Chapter 17: Additional Aspects
of Aqueous Equilibria
SC 132 CHEM 2
Chemistry: The Central Science
CM Lamberty
The Common Ion Effect

Consider the given equilibrium
The extent weak acid, HA, dissociates to form
H3O+ and determine pH is based on its Ka
 Until now in acid-base equilibria, we’ve only
started with reactant
 What happens if we start with some acid/base
AND some of its conjugate


Think about LeChatelier’s Principle before you answer
Buffers

Solns composed of



Initial concentration of weak species (acid or base)
Initial concentration of its conjugate
Main characteristics of buffers


Lower dissociation percentage of the weak species (common
ion effect)
Resists changes to pH after addition of acid or base
Key Buffer Components

Consider the expression for Ka of a weak acid:

If we solve for [H3O+], we get
So, [H3O+] can only change as the buffer-component
ration concentrations change
Addition of acid or base changes both factors in the
ratio, but the numerical value of the ratio changes very
little!


Calculating Buffer pH Practice
PROBLEM:
Calculate the pH of a buffer that consists 1.0M sodium phenolate
(C6H5COONa) and 1.2M phenol (C6H5COOH) (pKa of phenol is
10.00)
PROBLEM:
What is the buffer component concentration ratio, [NO2-]/[HNO2],
of a buffer that has a pH of 2.95 (Ka = 7.1x10-4 for HNO2)
The Henderson Hasselbach Eqtn
Remember, you NEVER have to use the
Henderson-Hasselbach eqtn…you can
always get the same info by constructing
an ICE table
 Convenience of Henderson-Hasselbach:

Get pH directly
 Simplifies buffer preparation problems

Buffer Capacity and Buffer Range
Buffer capacity is the ability to resist pH change.
The more concentrated the components of a buffer, the greater
the buffer capacity.
The pH of a buffer is distinct from its buffer capacity.
A buffer has the highest capacity when the component
concentrations are equal.
Buffer range is the pH range over which the buffer acts effectively.
Buffers have a usable range within ± 1 pH unit of the pKa of
its acid component.
Buffer ()pH After H3O+ or OH- Addition

There is an additional step to add to the
calculation process!
We must write the acid-base rxn b/t the H3O+
or OH- added and the buffer component it
reacts with
 Then, we recalculate the concentrations of
each buffer component


Plug these values into ICE table &
calculate what you’re being asked for
Practice
PROBLEM:
Calculate the pH:
(a) of a buffer solution consisting of 0.50M CH3COOH and 0.50M CH3COONa
(b) after adding 0.020mol of solid NaOH to 1.0L of the buffer solution in part (a)
(c) after adding 0.020mol of HCl to 1.0L of the buffer solution in part (a)
Ka of CH3COOH = 1.8x10-5. (Assume the additions cause negligible volume
changes.
Preparing a Buffer
1. Choose the conjugate acid-base pair.
2. Calculate the ratio of buffer component concentrations.
3. Determine the buffer concentration.
4. Mix the solution and adjust the pH.
Acid-Base Titrations

Controlled neutralization rxns can be
carried for a number of purposes
1.
2.

Determine concentration of an unknown acid
or base
Determine the Ka value of an unknown acid
or base
Our job will be to calculate pH values after
the addition of various volumes of added
acid or base
Acid/Base Titration Curves
•
•
•
You are dispensing one soln (from a buret) into a
known quantity of another soln (in an E. flask) or viceversa
Goal is to calculate pH after the addition of some amt
of titrant
Big Questions:
1. Which species are in soln @ indicated point of titration?
2. Do these species affect pH? If so, how?
a.
b.
c.
Only acids, bases, and conjugates of WEAK acids/bases influence pH
What type of titration is this (SA or WA/SSB or WB)
Is there an equil involved that must be taken into account?
Titrating Strong with Strong
4 main parts of titration curve:
1.
2.
3.
4.
Before any titrant added
–
Calculate pH from complete dssn of
starting material
–
pH dependent on if starting material is an
acid or base
Some titrant added, before EqP
–
See how much starting material used up
and what’s left
–
Calculate pH from complete dssn of left
over starting material
At EqP
–
No acid/base left; only have salt
–
Since both are strong species, salt DOES
NOT affect pH
*
pH = 7 @ EqP as salt of SA & SSB rxn is
NEUTRAL
Past EqP
–
Now have excess titrant
–
Determine amt of excess titrant
–
Calculate pH from complete dssn of
excess titrant
Steps to Titration Calculations

Before beginning:


Always determine the equivalence point in the titration
There are always 2 parts to a titration calculation
problem after some titrant has been added:
1.
2.
Write the acid-base reaction to determine which species
will be left in the container that affect pH
Write the reaction of the leftover acid or base with water
to determine [H3O+] or [OH-] in order to calculate pH


If species is weak, it will be an equilibrium calculation (and
probably a buffer)
If species is strong, just calculate its concnetration & take log
Practice
Calculate the pH of a solution if 25.00mL
of 0.200M HBr is mixed with 12.50mL of
0.100 M Ba(OH)2
 After 25.00mL of 0.100M Ba(OH)2
 After 50.00mL of Ba(OH)2

Equilibria of Slightly Soluble Ionic
Compounds
For the hypothetical compound, MpXq (s)
At equilibrium
Qsp = [Mn+]p [Xz-]q = Ksp
The magnitude of Ksp is a measure of the compound’s solubility in H2O
• Lower Ksp (more negative exponent) means lower solubility
Practice
PROBLEM: Write the ion-product expression for each of the following:
(a) Magnesium carbonate
(c) Calcium phosphate
(b) Iron (II) hydroxide
(d) Silver sulfide
Ksp Calculations

Again 2 basic types:
1.
2.



Calculate solubility (aka “X”) if Ksp is known

May require additional steps depending on what youre
being asked for
Calculate Ksp if solubility (X) is known
Remember, must convert all concentrations to
molarity
Because reactant is solid, ICE table starts under the
arrow
Common ion problems occur when soln starts w/
one of the solid’s ions to begin with

End result like in buffer is decreased dissociation (so
decreased solubility
Practice
PROBLEM:(a) Lead (II) sulfate is a key component in lead-acid car
batteries. Its solubility in water at 250C is 4.25x10-3g/100mL
solution. What is the Ksp of PbSO4?
(b) When lead (II) fluoride (PbF2) is shaken with pure water
at 250C, the solubility is found to be 0.64g/L. Calculate the
Ksp of PbF2.
PROBLEM: Calcium hydroxide (slaked lime) is a major component of
mortar, plaster, and cement, and solutions of Ca(OH)2 are
used in industry as a cheap, strong base. Calculate the
solubility of Ca(OH)2 in water if the Ksp is 6.5x10-6.
PROBLEM: In Sample Problem 19.6, we calculated the solubility of
Ca(OH)2 in water. What is its solubility in 0.10M
Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.
PROBLEM: A common laboratory method for preparing a precipitate is to
mix solutions of the component ions. Does a precipitate form
when 0.100L of 0.30M Ca(NO3)2 is mixed with 0.200L of
0.060M NaF?
Practice
PROBLEM:
An environmental chemist needs a carbonate buffer of pH 10.00
to study the effects of the acid rain on limsetone-rich soils. How
many grams of Na2CO3 must she add to 1.5L of freshly prepared
0.20M NaHCO3 to make the buffer? Ka of HCO3- is 4.7x10-11.
Factors Affecting Solubility
Temperature
 Presence of common ions

Common Ion Effect
 Solubility of slightly soluble salt is DECREASED by
the presence of a second solute that furnishes a
common ion


pH of solution
Addition of strong acid (H3O+) will INCREASE the
solubility of an insoluble salt if
 Anion is conjugate base of a weak acid
 Think about LeChatelier’s Principle

Factors Affecting Solubility

Presence of Complex Ions
Metal ions can act like Lew acids toward H2O or
other Lewis bases
 Both solubility of metal salt and metal ion and
Lewis base must be added together
 Need the formation constant (Kf) for complex ions
 Solubility of metal salts increases in the presence
of suitable Lewis bases (NH3, CN-, OH-)

Amphoterism
Amphoteric oxides and hydroxides are
substances soluble in SA and SB because
they can act as either acid or base
 Al+3, Cr+3, Zn+2, Sn+2
 Dissolve in acidic soln b/c contain basic
anions
 Dissolve in base b/c form complex ions

Precipitation and Separation of Ions

Use Q to determine direction of reaction
Q > Ksp ppt occurs until Q=Ksp
 Q= Ksp, equilibrium
 Q < Ksp, solid dissolves until Q=Ksp


Selective ppt of ions

Way of separating ions form one another
Qualitative
Analysis
Practice
PROBLEM: Write balanced equations to explain whether addition of H3O+ from a
strong acid affects the solubility of these ionic compounds:
(a) Lead (II) bromide
(b) Copper (II) hydroxide
(c) Iron (II) sulfide
Free-for-all Practice





Calculate the pH, pOH, [H3O+], and [OH-] of a solution
that is 0.0579M Ba(OH)2
Calculate the pH of a buffer solution that is 0.500M
NH3 and 0.675M NH4Cl. Ka for NH4+ is 1.75x10-5
If 1.36mg/L AgBr produces a saturated soln,
determine its solubility product constant. What would
its molar solubility be in a solution of 0.0750M MgBr2?
What is the maximum amount of solid sodium sulfate
that can be added to 1.00 L of 0.0020 M Ca(NO3)2
before precipitation of calcium sulfate begins? Ksp =
2.4  10-5 for calcium sulfate
A 0.050 M solution of the weak acid HA has [H3O+] =
3.77  10-4 M. What is the Ka for the acid? What is
the solution pH?
Free-for-all Practice


Buffer solutions with the component concentrations shown below
were prepared. Which of them should have the lowest pH?
A. [CH3COOH] = 0.25 M, [CH3COO-] = 0.25 M
B. [CH3COOH] = 0.75 M, [CH3COO-] = 0.75 M
C. [CH3COOH] = 0.75 M, [CH3COO-] = 0.25 M
D. [CH3COOH] = 0.25 M, [CH3COO-] = 0.75 M
E. [CH3COOH] = 1.00 M, [CH3COO-] = 1.00 M
What will be the effect of adding 0.5 mL of 0.1 M NaOH to 100 mL
of an acetate buffer in which [CH3COOH] = [CH3COO-] = 0.5 M?
A. The pH will increase slightly.
B. The pH will increase significantly.
C. The pH will decrease slightly.
D. The pH will decrease significantly.
E.
Since it is a buffer solution, the pH will not be affected.
Free-for-all Practice

Which of the following has the highest buffer capacity?
A. 0.10 M H2PO4-/0.10 M HPO42B. 0.50 M H2PO4-/0.10 M HPO42C. 0.10 M H2PO4-/0.50 M HPO42D. 0.50 M H2PO4-/0.50 M HPO42E.
They all have the same buffer capacity.

Which of the following acids should be used to prepare a buffer
with a pH of 4.5?
A. HOC6H4OCOOH, Ka = 1.0  10-3
B. C6H4(COOH)2, Ka = 2.9  10-4
C. CH3COOH, Ka = 1.8  10-5
D. C5H5O5COOH-2, Ka = 4.0  10-6
E.
HBrO, Ka = 2.3  10-9
Free-for-all Practice


A.
B.
C.
D.
E.
According to Brønsted and Lowry, which one of the following is not a
conjugate acid-base pair?
A.
H3O+/OHB.
CH3OH2+/CH3OH
C.
HI/ID.
HSO4-/SO42E.
H2/HThe acid dissociation constant Ka equals 1.26  10-2 for HSO4- and is 5.6 
10-10 for NH4+. Which statement about the following equilibrium is correct?
HSO4-(aq) + NH3(aq)
SO42-(aq) + NH4+(aq)
The reactants will be favored because ammonia is a stronger base than
the sulfate anion.
The products will be favored because the hydrogen sulfate ion is a
stronger acid than the ammonium ion.
Neither reactants or products will be favored because all of the species are
weak acids or bases.
The initial concentrations of the hydrogen sulfate ion and ammonia must
be known before any prediction can be made.
This reaction is impossible to predict, since the strong acid and the weak
base appear on the same side of the equation.