Download Aqueous Ionic Equilibrium

Chapter 16
“Aqueous
Equilibrium”
The Danger of Antifreeze
•
•
Each year, thousands of pets and wildlife die from consuming
antifreeze
Most brands of antifreeze contain ethylene glycol
•
Metabolized in the liver to glycolic acid
•
If present in high enough concentration in the bloodstream, it
overwhelms the buffering ability of HCO3−, causing the blood pH
to drop
When the blood pH is low, it ability to carry O2 is compromised
– sweet taste and initial effect is drunkenness
•
– HOCH2 COOH
– acidosis
•
The treatment is to give the patient ethyl alcohol, which has a
higher affinity for the enzyme that catalyzes the metabolism of
ethylene glycol
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
1
Making an Acid Buffer
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 H3 O+ ion concentration
2
Common Ion Effect
Practice
•
What is the pH of a buffer that is 0.100 M HC2H3 O 2 and 0.100 M
NaC2 H3O2?
Practice
•
What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and
0.071 M KF?
3
Henderson-Hasselbalch
Equation
• Calculating the pH of a buffer solution can be
simplified by using an equation derived from
the Ka expression called the HendersonHasselbalch Equation
• The equation calculates the pH of a buffer
from the Ka and initial concentrations of the
weak acid and salt of the conjugate base
– as long as the “x is small” approximation is valid
Henderson-Hasselbalch
Equation
& [A - ] #
!
pH = pK a + log$$
!
% [HA] "
Example
•
What is the pH of a buffer that is 0.050 M HC7H5 O 2 and 0.150 M
NaC7 H5O2?
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Example
•
What is the pH of a buffer that is 0.14 M HF (pK a = 3.15) and
0.071 M KF?
Using the HendersonHasselbalch Equation
• The Henderson-Hasselbalch equation is
generally good enough when the “x is small”
approximation is applicable
• Generally, the “x is small” approximation will
work when both of the following are true:
a)the initial concentrations of acid and salt are not very
dilute
b)the K a is fairly small
• For most problems, this means that the initial
acid and salt concentrations should be over
1000x larger than the value of Ka
Example
•
What is the pH of a buffer that has 0.100 mol HC2 H3 O 2 and 0.100 mol
NaC2H3O2 in 1.00 L that has 0.010 mol NaOH added to it
5
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−
Example
•
What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and
0.20 M NH4 Cl?
Buffering Effectiveness
• A good buffer should be able to neutralize
moderate amounts of added acid or base
– 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
• (2) the absolute concentrations of acid and base
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Effect of Relative Amounts of
Acid and Conjugate Base
• 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
Buffering Range
• 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
Lowest pH
Highest pH
pH = pK a + log(0.10 )
pH = pK a + log(10 )
pH = pK a ! 1
pH = pK a + 1
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
Example
•
Which of the following acids would be the best choice to
combine with its sodium salt to make a buffer with pH 4.25?
Chlorous Acid, HClO2
pKa = 1.95
Nitrous Acid, HNO2
Formic Acid, HCHO2
pKa = 3.34
pKa = 3.74
Hypochlorous Acid, HClO
pKa = 7.54
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Example
•
What ratio of NaCHO2 : HCHO2 would be required to make a
buffer with pH 4.25?
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]
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
8
Titration
Titration
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
9
Titration Curve: Strong Base
Added to Strong Acid
Example
• Show the titration of 25 mL of 0.100 M HCl
with 0.100 M NaOH
Titrating Weak Acid with a Strong
Base: Overview
•
•
•
•
1. Before any titrant (pH of weak acid)
2. After a small amount of titrant (~5mL)
3. At Half titration point (pH = pKa)
4. At the equivalence point
– (mol acid = mol base)
• 5. A drop beyond the endpoint (~0.5mL)
10
Titrating Weak Acid with a Strong
Base
• 1. The initial pH is that of the weak acid
solution
– calculate like a weak acid equilibrium problem
• 2. 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
• 3. Half-neutralization pH = pKa
Titrating Weak Acid with a Strong
Base
• 4. 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
• 5. Beyond equivalence point, the OH is in excess
– [OH−] = mol MOH xs/total liters
– [H3O+][OH−]=1 x 10-14
Example
•
A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH.
Calculate all 5 points.
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Titration Curve of a Weak
Base with a Strong Acid
Titration of a Polyprotic Acid
titration of 25.0 mL of
0.100 M H2 SO3 with
0.100 M NaOH
Monitoring pH During a
Titration
• The general method for monitoring the pH during the
course of a titration is to measure the conductivity of
the solution due to the [H3O+]
– using a probe that specifically measures just H3O +
• The endpoint of the titration is reached at the
equivalence point in the titration – at the inflection
point of the titration curve
• If you just need to know the amount of titrant added
to reach the endpoint, we often monitor the titration
with an indicator
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Monitoring pH During a
Titration
Phenolphthalein
Methyl Red
H
C
(CH3)2N
H
C
C
C
C
H
C
H
H
C
H
C
N
H3O+
(CH3)2N
C
C
C
H
N
N
OH-
N
N
H
H
C
H
C
C
C
H
H
C
H
C
C
C
H
C
CH
NaOOC
N
C
C
H
CH
NaOOC
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Monitoring a Titration with an
Indicator
• For most titrations, the titration curve shows a
very large change in pH for very small
additions of base near the equivalence point
• An indicator can therefore be used to
determine the endpoint of the titration if it
changes color within the same range as the
rapid change in pH
– pKa of HInd ≈ pH at equivalence point
Indicators
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
• The concepts of equilibrium can be applied to
salts dissolving, and then one can use the
equilibrium constant for the process to
measure relative solubilities in water
14
Solubility Product
• The equilibrium constant for the dissociation of a
solid salt into its aqueous ions is called the solubility
product, K sp
• For an ionic solid MnXm, the dissociation reaction is:
MnXm(s) ⇔ nMm+(aq) + mXn−(aq)
• The solubility product would be
K sp = [Mm+] n[Xn−] m
• For example, the dissociation reaction for PbCl2 is
PbCl2(s) ⇔ Pb2+(aq) + 2 Cl−(aq)
• Its equilibrium constant is
K sp = [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 = (n + m )
K sp
n
n
•m
m
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Example
•
Calculate the molar solubility of PbCl2 in pure water at 25°C
Example
•
Determine the K sp of PbBr2 if its molar solubility in water at 25°C
is 1.05 x 10-2 M
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
16
The Effect of Common Ion on
Solubility
• 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
Example
•
Calculate the molar solubility of CaF2 in 0.100 M NaF at 25°C
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) + nCO3 2− (aq)
H3 O+(aq) + CO3 2− (aq) ⇔ HCO3− (aq) + H2 O(l)
17
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 = K sp, the solution is saturated, no precipitation
• Q < K sp, the solution is unsaturated, no precipitation
• Q > K sp, 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
precipitation occurs
if Q > Ksp
a supersaturated solution will
precipitate if a seed crystal is added
Selective Precipitation
• 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
• A successful reagent can precipitate with
more than one of the cations, as long as their
Ksp values are significantly different
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Example
•
What is the minimum [OH−] necessary to just begin to precipitate Mg2+
(with [0.059]) from seawater?
Qualitative Analysis
• An analytical scheme that utilizes selective
precipitation to identify the ions present in a
solution is called a qualitative analysis
scheme
– wet chemistry
• 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
19
Group 1
• Group one cations are Ag+ , Pb2+, and Hg22+
• All these cations form compounds with Cl−
that are insoluble in water
– as long as the concentration is large enough
– PbCl2 may be borderline
• molar solubility of PbCl2 = 1.43 x 10-2 M
• Precipitated by the addition of HCl
Group 2
• Group two cations are Cd2+, Cu2+, Bi3+, Sn4+,
As3+, Pb2+, Sb3+, and Hg2+
• All these cations form compounds with HS−
and S2− that are insoluble in water at low pH
• Precipitated by the addition of H2S in HCl
20
Group 3
• Group three cations are Fe2+, Co2+, Zn2+,
Mn2+, Ni2+ precipitated as sulfides; as well as
Cr3+, Fe3+, and Al3+ precipitated as
hydroxides
• All these cations form compounds with S2−
that are insoluble in water at high pH
• Precipitated by the addition of H2S in NaOH
Group 4
• Group four cations are Mg2+, Ca2+, Ba2+
• All these cations form compounds with
PO43− that are insoluble in water at high
pH
• Precipitated by the addition of
(NH4)2HPO4
Group 5
• Group five cations are Na+, K+, NH4+
• All these cations form compounds that
are soluble in water – they do not
precipitate
• Identified by the color of their flame
21
Complex Ion Formation
• Transition metals tend to be good Lewis acids
• They often bond to one or more H2O molecules to
form a hydrated ion
– H2O is the Lewis base, donating electron pairs to form
coordinate covalent bonds
Ag+(aq) + 2 H2O(l) ⇔ Ag(H2 O)2+(aq)
• Ions that form by combining a cation with several
anions or neutral molecules are called complex ions
– e.g., Ag(H2O)2+
• The attached ions or molecules are called ligands
– e.g., H2O
Complex Ion Equilibria
• If a ligand is added to a solution that forms a
stronger bond than the current ligand, it will
replace the current ligand
Ag(H2 O)2 +(aq) + 2 NH3(aq) ⇔ Ag(NH3)2+(aq) + 2 H2O(l)
– Generally H2 O is not included, since its complex
ion is always present in aqueous solution
Ag+(aq) + 2 NH3(aq) ⇔ Ag(NH3)2+(aq)
Complex Ion Formation
• The reaction between an ion and
ligands to form a complex ion is called a
complex ion formation reaction
Ag+(aq) + 2 NH3(aq) ⇔ Ag(NH3)2+(aq)
• The equilibrium constant for the
formation reaction is called the
formation constant, Kf
22
Example
•
200.0 mL of 1.5 x 10-3 M Cu(NO3 ) 2 is mixed with 250.0 mL of
0.20 M NH3 . What is the [Cu2+] at equilibrium?
The Effect of Complex Ion
Formation on Solubility
• The solubility of an ionic compound that
contains a metal cation that forms a complex
ion increases in the presence of aqueous
ligands
AgCl(s) ⇔ Ag+(aq) + Cl−(aq)
Ksp = 1.77 x 10-10
Ag+(aq) + 2 NH3(aq) ⇔ Ag(NH3 )2+(aq)
K f = 1.7 x 107
•
Adding NH3 to a solution in equilibrium with
AgCl(s) increases the solubility of Ag+
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