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Chapter 16
Equilibria in Aqueous Systems
Buffers
•! buffers are solutions that resist changes in pH when an acid or base is added
•! they act by neutralizing the added acid or base
•! but just like everything else, there is a limit to what they can do, eventually the pH changes
•! many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion
Making a Buffer Solution from a Weak Acid
How Acid Buffers Work
HA(aq) + H2O(l) ! A!
(aq) + H3O+
(aq)
•! buffers work by applying Le Châtelier’s Principle to weak acid equilibrium
•! buffer solutions contain significant amounts of the weak acid molecules, HA – these molecules react with added base to neutralize it !!you can also think of the H3O
+ combining with the OH! to make H2O; the H3O
+ is then replaced by the shifting equilibrium
•! the buffer solutions also contain significant amounts of the conjugate base anion, A! - these ions combine with added acid to make more HA and keep the H3O
+ constant
Common Ion Effect
HA(aq) + H2O(l) ! A!
(aq) + H3O+
(aq)
•! adding a salt containing the anion, NaA, that is
the conjugate base of the acid (the common ion)
shifts the position of equilibrium to the left
•! this causes the pH to be higher than the pH of
the acid solution
!!lowering the H3O+ ion concentration
The Effect of Added Acetate Ion on the Dissociation of Acetic Acid
[CH3COOH]initial [CH3COO-]added % Dissociation* pH
* % Dissociation = [CH3COOH]dissoc
[CH3COOH]initial x 100
0.10 0.00
0.10 0.050
0.10
0.10 0.10
0.15
1.3
0.036
0.018
0.012
2.89
4.44
4.74
4.92
How a Buffer Works
Buffer with equal
concentrations of conjugate base and acid
OH- H3O+
Buffer after addition of H3O+
H2O + CH3COOH H3O+ + CH3COO-
Buffer after addition of OH-
CH3COOH + OH- H2O + CH3COO-
The Henderson-Hasselbalch Equation
HA + H2O H3O+ + A-
Ka = [H3O+] [A-]
[HA]
[H3O+] =
Ka [HA]
[A-]
- log[H3O+] = - log Ka + log
[A-]
[HA]
pH = pKa + log [base]
[acid]
Buffering Effectiveness •! a good buffer should be able to neutralize moderate
amounts of added acid or base
•! however, there is a limit to how much can be added before the pH changes significantly
•! the buffering capacity is the amount of acid or base a buffer can neutralize
•! the buffering range is the pH range the buffer can be effective
•! the effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base
Effectiveness of Buffers
•! a buffer will be most effective when the [base]:
[acid] = 1
!!equal concentrations of acid and base
•! effective when 0.1 < [base]:[acid] < 10
•! a buffer will be most effective when the [acid]
and the [base] are large
•! we have said that a buffer will be effective when
0.1 < [base]:[acid] < 10
•! substituting into the Henderson-Hasselbalch we can calculate the maximum and minimum pH at which the buffer will be effective
!!"
#$$%
&+=
][HA
][AlogppH
-
aK
Lowest pH
( )
1ppH
10.0logppH
!=
+=
a
a
K
K
Highest pH
( )
1ppH
10logppH
+=
+=
a
a
K
K
therefore, the effective pH range of a buffer is pKa ± 1
when choosing an acid to make a buffer, choose
one whose is pKa is closest to the pH of the buffer
Buffering Range
Buffering Capacity
•! buffering capacity is the amount of acid or base that can be added to a buffer without destroying its effectiveness
•! the buffering capacity increases with increasing absolute concentration of the buffer components
•! as the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves
•! buffers that need to work mainly with added acid generally have [base] > [acid]
•! buffers that need to work mainly with added base generally have [acid] > [base]
How Much Does the pH of a Buffer
Change When an Acid or Base Is Added?
•! though buffers do resist change in pH when acid or base are added to them, their pH does change
•! calculating the new pH after adding acid or base requires breaking the problem into 2 parts
1.! a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other !! added acid reacts with the A! to make more HA
!! added base reacts with the HA to make more A!
2.! an equilibrium calculation of [H3O+] using the new
initial values of [HA] and [A!]
Basic Buffers
B:(aq) + H2O(l) ! H:B+(aq) + OH!
(aq)
•! buffers can also be made by mixing a weak
base, (B:), with a soluble salt of its conjugate
acid, H:B+Cl!
H2O(l) + NH3 (aq) ! NH4+
(aq) + OH!
(aq)
The Common Ion Effect
Titration
•! in an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete
!!when the reaction is complete we have reached the endpoint of the titration
•! an indicator may be added to determine the endpoint
!!an indicator is a chemical that changes color when the pH changes
•! when the moles of H3O+ = moles of OH!, the titration
has reached its equivalence point
Titration
Color Changes for Various Acid-Base Indicators
Acid-Base Indicators
Titration Curve
•! a plot of pH vs. amount of added titrant
•! the inflection point of the curve is the equivalence point of the titration
•! prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH
•! the pH of the equivalence point depends on the pH of the salt solution !!equivalence point of neutral salt, pH = 7
!!equivalence point of acidic salt, pH < 7
!!equivalence point of basic salt, pH > 7
•! beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH
Curve for a strong acid-strong base titration
Acid Type
pKa1
Maximum pH
Minimum pH
Maximum Volume
Minimum Volume
Initial [Acid], M
Initial Acid Volume, mL
Initial [NaOH], M
Titration Curve of Weak Acid with NaOH
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40 45 50
Volume NaOH Added, mL
pH
pH
0
50
0
Show First Derivative
Curve for a
weak acid-
strong base
titration
Titration of 40.00mL of 0.1000M HPr
with 0.1000M NaOH
[HPr] = [Pr-]
pH = 8.80 at
equivalence point
pKa of HPr
= 4.89
methyl red!
Titrating Weak Acid with a Strong Base
•! the initial pH is that of the weak acid solution
!!calculate like a weak acid equilibrium problem
"!e.g., 15.5 and 15.6
•! before the equivalence point, the solution becomes a buffer
!!calculate mol HAinit and mol A!
init using reaction stoichiometry
!!calculate pH with Henderson-Hasselbalch using mol HAinit and mol A!
init
•! half-neutralization pH = pKa
•! at the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established
!!mol A! = original mole HA
"!calculate the volume of added base like Ex 4.8
!![A!]init = mol A!/total liters
!!calculate like a weak base equilibrium problem
"!e.g., 15.14
•! beyond equivalence point, the OH is in excess
!![OH!] = mol MOH xs/total liters
!![H3O+][OH!]=1 x 10-14
Titrating Weak Acid with a Strong Base
Comparison of Titrations
Various weak acid-strong base titrations
Curve for a
weak base-
strong acid
titration
Titration of 40.00mL of 0.1000M NH3
with 0.1000M HCl
pH = 5.27 at
equivalence point
pKa of NH4+ =
9.25
Weak base-strong acid titration
pKa = 7.19
pKa = 1.85
Curve for the titration of a weak polyprotic acid.
Titration of 40.00mL of 0.1000M
H2SO3 with 0.1000M NaOH
Titration of a polyprotic acid
Solubility Equilibria
•! all ionic compounds dissolve in water to some
degree
!!however, many compounds have such low solubility
in water that we classify them as insoluble
•! we can apply the concepts of equilibrium to
salts dissolving, and use the equilibrium
constant for the process to measure relative
solubilities in water
Solubility Product •! the equilibrium constant for the dissociation of a solid
salt into its aqueous ions is called the solubility product, Ksp
•! for an ionic solid MnXm, the dissociation reaction is:
MnXm(s) ! nMm+(aq) + mXn!(aq)
•! the solubility product would be
Ksp = [Mm+]n[Xn!]m
•! for example, the dissociation reaction for PbCl2 is
PbCl2(s) ! Pb2+(aq) + 2 Cl!(aq)
•! and its equilibrium constant is
Ksp = [Pb2+][Cl!]2
Molar Solubility
•! solubility is the amount of solute that will dissolve in a given amount of solution
!!at a particular temperature
•! the molar solubility is the number of moles of solute that will dissolve in a liter of solution
!!the molarity of the dissolved solute in a saturated solution
•! for the general reaction MnXm(s) ! nMm+(aq) + mXn!(aq)
Molar Solubility =
Ksp and Relative Solubility
•! molar solubility is related to Ksp
•! but you cannot always compare solubilities of
compounds by comparing their Ksps
•! in order to compare Ksps, the compounds must
have the same dissociation stoichiometry
Solubility-Product Constants (Ksp) of Selected Ionic
Compounds at 250C
Name, Formula Ksp
Aluminum hydroxide, Al(OH)3
Cobalt (II) carbonate, CoCO3
Iron (II) hydroxide, Fe(OH)2
Lead (II) fluoride, PbF2
Lead (II) sulfate, PbSO4
Silver sulfide, Ag2S
Zinc iodate, Zn(IO3)2
3 x 10-34
1.0 x 10-10
4.1 x 10-15
3.6 x 10-8
1.6 x 10-8
4.7 x 10-29
8 x 10-48
Mercury (I) iodide, Hg2I2
3.9 x 10-6
Relationship Between Ksp and Solubility at 250C
No. of Ions Formula Cation:Anion Ksp Solubility (M)
2 MgCO3 1:1 3.5 x 10-8 1.9 x 10-4
2 PbSO4 1:1 1.6 x 10-8 1.3 x 10-4
2 BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5
3 Ca(OH)2 1:2 5.5 x 10-6 1.2 x 10-2
3 BaF2 1:2 1.5 x 10-6 7.2 x 10-3
3 CaF2 1:2 3.2 x 10-11 2.0 x 10-4
3 Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5
•! addition of a soluble salt that contains one of the
ions of the “insoluble” salt, decreases the
solubility of the “insoluble” salt
•! for example, addition of NaCl to the solubility
equilibrium of solid PbCl2 decreases the
solubility of PbCl2
PbCl2(s) ! Pb2+(aq) + 2 Cl!(aq) addition of Cl! shifts the equilibrium to the left
The Effect of Common Ion on Solubility
The Effect of pH on Solubility
•! for insoluble ionic hydroxides, the higher the pH, the
lower the solubility of the ionic hydroxide
!!and the lower the pH, the higher the solubility
!!higher pH = increased [OH!]
M(OH)n(s) ! Mn+(aq) + nOH!(aq)
•! for insoluble ionic compounds that contain anions of
weak acids, the lower the pH, the higher the solubility
M2(CO3)n(s) ! 2 Mn+(aq) + nCO32!(aq)
H3O+(aq) + CO3
2! (aq) ! HCO3! (aq) + H2O(l)
Precipitation
•! precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound
•! if we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur
!!Q = Ksp, the solution is saturated, no precipitation
!!Q < Ksp, the solution is unsaturated, no precipitation
!!Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate
•! some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions
The effect of a common ion on solubility
PbCrO4(s) Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO4
2-(aq)
CrO42- added
Cr(NH3)63+, a typical complex ion.
M(H2O)42+
M(H2O)3(NH3)2+
M(NH3)42+
NH3
3NH3
The stepwise exchange of NH3 for H2O in M(H2O)42+.
The amphoteric behavior of aluminum hydroxide.
Al(H2O)3(OH)3(s) 3H2O(l) + Al(H2O)3(OH)3(s) Al(H2O)3(OH)4-(s) + H2O(l)
Solubility of amphoteric hydroxides
The general procedure for separating ions in qualitative analysis.
Add
precipitating ion
Cen
trifuge
Add
precipitating ion
Cen
trifuge