Additional Aspects of Aqueous Equilibria

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Additional Aspects of Aqueous Equilibria
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Aspects of Aqueous Equilibria The Common Ion
Effect
Salts like sodium acetate are strong electrolytes
NaC2H3O2(aq) ? Na(aq) C2H3O2-(aq)
The C2H3O2- ion is a conjugate base of a weak acid
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
H3O C2H3O2-
Ka
HC2H3O2
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The Common Ion Effect
Now, lets think about the problem from the
perspective of LeChateliers Principle
What would happen if the concentration of the
acetate ion were increased?
H3O C2H3O2-
Ka
HC2H3O2
Q gt K and the reaction favors reactant
Addition of C2H3O2- shifts equilibrium, reducing
H
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H3O C2H3O2-
The Common Ion Effect
Ka
HC2H3O2
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
Since the equilibrium has shifted to favor the
reactant, it would appear as if the dissociation
of the weak acid(weak electrolyte) had decreased.
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Aspects of Aqueous Equilibria The Common Ion
Effect
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
So where might the additional C2H3O2-(aq) come
from? Remember we are not adding H. So its not
like we can add more acetic acid.
How about from the sodium acetate?
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NaC2H3O2(aq) ? Na(aq) C2H3O2-(aq)

HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
In general, the dissociation of a weak
electrolyte (acetic acid) is decreased by adding
to the solution a strong electrolyte that has an
ion in common with the weak electrolyte
The shift in equilibrium which occurs is called
the COMMON ION EFFECT
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Lets explore the COMMON ION EFFECT in a little
more detail
Suppose that we add 8.20 g or 0.100 mol sodium
acetate, NaC2H3O2, to 1 L of a 0.100 M solution
of acetic acetic acid, HC2H3O2. What is the pH of
the resultant solution?
NaC2H3O2(aq) ? Na(aq)
C2H3O2-(aq)
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
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Calculate the pH of a solution containing 0.06 M
formic acid (HCH2O, Ka 1.8 x 10-4) and 0.03 M
potassium formate, KCH2O.
Now you try it!
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Calculate the fluoride ion concentration and pH
of a solution containing 0.10 mol of HCl and 0.20
mol HF in 1.0 L
HCl (aq) H2O ? H3O(aq) Cl-(aq)
HF (aq) H2O ? H3O (aq) F-(aq)
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Now, lets think about the problem from the
perspective of LeChateliers Principle
But this time lets deal with a weak base and a
salt containing its conjugate acid.
NH3(aq) H2O ? NH4(aq) OH- (aq)
Q gt K and the reaction favors reactant
NH4 OH-
Kb
NH3
Addition of NH4 shifts equilibrium, reducing OH-
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Calculate the pH of a solution produced by mixing
0.10 mol NH4Cl with 0.40 L of 0.10 M NH3(aq), pKb
4.74?
NH4Cl(aq) ? NH4(aq) Cl-(aq)
NH3(aq) H2O ? NH4(aq) OH-
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Common Ions Generated by Acid-Base Reactions
The common ion that affects a weak-acid or
weak-base equilibrium may be present because it
is added as a salt, or the common ion can be
generated by reacting an acid and base directly
(no salt would be necessary.which is kind of
convenient if you think about it)
Suppose we react 0.20 mol of acetic acid (weak)
with 0.10 mol of sodium hydroxide strong)
HC2H3O2 (aq) OH-(aq) ? H2O C2H3O2-(aq)
0
0.10 mol
0.20 mol
-0.10 mol
0.10 mol
-0.10 mol
0.10 mol
0
0.10 mol
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Suppose we react 0.20 mol of acetic acid with
0.10 mol of sodium hydroxide
HC2H3O2(aq) OH- (aq) ?H2O C2H3O2-aq)
0
0.20 mol
0.10 mol
-0.10 mol
-0.10 mol
0.10 mol
0.10 mol
0
0.10 mol
Lets suppose that all this is occurring in 1.0 L
of solution
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
0
0.10 M
0.10 M
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Sample problem Calculate the pH of a solution
produced by mixing 0.60 L of 0.10 M NH4Cl with
0.40 L of 0.10 M NaOH
NH4Cl(aq) ? NH4(aq) Cl- (aq)
NH4 OH- ? NH3
H2O
0
0.06 mol
0.04 mol
-0.04 mol
0.04 mol
-0.04 mol
0
0.02 mol
0.04 mol
NH4 H2O ? H3O NH3
Dont forget to convert to MOLARITIES
0.04 M
0
0.02 M
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Calculate the pH of a solution formed by mixing
0.50 L of 0.015 M NaOH with 0.50 L of 0.30 M
benzoic acid (HC7H5O2, Ka 6.5 x 10-5)
Now you try it!
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adding acid or base

• calculate the pH of a solution that has .2 mol of
  NaOH added to a solution that is .25 M HC2H3O2
  and .32M NaC2H3O2 HC2H3O2(aq) OH- (aq) ?H2O
  C2H3O2-aq
• ..25 .20 .32
• -.20 -.20
  .20
• .05 0 .52

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• HC2H3O2(aq) ? H C2H3O2-aq
• .05 0 .52
• -X X X
• X (.52X) 1.8 x 10-5
• .05-X

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BUFFERED SOLUTIONS
A buffered solution is a solution that resists
change in pH upon addition of small amounts of
acid or base.
Suppose we have a salt MX ? M(aq) X-(aq)
And weve added the salt to a weak acid
containing the same conjugate base as the salt,
HX
HX H2O ? H3O X-
And the equilibrium expression for this reaction
is
H X-
Ka
HX
Note that the concentration of the H is
dependent upon the Ka and the ratio between the
HX and X- (the conjugate acid-base pair)
HX
Ka
H
X-
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Two important characteristics of a buffer are
buffering capacity and pH. Buffering capacity is
the amount of acid or base the buffer can
neutralize before the pH begins to change to an
appreciable degree.

• The pH of the buffer depends upon the Ka

• This capacity depends on the amount of acid and
  base from which the buffer is made

• The greater the amounts of the conjugate
  acid-base pair, the more resistant the ratio of
  their concentrations, and
  hence the pH, to change

HX
H
Ka
X-
Henderson-Hasselbalch Equation
HX
-logH
-log Ka
X-
log
pKa
pH

X-
HX
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NaC2H3O2(aq) ? Na(aq) C2H3O2-(aq)
0.1 M
0.1 M
0.1 M
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-(aq)
0.1 M
0
0.1
-x
x
x
x
0.1 - x
0.1 x
x(0.1 x )
pH 4.74
x 1.8 x 10-5
1.8 x 10-5
0.1 - x
Using the Henderson-Hasselbalch Equation
Note that these are initial concentrations
.1

4.74
log
pH
.1
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A liter of solution containing 0.100 mol of
HC2H3O2 and 0.100 mol NaC2H3O2 forms a buffered
solution of pH 4.74. Calculate the pH of this
solution (a) after 0.020 mol NaOH is added, (b)
after adding 0.020 mol HCl is added.
NaC2H3O2(aq) ? Na(aq) C2H3O2-(aq)
0.1 M
0.1 M
0.1 M
Step 1
HC2H3O2(aq) OH- ? H2O C2H3O2-
(aq)
0.1 M
0.1 M
0.02 M
0.02 M
-0.02 M
-0.02 M
0.12 M
0.00 M
0.08 M
Henderson-Hasselbalch Equation
Step 2
Note that these are initial concentrations
.12
log

4.74
pH
.08
pH 4.92
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A liter of solution containing 0.100 mol of
HC2H3O2 and 0.100 mol NaC2H3O2 forms a buffered
solution of pH 4.74. Calculate the pH of this
solution (a) after 0.020 mol NaOH is added, (b)
after adding 0.020 mol HCl is added.
C2H3O2-(aq) H? HC2H3O2
0.10 M
0.10 M
0.02 M
Step 1
-0.02 M
-0.02 M
0.02 M
0.12 M
0.08M
0.00 M
Henderson-Hasselbalch Equation
Note that these are initial concentrations
.08
log
Step 2
pH
4.74

.12
pH 4.56
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Now consider, for a moment, what would have
happened if I had added 0.020 mol of NaOH or
0.02 mol HCl to .1 M HC2H3O2.
HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-aq
0
0.1 M
0
x
x
-x
0.1 - x
x
x
x2
1.8 x 10-5
x 0.0013
0.1 - x
pH 2.9
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HC2H3O2(aq) OH- ? H2O C2H3O2-(aq)
0.02 M
0.1 M
0.00
-0.02 M
-0.02 M
0.02 M
0.08 M
0.00 M
0.02 M
Henderson-Hasselbalch Equation
.02
pH
4.74
log

.08
pH 4.13
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HC2H3O2(aq) H2O ? H3O(aq) C2H3O2-aq
0.10
0.02 M
0
H from HCl Completely dissociates therefore the
pH is calculated without regard for the weak acid
pH -log 0.02
pH 1.7
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add 2 ml 10 M HCl to .1M HC2H3O2 .1M NaC2H3O2
pH 4.74
add 2 ml 10 M HCl to 1.8 x 10-5M HCl pH 4.74
to .12M HC2H3O2 .06M NaC2H3O2 pH 4.56 a
drop of .18
.02 M HCl pH 1.7 a drop of 3.04
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• Sample exercise Consider a litter of buffered
  solution that is 0.110 M in formic acid (HC2H3O)
  and 0.100 M in sodium formate (NaC2H3O ).
  Calculate the pH of the buffer
• before any acid or base are added,
• (b) after the addition of 0.015 mol HNO3,
• (c) after the addition of 0.015 mol KOH

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Titration Curves
End points
Stoichiometrically equivalent quantities of acid
and base have reacted
HCl(aq) NaOH(aq) ? H2O NaCl
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Titration of a weak acid and a strong base
results in pH curves that look similar to those
of a strong acid-strong base curve except that
the curve (a) begins at a higher pH, (b) the pH
rises more rapidly in the early part of the
titration, but more slowly near the equivalence
point, and (3) the equivalence point pH is not 7.0
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Calculating pHs from Titrations
Calculate the pH in the titration of acetic acid
by NaOH after 30.0 ml of 0.100 M NaOH has been
added to 50 mL of 0.100 acetic acid
HC2H3O2(aq) OH- ? H2O(l) C2H3O2-
0.003 mol
0.005 mol
0
0.003 mol
-0.003 mol
-0.003 mol
-0.002 mol
0.003 mol
0
.0370
log
4.74

pH
.0250
pH 4.91
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Determining the Ka From the Titration Curve
pKa pH 4.74
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Solubility Equilibria Ksp

• The equilibrium expression for the following
  reaction is CaF2(s) ? Ca2(aq) 2F-(aq)
• Ca2 F-2
• CaF2
• Look at the table on page 759 or the appendices
  A26, where you will find the value of the ksp to
  be 4.1 x 10-11

• Ksp

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Calculating Ksp from solubility

• The solubility of Bi2S3 is 1.0 x 10-15M what
  would be the Ksp?
• Bi2S3(s) ? 2Bi3(aq) 3S2-(aq)
• 1.0 x 10-15 2(1.0 x 10-15) 3(1.0x 10-15)
• Ksp Bi32 S2-3
• Ksp (2.0 x 10-15)2(3.0 x 10-15)3 1 x 10-73

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Calculating solubility from Ksp

• The Ksp of Cu(IO3)2 is 1.4 x 10-7 what would be
  the solubility?
• Cu(IO3)2(s) ? Cu2(aq) 2IO31-(aq)
• X X 2X
• Ksp 1.4 x 10-7 (X)(2X)2 4X3
• X 3.3 x 10-3 mol/L

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Common ion effect

• What would be the solubility of Ag2CrO4 in a
  solution that is .1M AgNO3,
• the Ksp 9.0 x 10-12
• Ag2CrO4 ? 2Ag1 CrO42-
• X 2X .1 X
• 9.0 x 10-12 (2X.1)2(X)
• Drop the 2X as insignificant
• X 9.0 x 10-10 mol/L

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pH and Solubility

• Mg(OH)2 ? Mg2 2OH-
• If the pH is raised (the OH- is raised) then we
  have the common ion effect and the solubility is
  decreased.
• If the pH is lowered (the H is raised) then OH-
  is reacted with H to make water and the
  solubility is increased.

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pH and solubility of salts

• Ag3PO4 ? 3Ag1 PO43-
• If the pH is lowered (the H is raised) the PO43-
  reacts with H to make HPO42- , which
  essentially removes PO43-
• from the equation, shifting the reaction to the
  right and the solubility is increased.

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Will a precipitate form?

• If 750 mL of 4.00E-3M Ce(NO3)3 is mixed with 300
  mL of 2.00E-2M KIO3, will a precipitate form?
• Ce(NO3)3(aq) KIO3 (aq) ? Ce(IO3)3(s) KNO3(aq)
• (750)(4 x 10-3) (1050) X X 2.86 x 10-3M for
  Ce3
• (300)(2 x 10-2) (1050) Y Y 5.71 x 10-3 M for
  IO31-
• Ce(IO3)3(s) ? Ce3 (aq) 3IO31- (aq)
• Ksp 1.9 x 10-10
• Q (2.86 x 10-3)(5.71 x 10-3)3 5.32 x 10-10
• Q is greater than K so a precipitate will form