III. Titrations
• In an acid-base titration, a basic (or
acidic) solution of unknown [ ] is reacted
with an acidic (or basic) solution of
known [ ].
• An indicator is a substance used to
visualize the endpoint of the titration.
• At the equivalence point, the number of
moles of acid equals the number of
moles of base.
III. Visual of a Titration
III. Titration/pH Curves
• A titration or pH curve is a plot of how
the pH changes as the titrant is added.
• It is possible to calculate the pH at any
point during a titration.
• Multiple pH’s can be calculated, and the
results plotted to create the theoretical
titration curve.
III. Find the Equivalence Point!
•
•
The keys to these types of problems
are writing the titration equation and
finding the equivalence point of the
titration.
The calculation then depends on what
region of the titration curve you are in:
1)
2)
3)
4)
Before titration begins
Pre-equivalence
Equivalence point
Post-equivalence
III. Illustrative Problem
• Sketch the pH curve for the titration of
25.0 mL of 0.100 M HCl with 0.100 M
NaOH.
III. Illustrative Problem Solution
1) Write the titration equation.

HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
2) Calculate the equivalence point. What
volume of NaOH is needed to completely
react with HCl?
III. Illustrative Problem Solution
3) Calculate initial pH before titration.

Since HCl is strong, 0.100 M HCl has
[H3O+] = 0.100 M, and pH = 1.000.
4) Calculate the pH of some points in the
pre-equivalence region.


As NaOH is added, the neutralization
reaction OH-(aq) + H3O+(aq)  H2O(l) takes
place.
We calculate the pH after addition of 5.00
mL of NaOH.
III. Illustrative Problem Solution
• To calculate the pH after 5.00 mL, we need to
calculate initial moles of acid and the number
of moles of base.
• We put these moles into a reaction chart.
III. Illustrative Problem Solution
H3O+
Initial
0.00250 mol
+
OH0.000500 mol

2H2O
0 mol
Change -0.000500 mol -0.000500 mol +0.001000 mol
Final
0.00200 mol
0 mol
0.001000 mol
The H3O+ leftover is in a larger volume so
we calculate its concentration.
III. Illustrative Problem Solution
• We do the same thing for some other
points: 10.0 mL, 15.0 mL, and 20.0 mL.
• Results summarized below.
Volume (ml)
5.00
10.0
15.0
20.0
Mol H3O+ [H3O+] (M)
0.00200
0.06667
0.00150
0.04286
0.00100
0.02500
0.00050
0.01111
pH
1.176
1.368
1.602
1.954
III. Illustrative Problem Solution
5) Calculate the pH at the equivalence
point.

For a strong-strong titration, pH always
equals 7.00 at the equivalence point!
6) Calculate the pH of some points in the
post-equivalence region.

In this region, the pH depends on the
excess OH- added.
III. Illustrative Problem Solution
• To find the excess added, calculate how
many mL past the equivalence point
have been added, convert to moles, and
divide by total volume. For 30.0 mL:
pH can then be found from pOH.
III. Illustrative Problem Solution
• Again, calculate for additional points like
35.0, 40.0, and 50.0 mL. Results
summarize below.
Volume (mL)
[OH-] (M)
pOH
pH
30.0
35.0
40.0
50.0
0.009091
0.01667
0.02307
0.03333
2.0414
1.7781
1.6370
1.4771
11.959
12.222
12.363
12.523
III. Illustrative Problem Solution
• Now we plot the data points and sketch the
pH titration curve!
III. Sample Problem
• A 0.0500 L sample of 0.0200 M KOH is
being titrated with 0.0400 M HI. What is
the pH after 10.0 mL, 25.0 mL, and 30.0
mL of the titrant have been added?
III. Weak Acid/Base Titrations
• The situation becomes a little more
complicated when a weak acid/base is
titrated with a strong base/acid.
• Again, the keys are to identify the
titration reaction and the equivalence
point.
• The method of calculating the pH will
then depend on the region of the
titration curve.
III. Four Different Regions
•
For a weak acid titrated with a strong base,
there are 4 regions as well:
1) Before titration: only HA in solution, so it’s a weak
acid problem!
2) Pre-equivalence: a mixture of HA and A-, so it’s a
buffer!
3) Equivalence point: only A- in solution, so it’s a
weak base problem!
4) Post-equivalence: adding excess OH-, so it’s a
dilution problem!
•
For a weak base titrated with a strong acid,
everything is just rewritten w/ conjugates!
III. Illustrative Problem
• A 50.00 mL sample of 0.02000 M
CH3COOH is being titrated with 0.1000
M NaOH. Calculate the pH before the
titration begins, after 3.00 mL of the
titrant have been added, at the
equivalence point, and after 10.20 mL of
the titrant have been added. Note that
Ka = 1.75 x 10-5 for acetic acid.
III. Illustrative Problem Solution
1) First, we need to write the titration eqn.

CH3COOH(aq) + OH-(aq)  CH3COO-(aq) + H2O(l)
2) Next, we calculate the equivalence point.
III. Illustrative Problem Solution
3) Before titration begins, it’s just a weak
acid problem.
CH3COOH
Initial
Change
Equil.
0.02000 M
-x
0.02000-x
+ H2O  H3O+
-------
0
+x
x
+ CH3COO0
+x
x
Solving this (with simplification), we get [H3O+] =
5.916 x 10-4, so pH = 3.228.
III. Illustrative Problem Solution
4) After 3.00 mL of 0.1000 M NaOH have
been added, we will have a mixture of
CH3COOH and CH3COO- in solution.

Since it’s a buffer, we can use the
Henderson-Hasselbalch eqn.
III. Illustrative Problem Solution
• In the H-H equation, we need a ratio of
CH3COO- to CH3COOH.
• After adding 3.00 mL, we are 3.00/10.00
to equivalence. We can use a relative
concentration chart.
CH3COOH +
OH-
 CH3COO- + H2O
Relative Initial
1
0
0
Change
-3.00/10.00 +3.00/10.00 +3.00/10.00
Relative Final 7.00/10.00
0
3.00/10.00
-------
III. Illustrative Problem Solution
• Now we just plug the relative final row
into the H-H equation.
III. Illustrative Problem Solution
5) At the equivalence point, all CH3COOH has
been converted to CH3COO-.


Initial moles of CH3COOH = moles of CH3COOat the equivalence point, but the volume has
increased.
Must calculate [CH3COO-] at equiv. pt.
III. Illustrative Problem Solution
• Now we solve a weak base problem.
CH3COO-
+ H2O  OH-
+
CH3COOH
Initial
0.016667 M
---
0
0
Change
Equil.
-x
0.016667-x
-----
+x
x
+x
x
Using Kb = 5.714 x 10-10 and the simplification,
x = 3.089 x 10-6. Thus, pOH = 5.5102 and pH = 8.490.
Note that pH does not equal 7.00!!
III. Illustrative Problem Solution
6) At 10.20 mL of added titrant, we are
0.20 mL past equivalence, and the pH
depends only on excess OH-. Thus:
Of course, this means that pOH = 3.479 and
pH = 10.52.
III. Sample Problem
• A 25.00 mL sample of 0.08364 M
pyridine is being titrated with 0.1067 M
HCl. What’s the pH after 4.63 mL of the
HCl has been added? Note that
pyridine has a Kb of 1.69 x 10-9.
III. Sample Weak/Strong Curves
• Important aspects about pH curves for
weak/strong titrations:
 At equivalence, pH does not equal 7.00.
 At ½ equivalence, pH = pKa.
III. Polyprotic Acid Titration
• If the Ka’s are
different enough,
you will see multiple
equivalence points.
• Since protons come
off one at a time, 1st
equiv. pt. refers to
the 1st proton, 2nd to
the 2nd, etc.
III. Detecting the Equiv. Pt.
• During a titration, the equivalence point
can be detected with a pH meter or an
indicator.
• The point where the indicator changes
color is called the endpoint.
• An indicator is itself a weak acid that
has a different color than its conjugate
base.
III. Phenolphthalein
III. Indicators
• The indicator has its own equilibrium:
 HIn(aq) + H2O(l)  In-(aq) + H3O+(aq)
• The color of an indicator depends on
the relative [ ]’s of its protonated and
deprotonated forms.
 If pH > pKa of HIn, color will be In-.
 If pH = pKa of HIn, color will be in between.
 If pH < pKa of HIn, color will be HIn.
III. Selecting an Indicator
IV. Solubility
• In 1st semester G-chem, you memorized
solubility rules and regarded
compounds as either soluble or
insoluble.
• Reality is not as clear cut – there are
degrees of solubility.
• We examine solubility again from an
equilibrium point of view.
IV. Solubility Equilibrium
• If we apply the equilibrium concept to
the dissolution of CaF2(s), we get:
 CaF2(s)  Ca2+(aq) + 2F-(aq)
• The equilibrium expression is then:
 Ksp = [Ca2+][F-]2
• Ksp is the solubility product constant,
and just like any other K, it tells you how
far the reaction goes towards products.
IV. Some Ksp Values
IV. Calculating Solubility
• Recall that solubility is defined as the
amount of a compound that dissolves in
a certain amount of liquid (g/100 g water
is common).
• The molar solubility is obviously the
number of moles of a compound that
dissolves in a liter of liquid.
• Molar solubilities can easily be
calculated using Ksp values.
IV. Ksp Equilibrium Problems
• Calculating molar solubility is essentially
just another type of equilibrium problem.
• You still set up an equilibrium chart and
solve for an unknown. Pay attention to
stoichiometry!
Fe(OH)3(s)  Fe3+(aq) + 3OH-(aq)
Initial
---
0
0
Change
---
S
3S
Equil.
---
S
3S
IV. Sample Problem
• Which is more soluble: calcium
carbonate (Ksp = 4.96 x 10-9) or
magnesium fluoride (Ksp = 5.16 x 10-11)?
IV. The Common Ion Effect
• The solubility of Fe(OH)2 is lower when
the pH is high. Why?
 Fe(OH)2(s)  Fe2+(aq) + 2OH-(aq)
 Le Châtelier’s Principle!
• common ion effect: the solubility of an
ionic compound is lowered in a solution
containing a common ion than in pure
water.
IV. Sample Problem
• Calculate the molar solubility of lead(II)
chloride (Ksp = 1.2 x 10-5) in pure water
and in a solution of 0.060 M NaCl.
IV. pH and Solubility
• As seen with Fe(OH)2, pH can have an
influence on solubility.
• In acidic solutions, need to consider if
H3O+ will react with cation or anion.
• In basic solutions, need to consider if
OH- will react with cation or anion.
IV. Sample Problems
a) Which compound, FeCO3 or PbBr2, is
more soluble in acid than in base?
Why?
b) Will copper(I) cyanide be more soluble
in acid or base? Why?
c) In which type of solution is AgCl most
soluble: acidic, basic, or neutral?
IV. Precipitation
• Ksp values can be used to predict when
precipitation will occur.
• Again, we use a Q calculation.
 If Q < Ksp, solution is unsaturated. Solution
dissolve additional solid.
 If Q = Ksp, solution is saturated. No more
solid will dissolve.
 If Q > Ksp, solution is supersaturated, and
precipitation is expected.
IV. Sample Problems
a) Will a precipitate form if 100.0 mL
0.0010 M Pb(NO3)2 is mixed with
100.0 mL 0.0020 M MgSO4?
b) The concentration of Ag+ in a certain
solution is 0.025 M. What
concentration of SO42- is needed to
precipitate out the Ag+? Note that Ksp
= 1.2 x 10-5 for silver(I) sulfate.
V. Complex Ions
• In aqueous solution, transition metal
cations are usually hydrated.
 e.g. Ag+(aq) is really Ag(H2O)2+(aq).
 The Lewis acid Ag+ reacts with the Lewis
base H2O.
• Ag(H2O)2+(aq) is a complex ion.
 A complex ion has a central metal bound to
one or more ligands.
 A ligand is a neutral molecule or an ion that
acts as a Lewis base with the central
metal.
V. Formation Constants
• Stronger Lewis bases will replace
weaker ones in a complex ion.
 e.g. Ag(H2O)2+(aq) + 2NH3(aq) 
Ag(NH3)2+(aq) + 2H2O(l)
 For simplicity, it’s common to write
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
 Since this is an equilibrium, we can write
an equilibrium expression for it.
V. Formation Constants
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
• Kf is called a formation constant.
• Unlike other equilibrium constants we’ve
seen, Kf’s are large, indicating favorable
formation of the complex ion.
V. Sample Formation Constants
V. Calculations w/ Kf’s
• Since Kf’s are so large, calculations with
them are slightly different.
• We assume the equilibrium lies
essentially all the way to the right.
• This changes how we set up our
equilibrium chart.
V. Illustrative Problem
• Calculate the concentration of Ag+ ion in
solution when 0.085 g silver(I) nitrate is
added to a 250.0 mL solution that is
0.20 M in KCN.
V. Illustrative Problem Solution
1) First, we must identify the complex
ion.



In solution, we will have Ag+, NO3-, and
K+, and CN-.
The complex ion must be made from Ag+
and CN-.
Looking at table of Kf’s, we find that
Ag(CN)2- has Kf = 1 x 1021.
V. Illustrative Problem Solution
2) Next, we need concentrations.


Already know that [CN-] = 0.20 M.
We calculate the [Ag+].
V. Illustrative Problem Solution
3) Now we set up our equilibrium chart.

Since Kf is so big, we assume the
reaction essentially goes to completion.
Ag+(aq)
+ 2CN-(aq)  Ag(CN)2-(aq)
Initial
0.00200 M
0.20 M
0
Change
≈ -0.00200
≈ -0.00400
≈ +0.00200
x
0.196
0.00200
Equil.
V. Illustrative Problem Solution
4) Finally, we solve for x.
Thus, [Ag+] = 5 x 10-23. It is very small, so our
approximation is valid. Note that book would
use 0.20 for [CN-] in the calculation.
V. Sample Problem
• A 125.0-mL sample of a solution that is
0.0117 M in NiCl2 is mixed with a 175.0mL sample of a solution that is 0.250 M
in NH3. After the solution reaches
equilibrium, what concentration of
Ni2+(aq) remains?
V. Complex Ions & Solubility
• Formation of complex ions enhances the
solubility of some normally insoluble ionic
compounds.
• Typically, Lewis bases will enhance solubility.
• e.g. Adding NH3 to a solution containing
AgCl(s) will cause more AgCl(s) to dissolve.
AgCl(s)  Ag+(aq) + Cl-(aq)
Ksp = 1.77 x 10-10
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq) Kf = 1.7 x 107
V. Complex Ion Formation
V. Metal Hydroxides
• All metal hydroxides can act as bases.
 e.g. Fe(OH)3(s) + 3H3O+(aq)  Fe3+(aq) +
6H2O(l)
• Some metal hydroxides can act as
acids and bases; they are amphoteric.
 In addition to the above, Al(OH)3(s) can also
absorb hydroxide.
 Al(OH)3(s) + OH-(aq)  Al(OH)4-(aq)
V. Aluminum Hydroxide
• In acidic solutions:
 Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+(aq) + H3O+(aq)
• As OH- is added, solution becomes neutral.
 Al(H2O)5(OH)2+(aq) + 2OH-(aq)  Al(H2O)3(OH)3(s) +
2H2O(l)
• In basic solutions:
 Al(H2O)3(OH)3(s) + OH-(aq)  Al(H2O)2(OH)4-(aq)
• Thus, solubility is very pH dependent.
V. Aluminum Hydroxide
V. Amphoteric Hydroxides
• There are not that many metal
hydroxides that are amphoteric.
• Only Al3+, Cr3+, Zn2+, Pb2+, and Sn2+.