Ch. 16: Ionic Equilibria
Buffer Solution An acid/base equilibrium system that is
capable of maintaining a relatively constant pH even if a
small amount of strong acid or base is added.
a) Components of a buffer solution:
• a mixture of a weak acid and its conjugate base
e.g. acetic acid & sodium acetate (HA & A–)
HA(aq)
H+(aq) + A–(aq)
or, ammonium chloride & ammonia (NH4+ & NH3)
NH4+(aq)
H+(aq) + NH3(aq)
pH of a Buffer Solution
(b) pH of buffer solution: (pH ≈ pKa of HA)
HA(aq)
H+
Initial
“M”
0
Change
–x
+x
Equil
M–x
x
+
A–(aq)
“A”
+x
A+x
[H+][A–]
x(A + x)
Ka =
=
[HA]
(M – x)
Since Ka is usually small (< ~ 10–4):
M – x ≈ M and A + x ≈ A
thus, Ka ≈ x(A)/M or simply,
[H+] ≈
and
Ka[HA]
[A–]
(general buffer system)
[H+] ≈ Ka moles HA
moles A–
Henderson-Hasselbach Eqn
[H+] ≈
Ka[HA]
[A–]
(general buffer system)
Again assuming that x is small, you can take the log of both sides and
rearrange to get the Henderson-Hasselbach equation;
pH = pKa + log
[base ]
[acid]
The HH equation is optional; if x is small approximation does not
work, you need to use the more general method anyway...
Buffering Action
(LeChatelier’s Principle)
– The major equilibrium in a buffer system is:
HA(aq)
H+(aq) + A–(aq)
1. What if some acid [H+] is added?
Since H+ is a product of above equilibrium, the reaction will shift
in reverse, so that
[HA] will increase and [A–] will decrease
But, as long as some A– remains, the system is still a mixture of
HA and A–. Therefore, the general equation,
[H+] ≈ Ka[HA]/[A–]
still applies. Thus,
[H+] is still ≈ Ka.
Buffering Action (cont)
2. What if some base (OH-) is added instead?
The added base (OH-) will neutralize some of the acid HA and
produce more of the conjugate base A–.
HA + OH– --> H2O + A–
[HA] will decrease and [A-] will increase
But, as long as some HA remains, the system is still a mixture of
HA and A–, and ∴ still a buffer solution.
buffer capacity: is the amount of acid or base you can add without causing
a large change in pH.
buffer range: the pH values for which a buffer system is the most effective.
(usually ±1 pH unit of either side of pKa)
Sample Buffer Solution Problems
1.
(a) The pKa of HF is 3.17. Calculate the pH of a 1.00-L
solution that is 1.00 M HF and 1.50 M NaF.
(b) What is the pH of this solution after addition of 50.0 mL
of 10.0 M HCl?
2.
What mass of solid NaCN must be added to 500 mL (3 sig
fig) of 0.50 M HCN to prepare a buffer solution with a pH
of 9.50? pKa for HCN is 9.21.
Sample Buffer Solution Problems
1.
(a) The pKa of HF is 3.17. Calculate the pH of a 1.00-L
solution that is 1.00 M HF and 1.50 M NaF.
(b) What is the pH of this solution after addition of 50.0 mL
of 10.0 M HCl?
Answer: (a) pH = 3.35
(b) pH = 2.99
2.
What mass of solid NaCN must be added to 500 mL (3 sig
fig) of 0.50 M HCN to prepare a buffer solution with a pH
of 9.50? pKa for HCN is 9.21.
Answer: 24 g NaCN
Another Sample Problem
• Write the balanced chemical equation for the important
equilibrium that is occurring in an aqueous solution of
NaNO2 and HNO2.
• Write the appropriate equilibrium constant expression for
the reaction.
• Determine the pH of a solution of 0.100 M of each
compound. (Ka HNO2 = 4.6 x 10–4)
Another Sample Problem
• Write the balanced chemical equation for the important
equilibrium that is occurring in an aqueous solution of
NaNO2 and HNO2.
• HNO2(aq)
H+(aq) + NO2–(aq)
• Write the appropriate equilibrium constant expression for
the reaction.
• Ka = [H+][NO2–]/[HNO2]
• Determine the pH of a solution of 0.100 M of each
compound. (Ka HNO2 = 4.6 x 10–4)
• 3.34
Titrations
Titration
End-point
Indicator
An unknown amount of one reactant is combined
exactly with a precisely measured volume of a
standard solution of the other.
When exactly stoichiometric amounts of two
reactants have been combined.
Substance added to aid in detection of the endpoint
(usually via a color change)
Example Problem
Vinegar is an aqueous solution of acetic acid, HC2H3O2. A 12.5
mL sample of vinegar was titrated with a 0.504 M solution of
NaOH. The titration required 15.40 mL of the base solution
in order to reach the endpoint. What is the molar
concentration of HC2H3O2 in vinegar?
NaOH(aq) + HC2H3O2(aq) --> NaC2H3O2(aq) + H2O(l)
(15.40 mL NaOH) x
0.504 mol NaOH
1000 mL NaOH soln
x
1 mole HC2H3O2
1 mole NaOH
= 0.007762 mol HC2H3O2
M=
0.007762 mol HC2H3O2
0.0125 L vinegar
= 0.621 M acetic acid
Titration of a Weak Acid by a Strong Base
e.g. 25.0 mL of 0.200 M HA (acetic acid) is titrated with
0.200 M NaOH (Ka = 1.8 x 10–5)
Example
e.g. 25.0 mL of 0.200 M HA (acetic acid) is titrated with 0.200
M NaOH.
Determine pH in each of four regions:
(a) at the start -- before any NaOH is added
Just a simple weak acid: pH = 2.72
(b) before the equivalence point:
The acid is partially neutralized by the added OH–
HA + OH– --> H2O + A–
Since both HA and A– are now present, the mixture is
a buffer solution !
and [H+] ≅ Ka (mole HA left) / (mole A– produced)
e.g. after addition of 10.0 mL of NaOH, pH = 4.57
Example, cont.
(c) at the equivalence point:
The acid (HA) is completely neutralized, and only the anion A–
is present
∴ the solution is now the salt of a weak acid!
Total volume is now 50.0 mL,
so [A–] = 0.10 M and pH = 8.88
(d) after the equivalence point:
All acid (HA) is gone and excess OH– is present
∴ the pH is determined by how much OH– is left after all the
acid has been neutralized
e.g. after 40.0 mL of NaOH added: pH = 12.66
Titration Curve
Indicators: Phenolphthalein
16
Choosing an Indicator
SUMMARY of Acid-Base Equilibrium Concepts
(Chapters 15 and 16)
1.
2.
3.
4.
5.
6.
pH scale and the autoionization of water (Kw)
Strong acids and bases
Weak acids and bases (Ka or Kb)
Salts of weak acids and salts of weak bases (conjugate
pairs: Kw ≅ KaKb)
Polyprotic acids (Ka1 and Ka2 etc)
Buffer solutions (Ka)
(mixture of weak acid and its conjugate base)
7.
Acid-Base Titrations
(especially weak acid by strong base, or weak base by strong
acid)
Sample Problems
1. A 0.625 g sample of an unknown weak acid (call it HA for
short) is dissolved in enough water to make 25.0 mL of
solution. This weak acid solution is then titrated with 0.100
M NaOH, and 45.0 mL of the NaOH solution is required to
reach the equivalence point. Using a pH meter, the pH of
the solution at the equivalence point is found to be 8.25.
(a) Determine the molecular mass of the unknown acid.
(b) Determine the pKa value of the unknown acid.
2. The pKb value for ammonia (NH3) is 4.74. Calculate the
mass (in grams) of solid ammonium chloride (NH4Cl, 53.5
g/mole) that must be added to 300 mL of 0.25 M NH3 solution
to make a buffer solution with a pH equal to 10.00.
Sample Problems
1. A 0.625 g sample of an unknown weak acid (call it HA for
short) is dissolved in enough water to make 25.0 mL of
solution. This weak acid solution is then titrated with 0.100
M NaOH, and 45.0 mL of the NaOH solution is required to
reach the equivalence point. Using a pH meter, the pH of
the solution at the equivalence point is found to be 8.25.
(a) Determine the molecular mass of the unknown acid.
(b) Determine the pKa value of the unknown acid.
Answer: (a) 139 g/mol HA
(b) pKa = 3.70
2. The pKb value for ammonia (NH3) is 4.74. Calculate the
mass (in grams) of solid ammonium chloride (NH4Cl, 53.5
g/mole) that must be added to 300 mL of 0.25 M NH3 solution
to make a buffer solution with a pH equal to 10.00.
Answer: 0.73 g NH4Cl
Solubility and Complex Ion Equilibria
Solubility Equilibria
(a) “Insoluble” salts are actually sparingly soluble
equilibrium:
solid salt
ions in solution
e.g. in a saturated solution of Ag2CO3, the following equilibrium is
occurring.
Ag2CO3(s)
2 Ag+(aq) + CO32–(aq)
Solubility Product Constant = Ksp
Ksp = [Ag+]2[CO32–] = 8.1 x 10–12
Molar Solubility
Ag2CO3(s)
2 Ag+(aq) + CO32–(aq)
Ksp = [Ag+]2[CO32–] = 8.1 x 10–12
• Given the Ksp value, determine the “molar solubility”
Let x = molar solubility !
= moles Ag2CO3 dissolved/liter
[CO32–] = x and [Ag+] = 2x
Ksp = 8.1 x 10–12 = [Ag+]2[CO32–] = (2x)2(x)
Ksp = 4x3 ∴ x = 1.3 x 10–4 M
Common Ion Effect
Common ion effect:
Sparingly soluble salts are even less soluble when a
“common” ion is present.
Le Châtelier’s Principle applies here:
salt
cation + anion
If an excess of one of the ions is present, the equilibrium will
shift in reverse and less salt will dissolve!
Example Problem
• Molar solubility of Ag2CO3 in 0.10 M Na2CO3?
Ag2CO3(s)
2 Ag+(aq) + CO32–(aq)
Ksp = 8.1 x 10–12
Let x = molar solubility of Ag2CO3
[Ag+] = 2x and
[CO32–] = 0.10 + x
Ksp = 8.1 x 10–12 = [Ag+]2[CO32–]
= (2x)2(0.10 + x)
• But since Ag2CO3 is not very soluble,
assume x << 0.1
Ksp ≈ (2x)2(0.10) ≈ 8.1 x 10–12
∴ x ≈ 4.5 x 10–6 M
Precipitation
When will a precipitate form?
The solution must be saturated for a ppt to form. Thus, the
initial ion concentrations must be greater than the
equilibrium values.
e.g.
Ag2CO3(s)
2 Ag+(aq) + CO32–(aq)
If initial concentrations of the ions are given, calculate the
ion product:
Q = [Ag+]2[CO32–]
using the initial conc values
If Q > Ksp then a precipitate will form!
In some cases where Q > Ksp there is no precipitate unless it is
disturbed; these are supersaturated.
Precipitation
a supersaturated solution will
precipitate if a seed crystal is added
precipitation occurs if
Q > Ksp
Sample Problem
A saturated solution of Mg3(PO4)2 has a concentration of 0.939
mg per liter. Calculate the solubility product constant for
Mg3(PO4)2.
Sample Problem
A saturated solution of Mg3(PO4)2 has a concentration of 0.939
mg per liter. Calculate the solubility product constant for
Mg3(PO4)2.
Answer: Ksp = 6.28 x 10–26
Qualitative analysis
• A solution containing several different cations can often be
separated by addition of a reagent that will form an
insoluble salt with one of the ions, but not the others
• an analytical scheme that utilizes selective precipitation to
identify the ions present in a solution is called a qualitative
analysis scheme
• a sample containing several ions is subjected to the
addition of several precipitating agents
• addition of each reagent causes one of the ions present to
precipitate out
Qualitative Analysis
30
Multiple Equilibria
(d) “Insoluble” salts of weak acids are more soluble in acidic
solutions
e.g. Ag2S(s)
2 Ag+(aq) + S2–(aq)
• If the solution is acidic, some S2– will be converted to HS–
and H2S, thus causing more Ag2S to dissolve
The process can be viewed as a combination of multiple
equilibria, occurring simultaneously:
Ag2S(s)
H+(aq) + S2–(aq)
H+(aq) + HS–(aq)
Net: Ag2S(s) + 2 H+(aq)
2 Ag+(aq) + S2–(aq)
HS–(aq)
H2S(aq)
2 Ag+(aq) + H2S(aq)
Complex Ion Equilibria
2. Complex Ion Equilibria (increases solubility)
• Metal ions form complex ions with Lewis bases, such as:
Cu2+(aq) + 4 NH3(aq)
Cu(NH3)42+(aq)
The reaction is generally very favorable so…K is very large!
The equilibrium constant is called a Formation Constant
[Cu(NH3)42+]
Kf =
= 1.1 x 1013 (large!)
2+
4
[Cu ][NH3]
Another example:
Ag+(aq) + 2 NH3(aq)
Kf =
Ag(NH3)2+(aq)
[Ag(NH3)2+]
= 1.6 x 107
[Ag+][NH3]2
• If a complex ion can form, then an “insoluble” salt will be much
more soluble (Le Châtelier’s Principle again!).
Example Problem
Given the above formation constant for Ag(NH3)2+ and the Ksp
value for AgBr of 5.0 x 10–13, calculate the molar solubility of
AgBr in (a) water, and (b) 0.10 M NH3.
(a) in water:
AgBr(s)
Ag+(aq) + Br–(aq)
Ksp = [Ag+][Br–] = x2 = 5.0 x 10–13
∴ x = molar solubility = 7.1 x 10–7 M
Example Problem, cont.
(b) in NH3 solution, two equilibria must be considered:
Ksp:
AgBr(s)
Ag+(aq) + Br–(aq)
K f:
Ag+(aq) + 2 NH3(aq)
Ag(NH3)2+(aq)
Knet: AgBr(s) + 2 NH3(aq)
Ag(NH3)2+(aq) + Br–(aq)
Knet = [Ag(NH3)2+][Br–]/[NH3]2
= KspKf = (5.0 x 10–13)(1.6 x 107) = 8.0 x 10–6
• Let x = molar solubility of AgBr
[Ag(NH3)2+] = [Br–] = x
[NH3] = 0.10 – 2x
Knet = x2/(0.10 – 2x)2 = 8.0 x 10–6
Since Knet is small in this case, assume 2x << 0.10
x2 / (0.10)2 ≈ 8.0 x 10–6 ∴ x ≈ 2.8 x 10–4
Thus, AgBr is about 1,000 times more soluble in NH3 solution
than in water alone.
Amphoteric Metal Hydroxides
• All metal hydroxides become more soluble in acidic solution
(like Ag2S), but some are also more soluble in basic
solution, e.g.
Al(H2O)63+(aq) + H2O(l)
Al(H2O)5(OH)2+(aq) + H3O+(aq)
• addition of OH− drives the equilibrium to the right and
continues to remove H+ from the molecules
Al(H2O)5(OH)2+(aq) + OH−(aq)
Al(H2O)4(OH)2+(aq) + OH−(aq)
Al(H2O)4(OH)2+(aq) + H2O (l)
Al(H2O)3(OH)3(s) + H2O (l)
• Others include Cr3+, Zn2+, Pb2+, Sb2+…
Sample Problem
Silver chromate, Ag2CrO4, is an “insoluble” substance with a
Ksp value of 1.2 x 10–12. Silver ion forms a stable complex
ion with cyanide ion that has the formula Ag(CN)2– and a
formation constant (Kf) of 5.3 x 1018. Calculate the molar
solubility of Ag2CrO4 in each of the following solutions.
Write balanced chemical equations for any important
equilibrium reactions that are occurring.
(a) in water
(b) in 2.00 M Na2CrO4
(c) in 2.00 M NaCN
Answers
(a) Ag2CrO4
2 Ag+(aq) + CrO42–(aq)
6.7 x 10–5 M
(b) Ag2CrO4
2 Ag+(aq) + CrO42–(aq)
3.9 x 10–7 M
(c) Ag2CrO4(s)
2 Ag+(aq) + CrO42–(aq)
2 Ag+(aq) + 4 CN–(aq)
2 Ag(CN)2–(aq)
Ag2CrO4(s) + 4 CN–(aq)
2 Ag(CN)2–(aq) + CrO42–(aq)
molar solubility ≈ 0.50 M