66.2
Standard 120
Count
100
80
Mean: Deviation: 13.9
60
40
20
0
0
10
20
30
40
50
Score
60
70
80
90
100
Another Scientific Pep Talk
to Start the Last Section of 2-B
Nope
Salk,
live virus?
John Enders, English and Germanic Language
Major. Later a Virologist. Nobel Laureate...
Sabine, oral vaccine?
Nope.
Last Lecture
common-ion effect; buffers,
natural dyes and pH indicators.
S, the molar solubility, is a measure of the
amount of a solid that can dissolve into
solution at equilibrium.
Keep track of the stoichiometry:
CaF2(s) = Ca2+(aq) + 2 F-(aq)
Titrations
Used to determine Ka values
Two components
analyte what is being investigated
titrant what is being varied
Four variations
strong acid/strong base and viceversa
strong acid/weak base and viceversa
Neutralization Reactions and Titration Curves
Titration Curve:
The plot of pH vs. volume of titrant added.
Equivalence point: (or Stoichiometric point)
The point in the reaction at which both acid and
base have been consumed. Neither acid nor base is
present in excess.
(#moles of titrant = #moles of analyte)
End point of the indicator:
The point at which the indicator changes color.
Example Strong Acid / Strong Base
How much strong base titrant 0.34 M NaOH must be added to
neutralize a strong acid analyte (25 mL of 0.5 M HCl) - i.e. to reach
the equivalence point?
Then calculate the pH:
(a) Before the addition of any NaOH titrant,
(b) 1 mL before the equivalence point,
(c) 1 mL after the equivalence point.
Example Strong Acid / Strong Base
How much strong base titrant 0.34 M NaOH must be added to
neutralize a strong acid analyte (25 mL of 0.5 M HCl) - i.e. to reach
the equivalence point?
Solution
Initial How much analyte (HCl)?
25 mL 0.5 mmol/mL = 12.5 mmol
Equivalence equal moles of titrant and analyte: nacid = nbase
12.5 mmol = 0.34 mmol/mL Vtitrant
Vtitrant = 36.8 mL
Vtitrant
pH value calculations:
(a) Before titration: We have 25 ml of 0.5 M HCl analyte.
pH = -log(0.5) = 0.30
pH value calculations
(b) 1mL before equivalence point
(remember: the equivalence point is 36.8 mL)
When Vtitrant=35.8 mL: ntitrant=35.8 mL 0.34 M = 12.172 mmol
Stoichiometric Calculation
Init
Add base
Change
Final
H3 O+
12.5 mmol
+
-12.172 mmol
0.328 mmol
OH
0
12.172 mmol
-12.172 mmol
0
2H2O
Total volume = 25 mL + 35.8 mL = 60.8 mL
pH= 2.27
[H3O+] = 0.328 mmol/60.8 mL = 0.0054 M
(c) 1mL AFTER equivalence point
(the equivalence point is 36.8 mL; so 36.8+1=37.8 ml)
ntitrant=37.8 ml 0.34 M = 12.852 mmol
Init
Add base
Change
Final
H3 O+ +
12.5mmol
-12.5mmol
0
OH
0
12.852 mmol
-12.5 mmol
0.352 mmol
2H2O
Total volume = 25 mL + 37.8 mL = 62.8 mL
[OH-] = 0.352 mmol/62.8 mL = 0.0056M
pOH= 2.25
pH = 14 2.25 = 11.75
Strong Acid/Weak Base
Analyte weak base : Titrant strong acid
B(aq) + H3O+(aq) HB(aq) + H2O(l)
pH < 7 at the stoichiometric point.
With your acid you must titrate away:
(i) the base itself (rxn above), but also
(ii) any OH- generated by the weak base making
its conjugate acid:
B(aq) + H2O(aq) HB(aq) + OH-(aq)
Weak Acid/Strong Base
Analyte weak acid : Titrant strong base
HA(aq) + OH(aq) A(aq) + H2O(l)
pH > 7 at the stoichiometric point
With your base you must titrate away:
(i) the acid itself (rxn above), but also
(ii) any H3O+ generated by the weak acid
making its conjugate base:
HA(aq) + H2O(aq) A-(aq) + H3O+(aq)
pKa estimable by
the pH of this
inflection point,
here.
Weak base Strong Acid Titration
NH3: Starting pH, before the titration
NH3 + H2O NH4+ + OH
I
Mbase
0
0
C
-x
+x
+x
E
Mbase-x
x
x
[NH4+][OH]
Kb=
[NH3]
x2/(Mbasex)=Kb
pH
x = [OH-]
pOH= -log[OH-]
7
pH = 14-pOH
Volume of acid
After a little strong acid is added:
nbase(init)=Mbase V(init), V(acid added) = nacid(added)/ Macid
Init
Add
CHG
After
NH3
+ O+
nbase(init)
NH H3 4 +
0
nacid(added)
-nacid(added) -nacid(added)
nbase(init)-nacid(added)
0
0
nacid(added)
nacid(added)
V(total) = V(init) + V(acid added)
M(new) = n(new)/V(total)
NH3 +
I
C
E
pH
H2O NH4+ + OH
Mbase(new)
Macid(new)
Mbase(new)-x
[OH-]
0
+x
-x
x
+ H2 O
+x
Macid(new)+x
pOH
pH
7
x
Volume of acid
Half neutralization:
nbase(init)=M V(init),
nacid(added)= nbase(init)
V(acid added) = nacid(added)/ Macid
Init
Add
CHG
After
NH3
+ H 3 O+
NH 4 + + H 2 O
nbase(init)
nbase(init)
-nbase(init) -nbase(init) nbase(init)
0
nbase(init)
nbase(init)
Easiest is to recognize that you have
a pH buffer here, which allows pKa
estimation:
[NH3]
pH = pKa + log
[NH4+]
= pKa
pH
7
= 14 - pKb
Volume of acid
At equivalence point:
nacid(added) = nbase(init)
NH3
Init
Add
CHG
After
+ H3 O+
nbase(init)
NH 4 +
0
nbase(init)
-nbase(init)
0
-nbase(init)
0
0
+ H2 O
nbase(init)
nbase(init)
NH4+ concentration = MNH4+ = nbase(init) /V(total)
pH calculation:
NH4+ + H2O NH3 + H3O+
I
MNH4+
0
0
C
E
Ka=
-x
+x
x
MNH4+-x
[NH3] [H3O+]
[NH4
+x
x
=
+]
pH
Kw
Kb
7
x2/(MNH4+ -x) =Ka
[H3O+] =x
pH
Volume of acid
Add more acid: nacid(added) > nbase(init)
Init
Add
CHG
After
NH3
nbase(init)
-nbase(init)
0
+
H3O +
NH 4 +
nacid(added)
-nbase(init)
nacid(added)-nbase(init)
+
H2 O
(acid in excess)
nbase(init)
nbase(init)
Strong acid concentration = [nacid(added)-nbase(init)] /V(total)
[H3O+]
pH
pH
Dissociation of NH4+ contributes
little to pH in this case.
7
Volume of acid
Determination of Ka
HA(aq) + OH(aq) A(aq) + H2O(l)
Halfway to the stoichiometric point is where
the concentration of acid=conjugate base (or
base=conjugate acid). Look for inflections.
nOH = nA = nHA = initial nHA/2
using total volume : [HA] = [A]
Ka = [H3O+]
or
pKa = pH
Recall: Polyprotic Acids
Triprotic Three equilibria : Ka1, Ka2, Ka3
H3PO4 + H2O H3O+ + H2PO4-
H2PO4 + H2O H3
-
O+
+ HPO4
2-
HPO42- + H2O H3O+ + PO43-
Add NaOH
[H3O+] [H2PO4-]
Ka1=
[H3PO4]
= 7.110-3
[H3O+] [HPO42-]
Ka2=
[H2PO4-]
= 6.310-8
[H3O+] [PO43-]
Ka3=
[HPO42-]
= 4.210-13
neutralize all 3 available protons, eventually
3 equivalence points involving acid/base hydrolysis reactions!
Titration of a Weak Polyprotic Acid
H3PO4 + OH
H2PO4
H2PO4 + OH
HPO42
HPO42 + OH
PO43
+ H2O
H2PO4
Can donate or accept a proton!
H2PO4- + H2O H3O+ + HPO42
Ka2 =
H2PO4- + H2O
Kb1=
HPO42
6.310-8
H3PO4 + OH
1.410-12
HPO42- + H2O H3O+ + PO43
Ka3 = 4.210-13
HPO42- +H2O
H2PO4- + OH
Kb2 =
1.610-7
Titration of a Weak Polyprotic Acid
H3PO4 + OH
H2PO4
H2PO4 + OH
HPO42
HPO42 + OH
PO43
+ H2O
At this inflection point,
the pH=pKa2.
Here, difficult to see, are pH=pKa1 and pH=pKa3.
The inflection points are more difficult to see because there is so
much H+(aq)and OH-(aq), respectively, but this is how they
are measured experimentally.
The third equivalence point of phosphoric acid can only be reached
in a strongly basic solution. For example, if you make up 1.0 M
Na3PO4 solution:
Kb = 2.410-2
Initial
Changes
Eqm
Kb=
PO43 + H2O OH +
1.0 M
0M
-x M
+x M
(1.00 - x) M
xM
[OH-] [HPO42-]
[PO43-]
x = 0.14 M,
=
x2
(1.00 - x)
pOH = +0.85
HPO420M
+x M
xM
= 2.410-2
pH = 13.15
Example
What volume of a 0.010 mol/L solution of KOH is required to
reach the (a) first; (b) second stoichiometric point in a titration
of 25.00 mL of 0.010 mol/L H2SO4?
Solution to the Example
nH 2 SO4 = cH 2 SO4 VH 2 SO4 = 0.010 M 25mL = 0.25mmol
Sulfuric acid is a diprotic acid, 2 mol of base are required to neutralize 1
mol of sulfuric acid.
nKOH (first eqv point ) = nH 2 SO4 = 0.25mmol
VKOH
nKOH
0.25mmol
=
=
= 25mL
cKOH 0.010mmol / mL
nKOH (second eqv point ) = 2nH 2 SO4 = 0.50mmol
VKOH
nKOH
0.50mmol
=
=
= 50mL
cKOH 0.010mmol / mL