Transcript Chem 1202 - LSU Department of Chemistry

Additional Aspects of
Aqueous Equilibrium
Chapter 17
Equilibrium, Part III
Watkins
Chem 1422, Chapter 17
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Acid/Base Titration
Titration = quantitative addition of two solutions.
What happens when a basic solution and an acidic
solution are added together?
Acid(aq) + Base(aq) → Salt(aq) + Water
The solution in the buret is
called the titrant (acid or base)
A plot of pH versus volume of added
titrant is called a titration curve.
The pH of the solution changes
as titrant is added.
Watkins
Chem 1422, Chapter 17
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Titration: Strong Acid by Strong Base
Titration = quantitative addition of two solutions.
What happens when a basic solution and an acidic
solution are added together?
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Base
NaOH(aq)
pH >> 7
Acid
HCl(aq)
pH << 7
Watkins
To detect the pH
change, a small amount
of indicator
(phenolphthalein, a
dye) has been added,
but it is colorless in the
acidic solution.
Chem 1422, Chapter 17
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Titration: Strong Acid by Strong Base
• At the equivalence point (also called the end point):
moles OH- = moles H+
• the added base has “neutralized” the acid so the pH
in the flask is determined by the salt.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Base
NaCl is a neutral salt.
In a neutral solution, phenolthalein
turns slightly pink.
pH = 7
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Chem 1422, Chapter 17
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Titration: Strong Acid by Strong Base
• At the equivalence point (also called the end point):
moles OH- = moles H+
• excess base beyond the equivalence point raises the
pH (determined by the amount of base added).
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Base
In strongly basic solution,
phenolthalein is red.
pH > 7
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Chem 1422, Chapter 17
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Titration: Strong Acid by Strong Base
Summary
Titration = quantitative addition of two solutions.
The pH change can be followed with a pH meter. The
titration curve is a plot of pH versus volume of titrant.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Base
Acid
pH < 7
Watkins
Salt
pH = 7
Chem 1422, Chapter 17
Base
pH > 7
6
Titration: Strong Acid by Strong Base
Titration Curve
This “sigmoidal
curve” is common in
science, especially
biology. It occurs
whenever there is a
rapid change in a
logarithmic function.
It is named after the
lowercase word-final
greek letter “sigma”: .
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Chem 1422, Chapter 17
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Titration: Strong Acid by Strong Base
Titration Curve
End Point
1 drop = 0.02 mL
Watkins
Chem 1422, Chapter 17
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Titration: Strong Acid by Strong Base
Phenolthalein indicator
changes from colorless
to red in 1 drop
±1 drop in 25 mL is an
error of ±0.08%
Methyl Red indicator
changes from red to
yellow in 1 drop
Any indicator that changes
color in the range 5 < pH < 9
Watkins
Chem 1422, Chapter 17
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Titration: Strong Base by Strong Acid
If the base is initially in the flask and the acid is added from the
buret, the titration curve is just reversed.
Watkins
Chem 1422, Chapter 17
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Titration: Weak Acid by Strong Base
When a strong acid is titrated by a strong base
the pH changes very abruptly at the end point,
and the pH at the end point is 7.
When a weak acid is titrated by a strong base
a basic salt is produced
the end-point pH is > 7
the pH change at the end point is less abrupt
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Chem 1422, Chapter 17
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Titration: Weak Acid by Strong Base
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Chem 1422, Chapter 17
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Titration: Weak Acid by Strong Base
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Chem 1422, Chapter 17
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Titration: Weak Acid by Strong Base
Watkins
Chem 1422, Chapter 17
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Titration: Weak Acid by Strong Base
Watkins
Chem 1422, Chapter 17
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Titration: Weak Acid by Strong Base
1 drop
DpH = 2.5
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Chem 1422, Chapter 17
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Titration: Weak Acid by Strong Base
1 drop
DpH = 1.5
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Chem 1422, Chapter 17
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Titration: Weak Acid by Strong Base
1 drop
DpH = 0.8
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Chem 1422, Chapter 17
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Titration: Weak Acid by Strong Base
1 drop
DpH = 0.2
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Chem 1422, Chapter 17
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Titration - Quantitative Problems
Titration is the quantitative addition of one solution to
another. The general titration problem is:
Calculate the pH in the flask after an arbitrary
amount of titrant has been added.
The three most common calculations are:
1. Calculate the pH in the flask and/or buret before
any titrant is added.
2. Calculate the volume of titrant that must be added
to reach the equivalence point.
3. Calculate the pH at the equivalence point.
Watkins
Chem 1422, Chapter 17
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Titration - Quantitative Problems
1. What is the pH in the flask/buret before any
titrant is added?
Strong acid, [H+] = Ca
Weak acid, [H+] 
KaCa
Strong base, [OH–] = Cb or 2Cb
Weak base, [OH ] 
–
Watkins
K bCb
Chem 1422, Chapter 17
pH = -log [H+]
pH = 14 + log [OH–]
21
Titration - Quantitative Problems
2. What volume of titrant must be added to reach the
end point?
a Acid + b Base → s Salt + w Water
Ca is the initial molarity of the acid
Cb is the initial molarity of the base
Va is the volume of acid in the flask at the end point
Vb is the volume of base in the flask at the end point
CaVa
a
=
CbVb
b
End-Point Titration Equation
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Chem 1422, Chapter 17
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Titration - Quantitative Problems
3. What is the pH at the end point?
a Acid + b Base → s Salt + w Water
The pH is determined by the salt.
For a SBSA salt, pH = 7
For a SBWA salt, pH > 7 (conjugate weak base)
[OH-] ≈ KbCb where Kb = Kw/Ka
For a WBSA salt, pH < 7 (conjugate weak acid)
[H+] ≈ K aCa where Ka = Kw/Kb
For a WBWA salt ... forget it!
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Chem 1422, Chapter 17
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Titration: Polyprotic Acids/Bases
Polyprotic acids and bases
show multiple equivalence
points, one for each Ka.
H+ + CO32- → HCO3-
H+ + HCO3- → H2CO3
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Chem 1422, Chapter 17
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Titration Problem
35.00 mL of 0.257 M acetic acid is to be titrated by 0.149
M calcium hydroxide. Calculate (1) the pH of the acetic
acid before titration; (2) the volume of base added to reach
the end point; (3) the pH at the end point.
(1) HC2H3O2(aq) + H2O(l) ⇌ C2H3O2-(aq) + H3O+(aq)
Acetic acid, Ka = 1.8×10-5, Ca = 0.257 M
[H+] ≈ (KaCa)½ = 2.15×10-3, pH = 2.67
(2) 2HC2H3O2(aq) + Ca(OH)2(aq) → Ca(C2H3O2)2(aq) + 2H2O(l)
Ca = 0.257 M, Va = 35.00 mL, Cb = 0.149 M, Vb = ?
CaVa/2 = CbVb/1
Vb = (CaVa)/(2Cb)
Vb = 30.18 mL
(end point volume: Va + Vb = 65.18 mL)
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Chem 1422, Chapter 17
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Titration Problem
35.00 mL of 0.257 M acetic acid is to be titrated by 0.149
M calcium hydroxide. Calculate (1) the pH of the acetic
acid before titration; (2) the volume of base added to reach
the end point; (3) the pH at the end point.
(3) The end-point solution contains calcium acetate; calcium is a
spectator, but acetate is the conjugate base of acetic acid:
Kb = Kw/Ka = 5.57×10-10.
The amount of acetate is CaVa = (35)(0.297) = 10.40 mmol, and
the final volume of the solution is 65.18 mL;
[C2H3O2-] = (10.48 mmol)/(65.18 mL) = 0.161 M (>> Kb)
[OH-] ≈ [(5.57×10-10)(0.161)]½ = 9.47×10-6
pOH = 5.02, pH = 8.98
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Chem 1422, Chapter 17
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Titration Problem
18.73 mL of 0.116 M ammonia is to be titrated by 0.273
M nitric acid. Calculate (1) the pH of the ammonia before
titration; (2) the volume of acid added to reach the end
point; (3) the pH at the end point.
(1) NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
Ammonia, Kb = 1.8×10-5, Cb = 0.116 M
[OH-] ≈ (KbCb)½ = 1.44×10-3, pOH = 2.84, pH = 11.16
(2) NH3(aq) + HNO3(aq) → NH4NO3(aq) + H2O(l)
Ca = 0.273 M, Va = ?, Cb = 0.116 M, Vb = 18.73 mL
CaVa/1 = CbVb/1
Va = (CbVb)/Ca
Va = 7.96 mL
(end point volume: Va + Vb = 26.69 mL)
Watkins
Chem 1422, Chapter 17
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Titration Problem
18.73 mL of 0.116 M ammonia is to be titrated by 0.273
M nitric acid. Calculate (1) the pH of the ammonia before
titration; (2) the volume of acid added to reach the end
point; (3) the pH at the end point.
(3) The end-point solution contains ammonium nitrate; nitrate is
a spectator, but ammonium is the conjugate acid of ammonia:
Ka = Kw/Kb = 5.57×10-10.
The mmol of ammonium is CbVb = (18.73)(0.116) = 2.17 mmol,
and the final volume of the solution is 26.69 mL;
[NH4+] = (2.17 mmol)/(26.69 mL) = 0.0813 M = Ca.
[H+] ≈ [(5.57×10-10)(0.0813)]½ = 6.73×10-6 pH = 5.17
Watkins
Chem 1422, Chapter 17
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Titration Problems
Strong acid & strong base titrations → neutral salt.
Weak acid or weak base titrations:
At the beginning of the titration, either a weak acid or a
weak base.
At the end-point of the titration, either the conjugate
weak base or the conjugate weak acid.
What about a solution that has both a weak acid and its
conjugate weak base?
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Chem 1422, Chapter 17
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Buffered Solutions
Buffer = a solution containing comparable amounts of
both members of a conjugate weak acid/weak base pair
Ca ≈ Cb
HAz(aq) + H2O(l)
Ca Ka
Bz(aq) + H2O(l)
Cb Kb
H3O+(aq) + Az-1(aq)
Cb Kb = Kw/Ka
OH-(aq) + HBz+1(aq)
Ca Ka = Kw/Kb
Ions (Az-1 and HBz+1) are added in the form of salts with
spectator counterions such as Na+ and ClWatkins
Chem 1422, Chapter 17
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Buffered Solutions
Buffer = a solution containing comparable amounts of
both members of a conjugate weak acid/weak base pair
Ca ≈ Cb
A buffer resists changes in pH when a small amount of
OH– or H+ is added.
To understand how buffers work, consider a buffer
solution composed of approximately equal amounts
of HA and A-: Ca ≈ Cb where Ka << 1.
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Chem 1422, Chapter 17
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Buffered Solutions (Acidic)
Buffer = weak acid (HA) + conjugate weak base (A-)
H+(aq) + A-(aq)
HA(aq)
+
I
C
E
-
HA(aq) H (aq) A (aq)
Ca
0
Cb
-x
+x
+x
Ca-x
x
Cb+x
Ka << 1
x(Cb  x)
Ka 
(Ca  x)
(Ca  x)
x  Ka
(Cb  x)
If Ca & Cb >> Ka, then x is small
with respect to both Ca and Cb:
Ca
[H ]  K a
Cb
Ca–x ~ Ca and Cb+x ~ Cb
Buffer Equation
Watkins
Chem 1422, Chapter 17

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Buffered Solutions (Basic)
Buffer = weak base (B) + conjugate weak acid (HB+)
HB+(aq)
I
C
E
HB+(aq)
Ca
-x
Ca-x
B(aq) + H+(aq)
H+(aq)
0
+x
x
B(aq)
Cb
+x
Cb+x
Ka
K w xCb  x 
Ka 

Ca  x 
Kb
(Ca  x)
x  Ka
(Cb  x)
If Ca & Cb > Ka, then x is small
with respect to both Ca and Cb:
Ca
[H ]  K a
Cb
Ca-x ~ Ca and Cb+x ~ Cb
Buffer Equation
Watkins
Chem 1422, Chapter 17

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Buffered Solutions
Buffer capacity is the amount of H+ or OH- neutralized
by the buffer before there is a significant change in
pH.
Buffer capacity depends on the composition of the
buffer.
The higher the concentrations of conjugate weak acid
and weak base (Ca & Cb), the greater the buffer
capacity.
Watkins
Chem 1422, Chapter 17
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Buffered Solutions
Ca

[H ]  K a
Cb
C'a
Ka '
Cb
Add some
strong base
The ratio does not change
much, so [H+] does not
change much
Start with a 1:1
buffer: Ca = Cb
HA
A-
OH-
Ca'
Cb'
H+
Ca
HA + OH- → H2O + AWatkins
Add some
strong acid
Cb
Ca'
Cb'
A- + H+ → HA
Chem 1422, Chapter 17
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Strong Acid/Base + Buffer
Problem Solving Strategy
To calcuate how the pH changes when a strong acid or a
strong base is added to a buffered solution:
1. Calculate the pH before addition.
2. Calculate the stoichiometric change in Ca and Cb.
3. Calculate the pH after addition.
Watkins
Chem 1422, Chapter 17
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Sample Problem I
1. 100 mL of any neutral unbuffered solution
e.g., water
(pH 7.00)
Add 1 mL of 1.0M HCl (pH 0)
2. Millimoles HCl added
= (1.0 mmol/mL)•(1 mL)
= 1 mmol
Volume of solution after addition
= 101 mL
3. [HCl] = [H+] in final solution
= (1 mmol)/(101 mL)
= 9.9×10-3 M
pH after addition = 2.00
Watkins
Chem 1422, Chapter 17
DpH = 5
37
Sample Problem II
100 mL of a buffer solution containing
0.5M acetic acid (Ka = 1.8×10-5) and
0.5M sodium acetate
+ 1 mL of 1.0M HCl
1. pH before addition:
[H+] = Ka(Ca/Cb)
= (1.8×10-5)(0.5/0.5) = 1.8×10-5
pH = 4.745
note that for a one-to-one buffer
[H+] = Ka and pH = pKa
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Chem 1422, Chapter 17
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Sample Problem II
2. Stoichiometric changes in Ca and Cb:
na = (0.5 mmol/mL)•(100 mL) = 50 mmol HC2H3O2
nb = (0.5 mmol/L)•(100 mL) = 50 mol C2H3O2–
C2H3O2-(aq) + H+(aq) → HC2H3O2(aq)
n(HCl) = (1.0 mmol/mL)•(1 mL) = 1 mmol HCl
na' = 50 + 1 = 51 mmol
nb' = 50 – 1 = 49 mol
Ca' = 51/101 = 0.505 M
Cb' = 49/101 = 0.485 M
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Chem 1422, Chapter 17
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Sample Problem II
3. pH after addition:
[H+]  Ka(Ca'/Cb')
= (1.8×10-5)(0.505/0.485) = 1.87×10-5
pH = 4.727
pH before addition was 4.745 so for this
buffered solution, DpH = 0.018
(instead of DpH = 5 for the unbuffered system)
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Chem 1422, Chapter 17
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Sample Problem II
Ca’ = na’/V and Cb’ = na’/V
where V is the volume in which both members
of the conjugate pairs are dissolved.
Therefore:
Ca’/Cb’ = na’/nb’
[H+]
Ca
na
≈ Ka
= Ka
nb
Cb
Buffer Equation
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Chem 1422, Chapter 17
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Sample Problems I & II
Add 1 mL
of 1 M HCl
pH 7
pH 2
DpH = 5.00
to 100 mL
water
Add 1 mL
of 1 M HCl
pH 4.745
DpH = 0.02
to 100 mL
of 1:1
HAc/Ac
buffer
Watkins
pH 4.727
Chem 1422, Chapter 17
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Buffers
[H+]
Ca
na
≈ Ka
= Ka
nb
Cb
Buffer Equation
Blood and other physiological solutions are
highly buffered, with multiple acid/salt
combinations (phosphate, acetate, carbonate,
etc.)
Thus, drinking Union coffee or taking an aspirin
does not change the pH of the blood significantly.
Too much aspirin can kill, by lowering blood pH,
a condition known as acidosis.
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Chem 1422, Chapter 17
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Buffer Examples
A buffer is also produced when a weak acid is
titrated with a strong base:
half-way
100% salt
0% acid
1:1 buffer
0% salt
100% acid
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Chem 1422, Chapter 17
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20 mL 0.416 M NaOH
Buffer Examples
50 mL of 0.367 M acetic acid (HAc, Ka = 1.8×10-5) is
titrated with 0.416 M sodium hydroxide. What is the
pH of the solution after 20 mL of the titrant is added?
First, calculate the inital pH ...
There are two steps to a general titration problem:
1. Stoichiometry: the strong base titrant reacts
completely with the weak acid to produce the
weak conjugate acid/base pair (net ionic reaction):
HAc + OH– → Ac– + H2O
2. Equilibrium: the conjugate pair forms a buffer:
HAc ⇌ H+ + Ac–
50 mL 0.367 M HAc
Initial pH: 2.59
One way to organize these calculations is to use a Reaction
Table for stoichiometry, followed by an ICE Table for
equilibrium.
Watkins
Chem 1422, Chapter 17
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20 mL 0.416 M NaOH
Buffer Examples
50 mL of 0.367 M acetic acid (HAc, Ka = 1.8×10-5) is
titrated with 0.416 M sodium hydroxide. What is the
pH of the solution after 20 mL of the titrant is added?
Reaction Table
1.
HAc + OH– → Ac– + H2O
iM
HAc
0.367
OH–
0.416
Ac–
0
H2 O
--
iV
50
20
0
--
in
Dn
18.35
–x
8.32
–x
0
+x
---
fn
18.35–x
8.32–x
x
--
fn
10.03
0
8.32
--
50 mL 0.367 M HAc
Initial pH: 2.59
mmol/mL
mL
mmol
The limiting reagent is OH–,
so x = 8.32
Approximate solution:
[H+] ≈ (1.8×10-5)(10.03/8.32) = 2.17×10-5, pH ≈ 4.66
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Chem 1422, Chapter 17
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20 mL 0.416 M NaOH
Buffer Examples
50 mL of 0.367 M acetic acid (HAc, Ka = 1.8×10-5) is
titrated with 0.416 M sodium hydroxide. What is the
pH of the solution after 20 mL of the titrant is added?
ICE Table
2.
HAc ⇌ H+ + Ac–
HAc
H+
Ac–
C
10.03/70
–x
0
+x
8.32/70
+x
E
0.1433–x
x
0.1189+x
I
Exact solution (x = mmol
x = 2.17×10-5, pH = 4.66
H+):
mmol/mL
Qi = 0, rx →
1.8×10-5
50 mL 0.367 M HAc
Initial pH: 2.59
x(0.1189+x)
=
(0.1433–x)
Approximate solution:
[H+] ≈ (1.8×10-5)(10.03/8.32) = 2.17×10-5, pH ≈ 4.66
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Chem 1422, Chapter 17
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Solubility Equilibria (Review Chapter 4)
All ionic solids (salts) dissolve to some extent in
water.
When the maximum amount of salt has dissolved,
the solution is saturated.
If the saturation limit is large, the salt is soluble.
If the saturation limit is small, the salt is slightly
soluble (“insoluble”).
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Solubility Equilibria (Review Chapter 4)
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Solubility Equilibria
In a saturated solution, dissolved ions are in equilibrium
with undissolved solid.
NaCl(s) ⇌ Na+(aq) + Cl–(aq)
Kc = Ksp = [Na+][Cl–]
(sp = “solubility product”)
diss. →
For soluble salts (NaCl),
Ksp > 1; equilibrium lies
to the right
For insoluble salts
(CaF2), Ksp < 1;
equilibrium lies to the
left
Watkins
ppt. ←
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Solubility Equilibria
Solubility is the number of grams of a salt which
dissolve in 1 liter of pure water (g/L = mg/mL)
Molar solubility is the number of moles of a salt which
dissolve in 1 liter of pure water (mol/L = mmol/mL)
Each of these can be converted to or from Ksp
Large Ksp == soluble salt
Small Ksp == insoluble salt
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Solubility Equilibria
Example: the solubility of NaCl is 357 g/L*
NaCl(s)
Na+(aq) + Cl-(aq)
(357 g/L)/(58.44 g/mol) = 6.109 mol/L
= [Na+] = [Cl-]
Ksp = [Na+][Cl-] = (6.109)(6.109) = 37.32
*Handbook
Watkins
of Chemistry and Physics
Chem 1422, Chapter 17
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Solubility Equilibria
Example: the solubility of CaF2 is 16.7 mg/L*
CaF2(s)
Ca2+(aq) + 2F–(aq)
(0.0167 g/L)/(78.08 g/mol) = 2.14×10-4 mol/L
[Ca2+] = 2.14×10-4 M
[F-] = 2[Ca2+] = 4.28×10-4 M
Ksp = [Ca2+][F–]2 = (2.14×10-4)(4.28×10-4)2 = 3.92×10-11
*Handbook
Watkins
of Chemistry and Physics
Chem 1422, Chapter 17
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Solubility Equilibria
The most common solubility problem is shown in the
following example:
What is the concentration of all ions in a saturated
solution of LaF3(s)?
La3+(aq) + 3F-(aq)
LaF3(s)
LaF3 La3+
F-
Ksp = 2×10-19
Ksp = [La3+][F-]3
I
XS
0
0
C
-x
+x
+3x
x = 9.28×10-6
x
3x
[La3+] = x = 9.28×10-6 M
E XS-x
= x(3x)3 = 27x4 = 2×10-19
[F-] = 3x = 2.78×10-5 M
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Solubility Equilibria
1.
2.
3.
4.
Failure to write down the equilibrium reaction;
Failure to write down the equilibrium expression;
Failure to fill out the ICE table;
Failure to place coefficients correctly.
1
3
La3+(aq) + 3F-(aq)
LaF3(s)
LaF3 La3+
F-
Ksp = 2×10-19
Ksp = [La3+][F-]3
I
XS
0
0
C
-x
+x
+3x
x = 9.28×10-6
x
3x
[La3+] = x = 9.28×10-6 M
E XS-x
Common Errors
Watkins
2
= x(3x)3 = 27x4 = 2×10-19
4
[F-] = 3x = 2.78×10-5 M
Chem 1422, Chapter 17
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Common Ion Effect
The solubility of a salt decreases when a common ion is
added (Le Châtelier’s principle)
CaF2(s)
Ca2+(aq) + 2F-(aq)
If NaF(s) is added to the saturated CaF2 solution,
additional F- appears in the solution. F- is called a
common ion (Na+ has no effect).
This shifts the equilibrium to the left, so precipitation
occurs and more CaF2(s) is formed.
Precipitation would also occur if Ca(NO3)2(s) were
added to the CaF2 solution.
Watkins
Chem 1422, Chapter 17
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Common Ion Problem
Calculate the molar solubility of CaF2(s) in a 0.05 M
CaF2(s)
Ca2+(aq) + 2F-(aq)
NaF(aq) solution.
CaF2 Ca2+
I
F-
XS
0
0.05
C
-x
+x
+2x
E
XS-x
x
0.05+2x
Qi = 0 so rx →
Ksp = 3.9×10-11 = [Ca2+][F-]2
3.9×10-11 = x(0.05+2x)2
1. Graphing Calculator
x = 1.56×10-8 M
2. Approximation: K << 1 so x << 1
3.9×10-11 ≈ x(0.05)2
x ≈ 1.56×10-8 M
x is the molar solubility
compare with the molar
solubility in water: 2.14×10-4 M
Watkins
Chem 1422, Chapter 17
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Common Ion Effect
The greater the
concentration of
common ion, the
less soluble the
salt.
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Chem 1422, Chapter 17
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Solubility & pH
The solubilities of salts with anions of weak acids
increase with decreasing pH (i.e., such salts dissolve in
strong acid). Example:
CaF2(s)
Ca2+(aq) + 2F–(aq) Ksp = 3.9×10-11
If a strong acid (e.g., HNO3) is added to the saturated
salt solution, a competing equilibrium occurs:
H+(aq) + F–(aq)
HF(aq)
Kc = Ka-1 = 1.5×103
The second reaction removes F– from solution and
perturbs the first equilibrium. According to
LeChatelier, the first reaction must proceed to the right
(dissolving more CaF2) to restore equilibrium.
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Chem 1422, Chapter 17
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Solubility & pH
There will be no calculations of competing equilibria in
this class.
Watkins
Chem 1422, Chapter 17
60
Solubility & pH
Iron Pyrite FeS2
(Fool's Gold)
Ksp = 6.7×10-9
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Chem 1422, Chapter 17
61
Dissolving vs. Precipitating
The solubility equilibrium summarizes two
different reactions, dissolution and precipitation.
Up to now, we have been concentrating on the
dissolving reaction and solubility problems:
Solid salt → Dissolved ions in solution
We now turn to the precipitation reaction and
precipitation problems:
Solid salt ← Dissolved ions in solution
Watkins
Chem 1422, Chapter 17
62
Precipitation of Metal Ions
To 1 L of 0.5 M solution of AgNO3, add 17.5 ng of NaCl(s).
Does a ppt form?
AgCl(s) ⇌ Ag+(aq) + Cl–(aq)
Ksp = [Ag+][Cl–] = 1.8×10-10
[Ag+]i = 0.5 M
[Cl–]i = 3.0×10-10 M
Qi = [Ag+]i[Cl–]i = (0.5)(3.0×10-10) = 1.5×10-10
Since Qi < Ksp, the equilibrium reaction would proceed
from left to right ...
... except there is no solid AgCl to begin with, so there is
no saturated solution and no equilibrium.
So nothing happens - there is no precipitate!
Watkins
Chem 1422, Chapter 17
63
Precipitation of Metal Ions
0.5 M AgNO3 + Cl–
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Ksp = [Ag+][Cl-] = 1.8×10-10
At what chloride concentration would pptn begin?
When Qi = Ksp:
Ksp = [Ag+]i[Cl–]i = (0.5)[Cl–] = 1.8×10-10
[Cl–]i = 3.6×10-10 M (21 ng NaCl/L)
As more chloride is added, more AgCl(s) precipitates.
When [Cl–]i reaches 0.5 M (29.2 g/L), most of the silver
in the initial solution would precipitate, because then
Ksp = [Ag+][Cl–]i = [Ag+](0.5) = 1.8×10-10
[Ag+] = 3.6×10-10 M
Watkins
Chem 1422, Chapter 17
64
Precipitation of Metal Ions
Metal ions can be removed from solution by precipitating
with an anion to form a slightly soluble salt.
If Qi < Ksp, no precipitation occurs
If Qi > Ksp, precipitation occurs until Q = Ksp.
Mixtures of metal ions can be separated by precipitating
salts with different solubilities.
Differential solubility problems
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Chem 1422, Chapter 17
65
Separation of Metal Ions
Consider an aqueous mixture of 0.5 M Zn(NO3)2 and
0.5 M Cu(NO3)2. Both metals form insoluble sulfides:
ZnS(s) ⇌ Zn2+(aq) + S2-(aq)
[Zn2+]i = 0.5 M
Ksp = 2×10-25
[S2-]sat = 4.0×10-25 M
CuS(s) ⇌ Cu2+(aq) + S2-(aq)
Ksp = 6×10-37
[Cu2+]i = 0.5 M
[S2-]sat = 1.2×10-36 M
[S2-]i = 0.
As Na2S(s) is added to the solution, [S2-] increases until
it reaches the saturation limit for the least soluble salt.
The [S2-] concentrations at the saturation limits for both
salts are [S2-]sat = Ksp/[metal ion]:
Watkins
Chem 1422, Chapter 17
66
Separation of Metal Ions
Consider an aqueous mixture of 0.5 M Zn(NO3)2 and
0.5 M Cu(NO3)2. Both metals form insoluble sulfides:
ZnS(s) ⇌ Zn2+(aq) + S2-(aq)
[Zn2+]i = 0.5 M
Ksp = 2×10-25
[S2-]sat = 4.0×10-25 M
CuS(s) ⇌ Cu2+(aq) + S2-(aq)
Ksp = 6×10-37
[Cu2+]i = 0.5 M
[S2-]sat = 1.2×10-36 M
[S2-]i = 0. S2- is added until it reaches 1.2×10-36 M,
CuS begins to precipitate;
however, since [S2-] has not yet reached the
saturation limit for ZnS,
ZnS does not precipitate.
Watkins
Chem 1422, Chapter 17
[S2-]
67
Separation of Metal Ions
Consider an aqueous mixture of 0.5 M Zn(NO3)2 and
0.5 M Cu(NO3)2. Both metals form insoluble sulfides:
ZnS(s) ⇌ Zn2+(aq) + S2-(aq)
[Zn2+]i = 0.5 M
Ksp = 2×10-25
[S2-]sat = 4.0×10-25 M
CuS(s) ⇌ Cu2+(aq) + S2-(aq)
Ksp = 6×10-37
[Cu2+]i = 0.5 M
[S2-]sat = 1.2×10-36 M
As more Na2S(s) is added to the solution, more CuS
precipitates and [S2-] increases until it reaches the
saturation limit for ZnS: 4.0×10-25 M.
At this concentration, ZnS would begin to precipitate.
[S2-]
Watkins
Chem 1422, Chapter 17
68
Separation of Metal Ions
Consider an aqueous mixture of 0.5 M Zn(NO3)2 and
0.5 M Cu(NO3)2. Both metals form insoluble sulfides:
ZnS(s) ⇌ Zn2+(aq) + S2-(aq)
[Zn2+]i = 0.5 M
Ksp = 2×10-25
[S2-]sat = 4.0×10-25 M
CuS(s) ⇌ Cu2+(aq) + S2-(aq)
Ksp = 6×10-37
[Cu2+]i = 0.5 M
[S2-]sat = 1.2×10-36 M
But when [S2-] reaches 4×10-25 M, [Cu2+] is now
[Cu2+] = (6×10-37)/(4×10-25) = 1.5×10-12 M
That is, essentially all of the copper ions (99.999... %)
have precipitated while the zinc ions remained in
solution!
Watkins
Chem 1422, Chapter 17
[S2-]
69