ACID BASE TITRATIONS
Chapter 15
Titration Curves
• A plot of pH versus amount of acid or base
added.
• At the EQUIVALENCE POINT on a titration
curve, the amount of acid = the amount of
base.
• The END POINT of a titration is determined by
a color change of an indicator.
• Ideally, the end point and equivalence point
will be within 1 drop of each other.
INDICATORS
• Select an indicator based on the pH range of the equivalence
point. pKa of indicator within +/- 1 pH unit of the equivalence
point.
• Indicators are organic dyes whose colors depend on the
[H3O+] or pH of a solution.

Most are produced synthetically.
– Ex.
• Phenolphthalein
• Universal indicator – mix of organic acids that indicate
over different ranges

Many are vegetable dyes.
– Ex.
• Litmus
• Purple cabbage indicator
INDI
CAT
ORS
• Generally they are WEAK ORGANIC ACIDS
– Symbolized
HIn
– Dissociation reaction:
HIn + H2O  H30+ + In-
Ex. Bromthymol blue
• HIn – yellow
• In- - blue
HIn + H20  In- + H30+
• Ka = [H30+] [In-1]
[HIn]
When the ratio goes to 1/10, a color change will occur.
• Adding an acid shifts the equilibrium left.
• Adding a base shifts the equilibrium right.
Indicator sample problem
• An indicator, HIn, has a Ka = 1.0 x 10-7 .
Determine the pH at which a color
change will occur given the following
scenarios:
• Acid titration
• Base titration
INDICATOR EXAMPLE
Titration of an acid
Titration of a base
• The solution is initially
acidic, [HIn] is dominant.
• The color change occurs
when [In]/[HIn] = 1/10
Ka = 1.0 x 10-7 = [H+] (1/10)
[H+] = 10 (1.0 x 10-7) = 1.0 x 10-
• The solution is initially
basic, [In] is dominant.
• The color change occurs
when [In]/[HIn] = 10/1
Ka = 1.0 x 10-7 = [H+] (10/1)
[H+] = (1.0 x 10-7)/10 = 1.0 x
10-8
pH = 8.0
6
pH = 6.0
STRONG ACID/ STRONG BASE
• Consider the
titration of 50.0 ml
of 0.2 M nitric acid
with 0.1 M NaOH
Calculate the pH @
Major species in soln?
• 0.0 ml base added
• 10.0 ml base added
• 20.0 ml base added
SA/SB
Major species in soln?
• 50.0 ml base added
• 100.0 ml base added
• 200.0 ml base added
STRONG BASE/ STRONG ACID
• The pH Curve for
the titration of
100.0 mL of 0.50
M NaOH with 1.0
M HCI
WEAK ACID/ STRONG BASE
• The pH Curve for the
Titration of 50.0 mL
of 0.100 M HC2H3O2
with 0.100 M NaOH
WEAK ACID/ STRONG BASE
• Before any BASE is added, the pH depends only on the
weak acid.
• After some base is added, but before the EQUIVALENCE
POINT, a series of weak acid/ salt buffer solutions
determine the pH.
• At the EQUIVALNECE POINT, hydrolysis of the anion of
the weak acid determines the pH.
• Beyond the equivalence point, EXCESS STRONG BASE
determines the pH.
WA/SB Sample problem:
• 30.0 ml of 0.10 M NaOH is added to 50.0 ml of 0.10 M
HF. What is the pH after all 30.0 ml are added? Ka =
7.2 x 10-4
1.
2.
3.
4.
5.
6.
Major species in soln?
Rxn?
HF initial? (Use mmol)
OH- added?
HF consumed?
F- formed?
WA/SB After equilibrium
1. Rxn?
2. Ka expression
3. Calculate concentrations using mmoles and
volumes M = mmol/ml
4. ICE
5. [H+] and pH
WEAK BASE/ STRONG ACID
• Calculate the pH at each of the following points in
the titration of 50.00 ml of a 0.01000M sodium
phenolate (NaOC6H5) solution with 1.000 M HCl
soltuion. Ka for HOC6H5 = 1.05 x 10-10.
• Initial
• Midpoint
• Equivalence point
WB/SA
• Initial – weak base Kb
• pH = pKa at the midpoint,
so pOH = pKb since
[BH+]/[B] = 1
WB/SA
• At equivalence.
• How many moles of HCl were needed to neutralize?
MaVa = MbVb
HCl + OC6H5- 
• New volume?
• Weak acid dissociation reaction.
HOC6H5- + H20 
• Ka expression.
• [H+]
• pH
POLYPROTIC ACIDS
• Consider 20.00ml
or 0.100 M
polyprotic acid H2A
titrated with 0.100
M NaOH.
• Ka1 = 1x10-3
• Ka2 = 1x10-7
• 2 equivalence
points are
expected.
POLY pH calculations @
• 0 ml base added
• H2A dissociation
• H2A  H+ + HA-
• 10.0 ml base added
• H2A/ HA- buffer
H2A + OH-  HA- + H20
POLY pH calculations @
• 20.0 ml base added –
first equivalence
point
• 30.0 ml base added – ½ way
between 1st and 2nd equivalence
point.
H2A + OH-  HA- + OH-  A2-
POLY pH calculations @
• 40.0 ml base added – 2nd • 50.0 ml NaOH added –
excess OH
equivalence point
H2A + OH-  HA- + OH-  A2-
A2- + H20  HA- + OH-
H2A + OH-  HA- + OH-  A2-
SOLUBILITY EQUILIBRIUM
Ksp
Ksp SOLUBILITY PRODUCT EQUILIBRIA
• Problems dealing with solubility of PARTIALLY
soluble ionic compounds (in other words, salts
that barely dissociate/dissolve in water)
• General Form, called the solubility product:
MX(s)  n M+ (aq) + p X – (aq)
• Ksp = [M+]n[X -]p
• Ex Ba(OH)2 (s)
Solubility
• Is NOT the solubility product.
• Uses an equilibrium problem to determine
how much can dissolve at a certain
temperature.
• Is related stoichiometrically to the initial
formula. Ex. Ba(OH)2 yields twice as many
OH-1 ions.
Calculate Ksp
• Given the solubility of FeC2O4 at eq. =
65.9mg/L
• Given the solubility of Li2CO3 is 5.48 g/L.
Calculate solubility
• Of SrSO4 with a Ksp of 3.2 x 10-7 in M and g/L
• Of Ag2CrO4 with Ksp of 9.0 x 10-12 in M and g/L
Relative solubilities
• Ksp can be used to compare the solubility
of solids that break apart into the same
number of ions.
• The bigger the Ksp, the more soluble.
• An ICE table is necessary if different
numbers of ions are produced.
Common Ion Effect
• If we try to dissolve the solid in solution with
either the cation or anion present, less will
dissolve.
• Calculate the solubility of strontium sulfate in
a 0.100M solution of Na2SO4
pH and solubility
• OH- can be a common ion.
• More soluble in acid, since OH- will be
removed from the reaction.
• For other anions, if they come from a weak
acid they are more soluble in acid than in
water.
Ex. CaC2O4  Ca+2 + C2O4 -2
H+ + C2O4-2  HC2O4Reduces the C2O4-2 in acidic solution
Precipitation
•
•
•
•
Ion product, Q = [M+]n[X -]p
If Q > Ksp, the precipitate forms
Q< Ksp, no precipitate
Q = Ksp at equilibrium
Sample PPT problem
• A solution of 750.0 ml of 4.00 x 10-3 M cerium
(III) nitrate is added to 300.0 ml of 2.00 x 10-2
M potassium iodate. Will cerium (III) iodate,
Ksp = 1.9 x 10-10 ppt. and if so what is the
concentration of ions in solution?
Selective precipitation
• Used to separate mixtures of metal ions
in solution.
• Add anions that will only ppt. certain
metals at a time.
• Used to purify mixtures.
• Often use H2S because in acidic solution,
Hg+2, Cd+2, Bi+3, Cu+2, and Sn+4 will ppt.
Selective precipitation
• In basic solution, adding OH- solution, S-2 will
increase so more soluble sulfides will ppt.
• Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3, Al(OH)3
Selective precipitation
• Follow the steps with first insoluble chlorides
(Ag, Pb, Ba)
• Then sulfides in acid
• Then sulfides in base
• Then insoluble carbonates (Ca, Ba, Mg)
• Alkali metals and NH4+ remain in solution
– Flame test, NH4+ yields an ammonia smell when
heated.
Complex ion equilibria
• A charged ion surrounded by LIGANDS.
• Ligands are LEWIS BASES using their lone pair
to stabilize the charged metal ions.
• Common ligands are NH3, H2O, CN-, Cl• COORDINATION NUMBER is the number of
attached ligands – usually twice the cations
charge.
– Ex. Cu(NH3)4+2 has a coordination # = 4
LIGAND attachment
• The addition of each ligand has its own
equilibrium.
• Usually the ligand is in large EXCESS.
• The complex ion will be the biggest ion in
solution.
• Complex formation helps dissolve otherwise
insoluble compounds.
Complex ion equilibrium
• Calculate the concentrations of Ag+ and
Ag(CN-)2-1 in a solution prepared by mixing
100.0 ml of 5.0 x 10-3 M AgNO3 with 100.0 ml
of 2.00 M KCN.
Ag+ + 2 CN-  Ag(CN)2-1 K1 = 1.3 x10-21