Transcript Chapter 16 Lecture 2 Titrations I. Strong Acid—Strong Base Titrations

Chapter 16 Lecture 2 Titrations
I.
Strong Acid—Strong Base Titrations
A.
Titration Basics
1) Titration = addition of a measurable volume of a known solution (titrant)
to an unknown solution until it is just consumed
2)
Use the stoichiometry of the reaction of the known and unknown to
calculate the concentration of the unknown solution
3)
A pH curve shows the change in pH versus volume of titrant as the
titration proceeds
a)
pH meter can be used to monitor
pH during the titration
b)
An acid-base indicator can be used
to signal reaching the equivalence point
4)
The unit mmol
a) Titrations often involve small amounts of a compound. Rather than
working with small numbers of moles, the millimole unit can be used.
b) 1 mmol 
c)
1
mol  1 x 10-3 mol
1000
Since 1 ml = 1 x 10-3 L, Molarity can be expressed either as:
mol 1000 mmol mmol
M


L
1000 ml
ml
B.
Titration of a strong acid with a strong base
1. Net reaction = H+ + OHH2O
Na+ + NO3- + H2O
2.
Example: HNO3(aq) + NaOH(aq)
3.
Let’s titrate 50.0 ml of 0.200 M HNO3 with 0.100 M NaOH.
a)
0.0 ml of NaOH added: pH = -log(0.2 M) = 0.699
b)
Initial
Equilibrium
10.0 ml of NaOH added:
H+
+
OH10 mmol
1 mmol
9 mmol
0 mmol
[H+] = 9 mmol / 60 ml = 0.15 M
c)
Initial
Equilibrium
Initial
Equilibrium
pH = 0.82
20.0 ml NaOH added:
H+
+
OH10 mmol
2 mmol
8 mmol
0 mmol
[H+] = 8 mmol / 70 ml = 0.11 M
d)
H2O
-------------
H2O
------------pH = 0.94
50.0 ml NaOH added:
H+
+
OH10 mmol
5 mmol
5 mmol
0 mmol
[H+] = 5 mmol / 100 ml = 0.05 M
H2O
------------pH = 1.30
e)
Initial
Equilibrium
100.0 ml of NaOH added: Equivalence Point
H+
+
OHH2O
10 mmol
10 mmol
------0 mmol
0 mmol
-------
[H+] = 1 x 10-7 M from water
f)
Initial
Equilibrium
150.0 ml NaOH added:
H+
+
OH10 mmol
15 mmol
0 mmol
5 mmol
[OH-] = 5 mmol / 200 ml = 0.025 M
g)
Initial
Equilibrium
200.0 ml NaOH added:
H+
+
OH10 mmol
20 mmol
0 mmol
10 mmol
[OH-] = 10 mmol / 250 ml = 0.04 M
pH = 7.00
H2O
------------pOH = 1.60
pH = 12.40
H2O
-------------
pOH = 1.40
pH = 12.60
4)
Important points:
a) pH increases slowly far from the equivalence point
b) pH changes quickly near the equivalence point
c) The equivalence point of a strong acid—strong base titration = 7.00
5)
The titration of a strong base with a strong acid is almost identical
II.
Titration of a Weak Acid with a Strong Base
A.
Addition of a strong base to a weak acid forms a Buffer Solution
1) HA + OHA- + H2O
2) If not enough base has been added to complete the reaction: HA/A- buffer
3) Problems can be worked either as equilibrium or buffer problems
4) Either way, you must consider the stoichiometry of the reaction first, then
consider the equilibrium or buffer problem
Example: Titrate 50.0 ml of 0.10 M HC2H3O2 (Ka = 1.8 x 10-5) with 0.10 M NaOH
1) 0.0 ml NaOH added: typical weak acid problem
HA
H+
+
Ax2
5
1.8 x 10 
 x  0.00134
Initial
0.1
0
0
0.1
Equilibrium 0.1 – x
x
x
pH  2.87
B.
2)
10.0 ml NaOH added:
OH+
Initial
1 mmol
Equilibrium 0
stoichiometry first, then buffer problem
HA
A+
H2O
5 mmol
0
-----4 mmol
1 mmol
------
pH = pKa + log([A-]/[HA]) = 4.74 + log(1/4) = 4.14
3)
25.0 ml NaOH added:
OH+
Initial
2.5 mmol
Equilibrium 0
stoichiometry first, then buffer problem
HA
A+
H2O
5 mmol
0
-----2.5 mmol
2.5 mmol
------
pH = 4.74 + log(2.5/2.5) = 4.74
When HA = A-, we are half-way to the equivalence point. pH = pKa
Titrations are a very precise way of finding pKa’s of acids and bases.
4)
40.0 ml NaOH added:
OH+
Initial
4 mmol
Equilibrium 0
stoichiometry first, then buffer problem
HA
A+
H2O
5 mmol
0
-----1 mmol
4 mmol
------
pH = 4.74 + log(4/1) = 5.35
5)
50.0 ml NaOH added: stoichiometry first, then weak base problem
OH+
HA
A+
H2O
Initial
5 mmol
5 mmol
0
-----Equilibrium 0
0 mmol
5 mmol
-----We are now at the equivalence point
A+
5 mmol
 0.05 M
Initial
100 ml
Equilibrium 0.05 - x
H2O
---------
HA
0
x
OH0
x
+
K W 1 x 10 14
x2
10
Kb 

 5.56 x 10 
5
K a 1.8 x 10
0.05
x  [OH  ]  5.27 x 10 6  pOH  5.28  pH  8.72
The major species at the equilibrium point of a weak acid titration, is its
conjugate base. Thus, the pH at the equivalence point > 7.00.
(The major species at the equilibrium point of a weak base titration, is its
conjugate acid. Thus, the pH at the equivalence point < 7.00)
6)
60.0 ml NaOH added:
OH+
Initial
6 mmol
Equilibrium 1 mmol
stoichiometry first, then strong base problem
HA
A+
H2O
5 mmol
0
-----0
5 mmol
------
pOH = -log(1mmol/110ml) = -log(9.1 x 10-3 M) = 2.04
pH = 11.96
7)
75.0 ml NaOH added:
OH+
Initial
7.5 mmol
Equilibrium 2.5 mmol
stoichiometry first, then strong base problem
HA
A+
H2O
5 mmol
0
-----0
5 mmol
------
pOH = -log(2.5mmol/125ml) = -log(0.02 M) = 1.70
C.
pH = 12.30
Important Points
1) pH increases more rapidly at the start than for a strong acid
2) pH levels off near pKa due to HA/A- buffering effect
3) Curve is steepest near equivalence point. Equivalence Point > 7.0
4) Curve is similar to strong acid—strong base after eq. pt. where OH- is major
D.
Example: Titrate 50.0 ml 0.1 M HCN (Ka = 6.2 x 10-10) with 0.10 M NaOH
1) pH after 8 ml NaOH added
2) pH at halfway point
3) pH at equivalence point
E.
Conclusions about weak acid titrations
1) The same amount of base is required
to reach the equivalence point of any acid.
It is not the Ka that matters, it is how much
acid is there.
2) pH at the equivalence point varies
depending on the Kb of the conjugate base
of the weak acid. The weaker the acid, the
higher the pH at the eq. pt.
3) Ka effects the shape of the titration
curve, but only before the eq. pt.
F.
Example: Find the Ka of an unknown acid. 2 mmol of the acid in 100 ml of
water was titrated with 0.05 M NaOH. After 20 ml NaOH was added, the pH
= 6.00. (Do stoichiometry, then buffer problem)
III. Titration of a Weak Base with a Strong Acid
A.
Similar problem to the titration of a weak acid with a strong base
1) Determine major species from the stoichiometry
2) Calculate pH from weak acid, buffer, or weak base accordingly
B.
Example: Titrate 100 ml of 0.10 M NH3 (Kb = 1.8 x 10-5) with 0.1 M HCl.
1) No HCl added: weak base pH calculation
2) HCl added, but before the equivalence point:
a) HCl will react to completion with NH3
b) Major species: NH3, NH4+
c) Buffer problem
3) At the equivalence point
a) All NH3 has been converted to NH4+
b) Weak acid problem, find Ka from Kb/KW
4) Past the equivalence point:
a) Excess HCl, is a strong acid
b) Weak acid NH4+ won’t effect pH
c) Strong acid problem
Titrations of Polyprotic Acids and Bases
1. Multiple Inflection Points = Multiple Equivalence Points will be seen
2. The volume required to reach each equivalence point will be the same
H2SO3
H+ + HSO3Ka1 = 1.6 x 10-2
HSO3H+ + SO32Ka2 = 6.4 x 10-8
IV. Acid-Base Indicators
A.
Finding the equivalence point of a titration
1) Use a pH meter
a) Plot pH versus titrant volume
b) Center vertical region = equivalence point
2)
Use an Acid-Base Indicator
a) Acid-Base Indicator = molecule that changes color based on pH
b) Choose an indicator that changes color at the equivalence point
c) End Point = when the indicator changes color. If you have chosen
the wrong indicator, the end point will be different than the eq. pt.
d) Indicators are often Weak Acids that lose a proton (causing the color
change) when [OH-] reaches a certain concentration
HIn + OHIn- + H2O
B.
Examine the behavior of a Hypothetical Indicator (Ka = 1 x 10-8)
1) HIn
H+ + Inred
blue
2) From the Ka equation, we can develop a formula showing how color
behaves depending on the pH
Ka
[H  ][In  ]
[In  ]
Ka 
  
[HIn]
[H ] [HIn]
a)
b)
c)
3)
The ratio of HIn and In- will determine the color
The ratio is a function of pH and a constant
We should be able to predict the color at different pH’s
If we add a few drops of HIn to a pH = 1.0 solution:
1 x 10-8 1 x 10-7 [In  ]
a) K a

[H ]
b)
4)

0.1

1

[HIn]
HIn is the dominant species so the color will be red
If we add OH- to the solution, eventually [In-] = [HIn]
The color will be equal parts red and blue = purple
5)
Usually, about 10% of the HIn must react before our eyes can detect a color
change.
[In - ] 1
Color change 
6)
[HIn]

10
Example: What is the pH at the color change of Bromthymol Blue (Ka = 1 x
10-7, HIn = yellow, In- = blue) starting in a strongly acidic solution?
Ka
[In  ] 1

6



10K

[H
]

1x10
 pH  6.00
a

[H ] [HIn] 10
C.
We can use the Henderson-Hasselbalch equation on Indicators as well
1) pH = pKa + log([In-]/[HIn])
2) pH = pKa + log(1/10) for a color change
a) log(1/10) = -1
b) pH for color change starting in acid is always pKa – 1 for any Indicator
3)
For a basic solution titrated with acid, [In-]/[HIn] = 10/1 for color change
a) Log(10/1) = +1, pH for color change will equal pKa + 1
b) Useful range for a pH Indicator is always pKa +/- 1
4)
The useful range and colors of common Indicators:
D.
Choosing an Indicator
1) We always want the end point (color change) to coincide with the eq. pt.
2) Strong Acid—Strong Base titrations have a large pH change at the
equivalence point, so you get sharp color change.
a) What indicator would work best for HCl/NaOH titration?
b) Eq. Pt. = 7.00
c) We will be going from acidic to basic, so we want [In-]/[HIn] = 1/10
d) pH = pKa + log(1/10) = pKa – 1
e) Since we know the pH at eq. pt. = 7.00, pKa of the indicator we need
should be 8.00 (Ka = 1 x 10-8)
f) If we were going from basic to acidic, we would want pKa = 6.00
3)
How crucial is titrating exactly to the equivalence point?
a) NaOH = 99.9 ml pH = 5.3
b) NaOH = 100.0 ml pH = 7.0
c) NaOH = 100.1 ml pH = 8.7
d) Wide range of Indicators would tell us the eq. pt. to within 0.1 ml
HCl titration
4)
HC2H3O2 titration
Indicators for weak acid titrations
a) The weaker the acid, the smaller the pH change at the eq. pt.
b) We have less flexibility in choosing our indicator
c) 0.1 M HC2H3O2 titrated with 0.1 M NaOH has pH = 8.7 at eq. pt.
i. Phenolphthalein ok (8.0—10.0 endpoints)
ii. Thymol Blue ok (8.0—9.2 endpoints)
iii. Methyl Red in not ok (4.2—6.2 endpoints)