.
25.0 mL (total) of 0.10 M HCl.
35.0 mL (total) of 0.10 M HCl.
Chapter 16
22
11
Polyprotic Acids
Chapter 16
23
Solubility Equilibria
• Many biological processed involve the dissolution or
precipitation of slightly soluble ionic compounds.
MmXx (s) ⇆ m Mx+ (aq) + x Xm- (aq)
• Rather than writing a Kc expression, we write a solubility
product constant (Ksp) for saturated solutions.
K SP = [M x + ]m [X m− ]x
• This expression can only be written for the insoluble
salts (which are slightly soluble)
Chapter 16
WHY?!
24
12
Solubility Product Equilibria
• We write the Ksp the same way we do the Kc
• Remember, we don’t include pure solids or liquids in equilibrium
constant expressions.
• If we add Ag2SO4 to water, some of the slightly soluble Ag2SO4
dissolves:
Ag2SO4(s) ⇆ 2 Ag+(aq) + SO42-(aq)
• We can write the Ksp expression for the reaction:
Ksp = [Ag+]2[SO42-]
• Write the KSP expression for the following:
AgCl (s)
Ca3(PO4)2 (aq)
PbI2 (s)
Cr(OH)3 (s)
Chapter 16
25
Calculation of Ksp
For the following equilibrium, the
concentrations are:
[Mg2+] = 0.00016 M
[OH-] = 0.00033 M.
What is Ksp?
Ksp = [Mg2+][OH-]2 = (0.00016)(0.00032)2
Chapter 16
Ksp = 1.6 × 10-11
26
13
Ksp Values
How do you think the KSP relates to a compound’s solubility?
Many other KSP values in Table C.4 in the Appendix
Chapter 16
27
Calculation of Solubility
• Once a KSP value is know for a compound, you can use
it to calculate the solubility of the compound.
• These solubilities are approximate values (as there can
be side reactions).
Ag2CrO4 (s) ⇆
2 Ag+(aq)
+ CrO42- (aq)
Initial
Excess
0
0
Change
-X
+ 2X
+X
Equilibrium
Excess
2X
X
Ksp = [Ag+]2 [CrO42- ]
1.1 × 10-11 = (2X)2 • (X)
Chapter 16
X = 1.4 x 10-4 M
28
14
Ion Solubility Product (QSP)
• The Ion Solubility Product (QSP) is the
solubility equivalent of the reaction quotient.
• It can be used to determine whether a precipitate
will form.
Q < Ksp
Unsaturated
Q = Ksp
Saturated
Q > Ksp
Supersaturated; ppt forms.
Chapter 16
29
Solubility Product Equilibria
• Exactly 200 mL of 0.0040 M BaCl2 are added to
exactly 600 mL of 0.0080 M K2SO4. Will a
precipitate form?
• If 2.00 mL of 0.200 M NaOH are added to 1.00
L of 0.100 M CaCl2, will precipitation occur?
Chapter 16
30
15
Fractional Precipitation
• Fractional precipitation is a method of removing one ion type while
leaving others in solution.
• Ions are added that will form an insoluble product with one ion and a
soluble one with others.
• When both products are insoluble, their relative Ksp values can be used
for separation.
Chapter 16
31
Fractional Precipitation
Chapter 16
32
16
Fractional Precipitation
• Silver nitrate is slowly added to a solution that is 0.020 M in
Cl– ions & 0.020 M in Br– ions. Calculate the concentration
of Ag+ ions (in mol/L) required to initiate (a) the pptn of
AgBr, and (b) the pptn of AgCl.
• Ksp’s of AgCl and Ag3PO4 are 1.6 x 10–10 and 1.8 x 10–18,
respectively. If Ag+ is added to 1.00 L of a solution
containing 0.10 mol Cl– and 0.10 mol PO43–, calculate the
concentration of Ag+ ions required to initiate (a) the pptn of
AgCl, and (b) the pptn of Ag3PO4.
Chapter 16
33
Common Ion Effect and Solubility
• As the solubility product (Ksp) is an equilibrium constant,
precipitation will occur only when the ion solubility product
(Q) exceeds the Ksp for a compound.
• This means that LeChatelier’s principle is in effect.
• Increasing the concentration of one of the ions by adding a
second solution puts stress on the equilibrium.
PbCl2 (s) ⇆ Pb2+ (aq) + 2 Cl- (aq)
• If NaCl is added to saturated PbCl2, what will happen?
• What about adding NaNO3?
Chapter 16
34
17
Common Ion Effect and Solubility
• Calculate the solubility of silver chloride in a 6.5
x 10–3 M silver chloride solution.
• Calculate the solubility of AgBr in:
(a) pure water
(b) 0.0010 M NaBr
Chapter 16
35
pH and Solubility
• An ionic compound
that contains a basic
anion becomes
more soluble as the
pH decreases.
CaCO3 (s) ⇆ Ca2+ (aq) + CO32- (aq)
Chapter 16
36
18
Complex Ions and Solubility
• A complex ion is an ion containing a central metal cation bonded to
one or more molecules or ions.
• Most metal cations in complex ions are transition metals because they
have more than one oxidation state.
• The formation constant (Kf) is the equilibrium constant for the
complex ion formation.
• The formation of a complex ion is a step-wise process and each step
has its own equilibrium constant value:
Ag+ (aq) + NH3 (aq) ⇆ Ag(NH3) + (aq) K1 = 2.1 x 103
Ag(NH3)+ (aq) + NH3 (aq) ⇆ Ag(NH3)2+ (aq) K2 = 8.1 x 103
Ag+ (aq) + 2 NH3 (aq) ⇆ Ag(NH3)2+ (aq) Kf = 1.7 x 107
Chapter 16
What does this large value tell us about the
formation of the complex ion?
37
Complex Ions and Solubility
• The net reaction for the dissolution of AgCl in ammonia is an addition
of the two net ionic equations:
AgCl (s) ⇆ Ag+ (aq) + Cl- (aq)
KSP = 1.8 x 10-10
Ag+ (aq) + 2 NH3 (aq) ⇆ Ag(NH3)2+ (aq)
Kf = 1.7 x 107
AgCl(s) + 2 NH3 (aq) ⇆ Ag(NH3)2+ (aq) + Cl- (aq)
K = KSP x Kf
The value of K is much higher than the KSP value for AgCl, indicating that
the solubility of the compound is greatly increased in ammonia versus water
Chapter 16
= 3.1 x 10-3
38
19
Complex Ions and Solubility
• Calculate the molar solubility of AgBr in a 1.0 M NH3
solution. The KSP of AgBr is 5.4 x 10-13. The Kf for the
complex ion is 1.7 x 107.
• A 0.20 mole quantity of CuSO4 is added to a liter of
1.20 M NH3 solution. What is the concentration of Cu2+
ions at equilibrium?
Several Kf values in Table C.6 in the Appendix
Chapter 16
39
20