Buffers, Titrations, and Aqueous Equilibria

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Buffers, Titrations, and Aqueous Equilibria
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Common Ion Effect

• The shift in equilibrium that occurs because of
  the addition of an ion already involved in the
  equilibrium reaction. (Le Chateliers principle)
• AgCl(s) ? Ag(aq) Cl?(aq)
• This affects the concentrations of other ions,
  notably H

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Calculations with ICE
What is the H and ionization of 1.0 M HF
mixed with 1.0 M NaF. (1.0 M HF alone, H2.7 x
10-2, ion2.7) HF(aq) ltgt H (aq) F-
(aq) I 1.0 0 1.0 C -x x x E
1.0-x x 1.0 x Ka 7.2 x 10-4 x
(1.0x) / (1.0 - x) x/1 xH 7.2 x 10-4
ion 7.2 x 10-4 /1.0 .072
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A Buffered Solution

• . . . resists change in its pH when either H or
  OH? are added.
• 1.0 L of 0.50 M H3CCOOH
• 0.50 M H3CCOONa
• pH 4.74
• Adding 0.010 mol solid NaOH raises the pH of the
  solution to 4.76, a very minor change.

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Key Points on Buffered Solutions

• 1. They are weak acids or bases containing a
  common ion.
• 2. After addition of strong acid or base, deal
  with stoichiometry first, then equilibrium.

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Demonstration of Buffer Action
A buffered solution of 1.0 L of 0.5 M HC2H3O2
Ka1.8 x 10-5 and 0.5 M NaC2H3O2 has 0.01 mol of
solid NaOH added. What is the new pH? HC2H3O2
(aq)ltgt C2H3O2-(aq) H (aq) I
0.5 0.5 0 C -x x x E 0.5
-x 0.5 x x Ka1.8 x 10-5 0.5(x) / 0.5
x 1.8 x 10-5 pH 4.74
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When OH- is added, it takes away an equal amount
of H ions, and will affect also the acid
concentration by the same amount. It will also
add to the common ion. HC2H3O2 (aq)ltgt
C2H3O2-(aq) H (aq) I 0.5 0.5 1.8 x
10-5 S -0.01 0.01 - 1.8 x 10-5 due
to adding base I 0.49 0.51 0 C -x x x E
0.49-x 0.51 x x Ka 1.8 x 10-5 x (.51)/ .49
x 1.73 x 10-5 pH 4.76 Practically no
change in pH after base is added.
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Henderson-Hasselbalch Equation

• Useful for calculating pH when the
  A?/HA ratios are known.

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Base Buffers With H/H Equation
pOH pKb log ( HB / B ) pKb log
(acid/ base) Base buffers can be calculated
in similar fashion to acid buffers. pH 14 - pOH
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Buffered Solution Characteristics

• Buffers contain relatively large amounts of weak
  acid and corresponding base.
• Added H reacts to completion with the weak base.
• Added OH? reacts to completion with the weak
  acid.
• The pH is determined by the ratio of the
  concentrations of the weak acid and weak base.

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H/H Calculations
From the previous example, HC2H3O2 .49 M
after the addition of base, and C2H3O2- .51
M. Using the H/H equation, pH pKa log
(.51)/(.49) 4.74 0.02 4.76 same as
using ICE, but easier A weak base buffer is made
with 0.25 M NH3 and 0.40 M NH4Cl. What is the pH
of this buffer? Kb 1.8 x 10-5 NH3 (aq) H2O
(l) ltgt NH4 (aq) OH- (aq) pOH pKb log
(NH4 / NH3) 4.74 log (.40/.25) 4.94 pH
14- 4.94 9.06
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H/H Calculations
The weak base buffer with pH of 9.06 from the
previous example has 0.10 mol of HCl added to it.
What is the new pH? NH3 (aq) H2O (l)
ltgt NH4 (aq) OH- (aq) I .25 .40
1.15 x 10-5 S - .10 .10 - 1.15
x 10-5 due to acid I .15 .50
0 Using H/H, pOH 4.74 log (.5)/(.15)
5.26 pH 14 - 5.26 8.74
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Buffering Capacity

• . . . represents the amount of H or OH? the
  buffer can absorb without a significant change in
  pH.

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Homework!!

• p. 759ff 12, 14 bd, 25 cd, 29, 30

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Titration (pH) Curve

• A plot of pH of the solution being analyzed as a
  function of the amount of titrant added.
• Equivalence (stoichiometric) point Enough
  titrant has been added to react exactly with the
  solution being analyzed.

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Strong Acid/Base Titration
In titrations, it is easier to calculate
millimoles(mmol), which would be Molarity x mL,
in stoichiometry. What would be the pH of 50.0 mL
of a 0.2 M solution of HCl after 20.0 mL of 0.1 M
NaOH have been added? HCl---50.0 x 0.2 10.0 mmol
NaOH--- 20.0 x 0.1 2.0 mmol H
OH- ----gt H2O volumes must be added I
10mmol 2 mmol 50 20 70 S -2
mmol -2mmol H 8 mmol/ 70 mL 0.11 M
End 8mmol 0 mmol pH 0.95
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Weak Acid - Strong Base Titration

• Step 1 - A stoichiometry problem - reaction is
  assumed to run to completion - then determine
  remaining species.
• Step 2 - An equilibrium problem - determine
  position of weak acid equilibrium and calculate
  pH.

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Weak Acid/Strong Base
First do a stoichiometry, then equilibrium using
H/H equation What is the pH of 50.0 mL of a
0.100M HCN solution after 8.00 mL of 0.100 M NaOH
has been added? Ka 6.2 x 10-10 HCN OH-
-----gt H2O CN- Volume I 5 mmol 0.8
mmol 0 mmol 50 8 58 S - 0.8 mmol
-0.8 mmol 0.8 mmol I 4.2 mmol /58mL
0.8 mmol /58 mL 0.072M 0.0138 M pH pKa
log (0.0138/0.072) 8.49
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Acid-Base Indicator

• . . . marks the end point of a titration by
  changing color.
• The equivalence point is not necessarily the same
  as the end point.

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pH
15_334
0
1
2
3
4
5
6
7
8
9
10
11
12
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Crystal Violet
Cresol Red
Thymol Blue
Erythrosin B
2,4-Dinitrophenol
Bromphenol Blue
Methyl Orange
Bromcresol Green
Methyl Red
Eriochrome Black T
Bromcresol Purple
Alizarin
Bromthymol Blue
Phenol Red
o
-Cresolphthalein
Phenolphthalein
Thymolphthalein
Alizarin Yellow R
The pH ranges shown are approximate. Specific
transition ranges depend on the indicator solvent
chosen.
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Homework !!

• p. 761 33, 35, 45 (for 35 only)

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Solubility Product

• For solids dissolving to form aqueous solutions.
• Bi2S3(s) ? 2Bi3(aq) 3S2?(aq)
• Ksp solubility product constant
• and
• Ksp Bi32S2?3

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Solubility Product

• Solubility s concentration of Bi2S3 that
  dissolves, which equals 1/2Bi3 and 1/3S2?.
• Bi2S3 (s) ltgt 2 Bi3 (aq) 3 S2?(aq)
• 2 s 3 s
• Note Ksp is constant (at a given temperature)
• s is variable (especially with a common
• ion present)

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Relation of Ksp to s

• Ionic compounds will dissociate according to set
  equations, which will give set Ksp to s relation.
  Use ICE to find it.
• A2B ltgt 2A B
• 2s s Ksp (2s)2(s)4s3
• AB3 ltgt A 3B
• s 3s Ksp (s)(3s)327s4

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Ksp from s
If 4.8 x 10-5 mol of CaC2O4 dissolve in 1.0 L of
solution, what is its Ksp? CaC2O4 mol / L
4.8 x 10-5 / 1.0 L 4.8 x 10-5 M CaC2O4 (s)
ltgt Ca2 (aq) C2O4-2 (aq) s s Ksp
(s)(s) s2 (4.8 x 10-5 )2 2.3 x 10-9 If the
molar solubility (s) of BiI3 is 1.32 x 10-5, find
its Ksp. BiI3 (s) ltgt Bi3 3I- s
3s Ksp (s)(3s)3 27 s4 Ksp 27(1.32 x
10-5)4 8.20 x 10-19
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s from Ksp
If Ksp for Ag2CO3 is 8.1 x 10-12, find its molar
solubility. Ag2CO3 (s) ltgt 2 Ag (aq) CO3-2
(aq) 2 s s Ksp (2s)2(s) 4 s3 8.1 x
10-12 s 1.27 x 10-4 If Ksp for Al(OH)3 is 2 x
10-32, find its molar solubility. Al(OH)3 (s)
ltgt Al3 (aq) 3 OH- (aq) s 3 s
Ksp (s)(3s)3 27s4 2 x 10-32 s 5.22 x
10-9
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Equilibria Involving Complex Ions

• Complex Ion A charged species consisting of a
  metal ion surrounded by ligands (Lewis bases).
• Coordination Number Number of ligands attached
  to a metal ion. (Most common are 6 and 4.)
• Formation (Stability) Constants The equilibrium
  constants characterizing the stepwise addition of
  ligands to metal ions.

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Complex Ion Examples

• Coordination number2
• Ag(NH3)2
• Coordination number4
• Al(OH)4-1 Cu(NH3)42
• Coordination number6
• Fe(CN)6-4 Fe(SCN)6-3

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Homework and XCR

• p. 762ff 50, 51, 59, 60
• XCR 72, 76, 86
• Super XCR 90