Transcript Solubility Product

Applications of Aqueous
Equilibria
“Common Ions”
• When we dissolve acetic acid in water, the
following equilibrium is established:
– CH3COOH  CH3COO- + H+
• If sodium acetate were dissolved in solution,
which way would this equilibrium shift?
“Common Ions”
• When we dissolve acetic acid in water, the
following equilibrium is established:
– CH3COOH  CH3COO- + H+
• If sodium acetate were dissolved in solution,
which way would this equilibrium shift?
– Adding acetate ions from a strong electrolyte would
shift this equilibrium left (Le Chatelier’s principle).
Common Ion Effect
• Whenever a weak electrolyte (acetic acid)
and a strong electrolyte (sodium acetate)
share a common ion, the weak electrolyte
ionizes less than it would if it were alone.
(Le Chatelier’s)
• This is called the common-ion effect.
Steps for Common-Ion Problems
• 1. Consider which solutes are strong electrolyte
and weak electrolytes.
• 2. Identify the important equilibrium (weak)
that is the source of H+ and therefore
determines pH.
• 3. Create an ICE chart using the equilibrium
and strong electrolyte concentrations.
• 4. Use the equilibrium constant expression to
calculuate [H+] and pH.
• What is the concentration of silver and chromate in a
solution with silver chromate in 0.1 M silver nitrate
Ag2CrO4(s)  2Ag+1 + CrO4-2
+1?
What
are
the
sources
of
Ag
I
0.1
0
C
+2x
+x
E
0.1 + 2x
x
Why is this a plus sign?
Ksp = 9.0 x 10-12 = [Ag+1]2 • [CrO4-2]
= 9.0 x 10-12 = [.1 +2x]2 • [x]
9.0 x 10-12 = 0.12 • x
[CrO4-2] = x = 9.0 x 10-10 M
[Ag+1] = .1 + 2x = .1M
[Ag2CrO4] = 9.0 x 10-10
Sample Problem
• What is the pH of a solution made by adding
0.30 mol of acetic acid and 0.30 mol of sodium
acetate to enough water to make 1.0 L of
solution?
• Ka for acetic acid = 1.8 x 10-5
pH = 4.74
Sample Problem
• Calculate the fluoride concentration and pH of
a solution that is 0.20 M in HF and 0.10 M in
HCl
• Ka for HF = 6.8 x 10-4.
• [F-] = 1.2x10-3M
• pH = 1.00
Buffers
Solutions that resist pH change when small
amounts of acid or base are added.
Two types
weak acid and its salt
weak base and its salt
HX(aq) + H2O(l)
Add OHshift to right
H+(aq) + X-(aq)
Add H+
shift to left
Based on the common ion effect.
Buffers
8
7
pH
6
5
buffered
4
3
unbuffered
2
1
0
0
20
40
60
80
ml HCl added
100
120
Buffers and blood
Control of blood pH
Oxygen is transported primarily by hemoglobin in the
red blood cells.
CO2 is transported both in plasma and the red blood
cells.
CO2 (aq) + H2O
H2CO3 (aq)
H+(aq) + HCO3-(aq)
Buffers
Composition and Action of Buffered Solutions
HX(aq)
The Ka expression is
H+(aq) + X-(aq)
[H  ][X - ]
Ka 
.
[HX]
[HX]
[H  ]  K a
.
[X ]
A buffer resists a change in pH when a small
amount of OH- or H+ is added.
Buffered Solutions
Addition of Strong Acids or Bases to Buffers
• With the concentrations of HX and X- (note the
change in volume of solution) we can calculate the pH
from the Henderson-Hasselbalch equation
pH  p K a
pH  p K a
[ X- ]
 log
[HX]
conjugate base
 log
conjugate acid
Equation not necessary, since you know
Ka = [X-]
[H+] [HX]
Buffers
Action of Buffered Solutions
Buffers
Buffer Capacity and pH
Buffer capacity is the amount of acid or base
neutralized by the buffer before there is a
significant change in pH.
The greater the amounts(molarity) of the
conjugate acid-base pair, the greater the
buffer capacity.
The pH of the buffer depends on Ka
Buffered Solutions
Addition of Strong Acids or Bases to Buffers
Buffer
Addition of Strong Acids or Bases to Buffers
Break the calculation into two parts:
stoichiometric and equilibrium.
The amount of strong acid or base added
results in a neutralization reaction:
X- + H+  HX + H2O
HX + OH-  X- + H2O.
By knowing how much H+ or OH- was added
(stoichiometry) we know how much HX or Xis formed.
Buffers
The final [HX] and [X-] after the neutralization
reaction are used as the initial concentrations for
the equilibrium reaction.
HX
H+
XInitial conc., M
Change, DM
Eq. Conc., M
Then the equilibrium constant expression is used to
find [H+] and pH = - log [H+]
Ka = [H+] [X-]
[HX]
Buffer Example
Determine the initial pH of a buffer solution
that has 0.10 M benzoic acid and 0.20 M
sodium benzoate at 25 oC. Ka = 6.5 x 10-5
H+(aq) + Bz-(aq)
HBz(aq) + H2O(l)
HBz
H+
Bz-
Initial conc., M
0.10
0.00
0.20
Change, DM
-x
Eq. Conc., M
0.10 - x
+x
x
+x
0. 20 + x
Buffer Example
Solve the equilibrium equation in terms of x
Ka = 6.5 x 10-5 =
x (0. 20 + x)
0.10 - x
= (6.5 x 10-5 )(0.10) / (0. 20)
(assuming x<<0.10)
= 3.2 x 10-5 M H+
pH = - log (3.2 x 10-5 M) = 4.5
initial pH
x
Buffer Example
Determine the pH of a buffer solution that has 0.10 M
benzoic acid and 0.20 M sodium benzoate at 25 oC after
0.05 moles of HCl is added.
First, find the concentrations of the HBz and Bzafter HCl is added.
Initial conc., M
Change, DM
Eq. Conc., M
HBz
0.15
-x
0.15 - x
H+
0.00
+x
x
Bz0.15
+x
0. 15 + x
The 0.05 mol HCl reacts completely with 0.05 mol Bz-(aq)
to form 0.05 mol HBz(aq)
Then equilibrium will be re-established based on the new
initial concentrations of 0.15 M HBz(aq) and 0.15 M Bz(aq).
Buffer Example
Solve the equilibrium equation in terms of x
Ka = 6.5 x 10-5 =
x (0. 15 + x)
0.15 - x
x
= (6.5 x 10-5 )(0.15) / (0. 15)
(assuming x<<0.15)
= 6.5 x 10-5 M H+
pH = - log (6.5 x 10-5 M) = 4.2
after 0.05 mole HCl added
Buffer Example
Determine the pH of a buffer solution that has 0.10 M benzoic acid
and 0.20 M sodium benzoate at 25 oC after 0.05 moles of NaOH is
added.
First, find the concentrations of HBz and Bz- after NaOH is
added.
HBz
H+
BzInitial conc., M
0.05
0.00
0.25
Change, DM
-x
+x
+x
Eq. Conc., M
0.05 - x
x
0. 25 + x
The 0.05 mol NaOH reacts completely with 0.05 mol
HBz(aq) to form 0.05 mol Bz-(aq)
Then equilibrium will be re-established based on the new
initial concentrations of 0.05 M HBz(aq) and 0.25 M Bz-(aq).
Buffer Example
Solve the equilibrium equation in terms of x
Ka = 6.5 x 10-5 =
x
x (0. 25 + x)
0.05 - x
= (6.5 x 10-5 )(0.05) / (0. 25)
(assuming x<<0.05)
= 1.3 x 10-5 M H+
pH = - log (1.3 x 10-5 M) = 4.9
after 0.05 mole NaOH added
Acid-Base Titrations
Strong Acid-Base Titrations
• The plot of pH
versus volume
during a titration
is a titration curve.
Remember the Acid-Base Titration Curves?
Buffer
Zone
HC2H3O2
HCl
Acetic Acid/Acetate Ion Buffer Lab
For this experiment, you will prepare a buffer that
contains acetic acid and its conjugate base, the
acetate ions. The equilibrium equation for the
reaction is shown below:
HC2H3O2(aq) + H2O(l) <=> H+ (aq)+ C2H3O2- (aq)
The equilibrium expression for this reaction, Ka, has
a value of 1.8 x 10-5 at 25ºC.
Acetic Acid/Acetate Ion Buffer Lab
 The ratio between the molarity of the acetate ions to the
molarity of the acetic acid in your buffer must equal the
ratio between the Ka value and 10- assigned pH.
 This ratio should be reduced , so that either the
[HC2H3O2] or [C2H3O2- ] has a concentration of 0.10 M,
and the concentration of the other component must fall
within a range from 0.10 M to 1.00 M.
 Complete the calculations only that are needed to
prepare 100.0 mL of your assigned buffer solution that
has these specific concentrations. Can you predict the
final pH when a strong acid or base is added to the
buffer solution?
Acetic Acid/Acetate Ion Buffer Lab
Bu ffer 1
Given: 1.00 M acetic acid
solid sodium acetate t rihydrate
Problem: Prepare 100.0 mL of a buffer solut ion
with a pH of 4.50 that has a concentration of each
component within the range of 0.10 M to 1.00 M.
Use the lowest possible concentrations for the acid
and conjugate base in the buffer.
SHOW ALL CALCULATIONS USED!
Bu ffer 3
Given: 1.00 M acetic acid
1.00 Msodium hydroxide
Problem: Prepare 100.0 mL of a buffer solut ion
with a pH of 4.50 that has a concentration of each
component within the range of 0.10 M to 1.00 M.
Use the lowest possible concentrations for the acid
and conjugate base in the buffer.
SHOW ALL CALCULATIONS USED!
Bu ffer 5
Given: 1.00 M hydrochloric acid
solid sodium acetate t rihydrate
Problem: Prepare 100.0 mL of a buffer solut ion
with a pH of 5.00 that has a concentration of each
component within the range of 0.10 M to 1.00 M.
Use the lowest possible concentrations for the acid
and conjugate base in the buffer.
SHOW ALL CALCULATIONS USED!
Bu ffer 2
Given: 1.00 M hydrochloric acid
solid sodium acetate t rihydrate
Problem: Prepare 100.0 mL of a buffer solut ion
with a pH of 4.50 that has a concentration of each
component within the range of 0.10 M to 1.00 M.
Use the lowest possible concentrations for the acid
and conjugate base in the buffer.
SHOW ALL CALCULATIONS USED!
Bu ffer 4
Given: 1.00 M acetic acid
solid sodium acetate t rihydrate
Problem: Prepare 100.0 mL of a buffer solut ion
with a pH of 5.00 that has a concentration of each
component within the range of 0.10 M to 1.00 M.
Use the lowest possible concentrations for the acid
and conjugate base in the buffer.
SHOW ALL CALCULATIONS USED!
Bu ffer 6
Given: 1.00 M acetic acid
1.00 Msodium hydroxide
Problem: Prepare 100.0 mL of a buffer solut ion
with a pH of 5.00 that has a concentration of each
component within the range of 0.10 M to 1.00 M.
Use the lowest possible concentrations for the acid
and conjugate base in the buffer.
SHOW ALL CALCULATIONS USED!
Acetic Acid/Acetate Ion Buffer Lab
Bu ffer 7
Given: 1.00 M acetic acid
solid sodium acetate t rihydrate
Problem: Prepare 100.0 mL of a buffer solut ion
with a pH of 4.00 that has a concentration of each
component within the range of 0.10 M to 1.00 M.
Use the lowest possible concentrations for the acid
and conjugate base in the buffer.
SHOW ALL CALCULATIONS USED!
Bu ffer 9
Given: 1.00 M acetic acid
1.00 Msodium hydroxide
Problem: Prepare 100.0 mL of a buffer solut ion
with a pH of 4.00 that has a concentration of each
component within the range of 0.10 M to 1.00 M.
Use the lowest possible concentrations for the acid
and conjugate base in the buffer.
SHOW ALL CALCULATIONS USED!
Bu ffer 11
Given: 1.00 M hydrochloric acid
solid sodium acetate t rihydrate
Problem: Prepare 100.0 mL of a buffer solut ion
with a pH of 5.50 that has a concentration of each
component within the range of 0.10 M to 1.00 M.
Use the lowest possible concentrations for the acid
and conjugate base in the buffer.
SHOW ALL CALCULATIONS USED!
Bu ffer 8
Given: 1.00 M hydrochloric acid
solid sodium acetate t rihydrate
Problem: Prepare 100.0 mL of a buffer solut ion
with a pH of 4.00 that has a concentration of each
component within the range of 0.10 M to 1.00 M.
Use the lowest possible concentrations for the acid
and conjugate base in the buffer.
SHOW ALL CALCULATIONS USED!
Bu ffer 10
Given: 1.00 M acetic acid
solid sodium acetate trihydrate
Problem: Prepare 100.0 mL of a buffer solut ion
with a pH of 5.50 that has a concentration of each
component within the range of 0.10 M to 1.00 M.
Use the lowest possible concentrations for the acid
and conjugate base in the buffer.
SHOW ALL CALCULATIONS USED!
Making a Buffer Calculations
You want to prepare 500.0 mL of a buffer with a pH = 10.00, with
both the acid and conjugate base having molarities between 0.10 M to
1.00 M. You may choose from any of the acids listed below:
Name
acetic acid
tartaric acid
citric acid
malonic acid
carbonic acid
phosphoric acid
ammonium
Formula
HC2H3O2
H2C4H4O6
H3C6H5O7
H2C3H2O4
H2CO3
H3PO4
NH41+
Ka1
1.8 x 10-5
1.0 x 10-3
7.4 x 10-4
1.5 x 10-3
4.3 x 10-7
7.5 x 10-3
5.6 x 10-10
Ka2
4.6 x 10-5
1.7 x 10-5
2.0 x 10-6
5.6 x 10-11
6.2 x 10-8
Ka3
4.0 x 10-7
4.2 x 10-13
You must select an acid with a Ka value close to 10- assigned pH.
The only two options are ammonium or the hydrogen carbonate
ions.
2
+
+
H
NH
H
CO
 3


3
-10
-11
K a = 5.6 x 10 =
K a2 = 5.6 x 10 =
+
1NH4
HCO 3
 






Making a Buffer Calculations
K a = 5.6 x 10-10 =
+
H
 NH3 

NH4
5.6 x 10-10 NH 3 
=
-10
+
10
NH
 
4


+

5.6 x 10-10 NH 3 
=
+
+
H
NH
 
4


5.6 NH 3 
=
+
1
NH 4



0.56 NH3 
Divide both [ ] by 10.....
=
+
0.1
NH4



0.1 mol NH+ 53.5 g NH Cl 
4
4
x g NH4Cl = 500 mL
= 2.68 g NH4Cl

+ 
 1000 mL  1 mol NH4 


0.56 mol NH3 1000 mL conc NH
3
x mL conc NH3 = 500 mL buffer


1000 mL buff 14.8 mol NH3 
=18.9 mL 14.8 M NH
3
Making a Buffer Calculations
Place ~250 mL of distilled water in a 500 mL volumetric flask.
Add 2.68 g NH4Cl and dissolve. Then mix in 18.9 mL of 14.8 M
NH3. Fill with distilled water to the 500 mL mark on the flask.
If there is no concentrated NH3 available, the NH3 can be
produced by neutralizing additional NH4Cl with 1.00 M NaOH.
0.56 mol NH3 1 mol NH4 Cl 58.5 g NH4 Cl 
x g NH4 Cl = 500 mL buffer



1000 mL buff 1 mol NH3 1 mol NH4 Cl 
=16.38 g NH4 Cl
0.56 mol NH3 1 mol NaOH1000 mL NaOH
x mL NaOH= 500 mL buffer



1000 mL buff 1 mol NH3 1.00 mol NaOH
= 280. mL NaOH
Dissolve 19.06 g NH4Cl (2.68 g + 16.38 g) in 280. mL of 1.00 M NaOH.
Then dilute with distilled water and fill to the 500 mL mark on the flask.
Making a Buffer Calculations
Place ~250 mL of distilled water in a 500 mL volumetric flask.
Add 2.68 g NH4Cl and dissolve. Then mix in 18.9 mL of 14.8 M
NH3. Fill with distilled water to the 500 mL mark on the flask.
If there is no NH4Cl available, the NH4+ can be produced by
neutralizing additional NH3 with 1.00 M HCl.
0.10 mol NH 1 mol NH 1000 mL NH
4
3
3
x mL NH3 = 500 mL buffer



 
 1000 mL buff1 mol NH4 14.8 mol NH3 
= 3.38 mL NH3
0.10 mol NH  1 mol HCl 1000 mL HCl
4
x mL HCl= 500 mL buffer



 
 1000 mL buff1 mol NH4 1.00 mol HCl
= 50.0 mL HCl
Place ~250 mL distilled water in volumetric flask. Add 50.0 mL of 1.00 M
HCl and mix. Then add 22.3 mL of concentrated NH3 (18.9 mL + 3.4 mL).
Mix and fill with distilled water to the 500 mL mark on the flask.
Making a Buffer Calculations
-11
K a2 = 5.6 x 10 =
-11



H +  CO 3

1-
HCO 3
CO 3
2


5.6 x 10
=
-10
110
HCO
 

3
2




CO 3
-11
2


5.6 x 10
=
+
1H
HCO
 
3


CO 3
2


0.56
=
11
HCO3


CO 3
2-


0.100
Multiply both [ ] by 0.179.....
=
0.179 HCO 31
Prepare 500. mL of the buffer that has [CO32] = 0.100 M and [HCO31-] = 0.179 M.

Making a Buffer Calculations
Prepare 500. mL of the buffer that has [CO32-] = 0.100 M
and [HCO31-] = 0.179 M.
0.100 mol CO2- 106 g Na CO 
3
2
3
x g Na2CO 3 = 500 mL

2- 
 1000 mL  1 mol CO3 
= 5.30 g Na2CO 3
0.179 mol HCO1- 84.0 g NaHCO 
3
3
x g NaHCO3 = 500 mL

1- 
1000 mL

 1 mol HCO3 
= 7.52 g NaHCO3
Place ~250 mL distilled water in a 500 mL volumetric flask.
Add 5.30 g Na2CO3 and 7.52 g NaHCO3 and dissolve. Fill
with distilled water to the 500 mL mark on the flask.
Acid-Base Titrations
Strong Acid-Base Titrations
• The plot of pH
versus volume
during a titration
is a titration curve.
Indicator examples
• Acid-base indicators are weak acids that
undergo a color change at a known pH.
pH
phenolphthalein
Indicator examples
Select the indicator that undergoes a color change closest
to the pH at the equivalence point, where all of the acid
has been neutralized by the base.
bromthymol blue
methyl red
Acid-Base Titrations
Strong Acid-Base Titrations
• Consider adding a strong base (e.g. NaOH) to a
solution of a strong acid (e.g. HCl).
– Before any base is added, the pH is given by the strong acid
solution. Therefore, pH < 7.
– When base is added, before the equivalence point, the pH is
given by the amount of strong acid in excess. Therefore, pH
< 7.
– At equivalence point, the amount of base added is
stoichiometrically equivalent to the amount of acid
originally present. Therefore, the pH is determined by the
salt solution. Therefore, pH = 7.
Acid-Base Titrations
Strong Acid-Base Titrations
• Consider adding a strong base (e.g. NaOH) to a
solution of a strong acid (e.g. HCl).
Acid-Base Titrations
Strong Acid-Base Titrations
• We know the pH at equivalent point is 7.00.
• To detect the equivalent point, we use an indicator
that changes color somewhere near 7.00.
– Usually, we use phenolphthalein that changes color between
pH 8.3 to 10.0.
– In acid, phenolphthalein is colorless.
– As NaOH is added, there is a slight pink color at the
addition point.
– When the flask is swirled and the reagents mixed, the pink
color disappears.
– At the end point, the solution is light pink.
– If more base is added, the solution turns darker pink.
Acid-Base Titrations
Strong Acid-Base Titrations
• The equivalence point in a titration is the point at
which the acid and base are present in stoichiometric
quantities.
• The end point in a titration is the observed point.
• The difference between equivalence point and end
point is called the titration error.
• The shape of a strong base-strong acid titration curve
is very similar to a strong acid-strong base titration
curve.
Acid-Base Titrations
Strong Acid-Base Titrations
Acid-Base Titrations
Strong Acid-Base Titrations
• Initially, the strong base is in excess, so the pH > 7.
• As acid is added, the pH decreases but is still greater
than 7.
• At equivalence point, the pH is given by the salt
solution (i.e. pH = 7).
• After equivalence point, the pH is given by the strong
acid in excess, so pH < 7.
Acid-Base Titrations
Weak Acid-Strong Base Titrations
• Consider the titration of acetic acid, HC2H3O2 and
NaOH.
• Before any base is added, the solution contains only
weak acid. Therefore, pH is given by the equilibrium
calculation.
• As strong base is added, the strong base consumes a
stoichiometric quantity of weak acid:
HC2H3O2(aq) + NaOH(aq)  C2H3O2-(aq) + H2O(l)
Acid-Base Titrations
Weak Acid-Strong Base Titrations
Acid-Base Titrations
Weak Acid-Strong Base Titrations
• There is an excess of acetic acid before the
equivalence point.
• Therefore, we have a mixture of weak acid and its
conjugate base.
– The pH is given by the buffer calculation.
• First the amount of C2H3O2- generated is calculated,
as well as the amount of HC2H3O2 consumed.
(Stoichiometry.)
• Then the pH is calculated using equilibrium
conditions. (Henderson-Hasselbalch.)
Acid-Base Titrations
Weak Acid-Strong Base Titrations
• At the equivalence point, all the acetic acid has been
consumed and all the NaOH has been consumed.
However, C2H3O2- has been generated.
– Therefore, the pH is given by the C2H3O2- solution.
– This means pH > 7.
• More importantly, pH  7 for a weak acid-strong base titration.
• After the equivalence point, the pH is given by the
strong base in excess.
Acid-Base Titrations
Weak Acid-Strong Base Titrations
• For a strong acid-strong base titration, the pH begins
at less than 7 and gradually increases as base is
added.
• Near the equivalence point, the pH increases
dramatically.
• For a weak acid-strong base titration, the initial pH
rise is more steep than the strong acid-strong base
case.
• However, then there is a leveling off due to buffer
effects.
Acid-Base Titrations
Weak Acid-Strong Base Titrations
• The inflection point is not as steep for a weak acidstrong base titration.
• The shape of the two curves after equivalence point is
the same because pH is determined by the strong base
in excess.
• Two features of titration curves are affected by the
strength of the acid:
– the amount of the initial rise in pH, and
– the length of the inflection point at equivalence.
Acid-Base Titrations
Weak Acid-Strong Base Titrations
• The weaker the acid,
the smaller the
equivalence point
inflection.
• For very weak acids, it
is impossible to detect
the equivalence point.
Acid-Base Titrations
Weak Acid-Strong Base Titrations
• Titration of weak bases with strong acids have similar
features to weak acid-strong base titrations.
Acid-Base Titrations
Titrations of Polyprotic Acids
• In polyprotic acids, each ionizable proton dissociates
in steps.
• Therefore, in a titration there are n equivalence points
corresponding to each ionizable proton.
• In the titration of Na2CO3 with HCl there are two
equivalence points:
– one for the formation of HCO3– one for the formation of H2CO3.
Acid-Base Titrations
Titrations of Polyprotic Acids
Acid-Base Titrations
Titrations of Polyprotic Acids
Ka = [H+] [X-]
[HX]
At the equivalence point, [H+] = [OH-]
At ½ the equivalence point, [X-] = [HX]
SOOOO…. At ½ equivalence point, Ka = [H+]
So pKa = pH
Solubility
• What happens when an ionic compound is
dissolved in water?
• What forces does water have to overcome?
• Are all ionic compounds soluble?
– Remember the solubility rules?
– It’s time to review them!
Solubility Rules
for Common Ionic Compounds in Water
1.
2.
3.
4.
5.
6.
Most nitrates are soluble.
Most salts containing alkali metal ions and ammonium are soluble
Li+1, Na+1, K+1, Cs+1, NH4+1
Most chloride, bromide, and iodide salts are soluble
Notable exceptions are Hg2+2, Ag+1, Pb+2
Most sulfate (SO4) salts are soluble
Notable exceptions are Ca +2, Sr+2, Ba+2, Pb+2, Hg2 +2
Most hydroxides are only slightly soluble. Na+ and K+are soluble.
Ba +2, Sr +2 and Ca +2 are marginally soluble
Most sulfide (S-2), carbonate (CO3-2), chromate (CrO4-2) and
phosphate (PO4-3) are only slightly soluble.
Properties of aqueous solutions
There are two general classes of solutes.
Electrolytic
ionic compounds in polar solvents
dissociate in solution to make ions
conduct electricity
may be strong (100% dissociation) or weak (less than
10%)
Nonelectrolytic
do not conduct electricity
solute is dispersed but does not dissociate
Heterogeneous Equilibria
A(s) + H2O  B(aq) + C (aq)
Ksp = [B] • [C]
Why is this not divided by [A]?
Why is [H2O] not included?
Known as the Solubility Product = Ksp
Solubility Products, KSP
KSP expressions are used for ionic materials that
are only slightly soluble in water.
Their only means of dissolving is by
dissociation.
AgCl(s)
Ag+ (aq) + Cl- (aq)
KSP = [ Ag+] [ Cl-]
Solubility Products, KSP
At equilibrium, the system is a saturated solution of
silver and chloride ions.
The only way to know that it is saturated it to
observe some AgCl at the bottom of the solution.
As such, [AgCl] is a constant and KSP expressions
do not include the solid form in the equilibrium
expression. The [H2O] for solvation process is
also excluded from the KSP expression.
Solubility Products, KSP
Determine the solubility of AgCl in water at 20 oC
in terms of grams / 100 mL
KSP = [Ag+] [Cl-] = 1.7 x 10-10
At equilibrium, [Ag+] = [Cl-] so
1.7 x 10-10 = [x] 2
[Ag+]
= 1.3 x 10-5 M
g AgCl
= 1.3 x 10-5 mol/L * 0.10 L * 143.32 g/mol
Solubility = 1.9 x 10-4 g / 100 mL
Calculating Ksp
Calculate the Ksp for Bismuth Sulfide which
has a solubility of 1.0 x 10-15 .
Bi2S3 → 2Bi+3 + 3S-2
Ksp = [Bi+3]2 • [S-2]3
For every Bi2S3 dissolved
2 Bi+3 and 3S-2 are formed
Ksp = [2 x 1.0 x 10-15 ]2 • [3 x 1.0 x 10-15 ]3
= 1.1 x 10-73
Calculating Solubility
Calculate
themolarity
concentration
of ions
What is the
of a saturated
solution
of (II)
Cu(IO
for copper
iodate.
3)2?
ksp = 1.4 x 10-7 at 25 °C
Cu(IO3)2(s)  Cu+2 + 2 IO3-1
1.4 x 10-7 = [x] • [ 2x]2 = 4x3
x = 3.3 x 10-3 mol/L = [Cu+2]
-1]
3.3
= [Cu(IO
)
]
2x x=10
6.6-3 xmol/L
10-3 mol/L
= [IO
3 23
Solubility Products, KSP
Another example
Calculate the silver ion concentration when
excess silver chromate is added to a 0.010
M sodium chromate solution.
KSP Ag2CrO4 = 1.1 x 10-12
Ag2CrO4 (s)
2Ag+ + CrO42-
Solubility Products, KSP
KSP = 1.1 x 10-12 = [ Ag+ ]2 [ CrO42- ]
[CrO42-] = [CrO42-]Ag2CrO4 + [CrO42-]Na2CrO4
With such a small value for KSP, we can assume
that the [CrO42-] from Ag2CrO4 is negligible.
If we’re wrong, our silver concentration will be
significant (>1% of the chromate concentration.)
Then you’d use the quadratic approach.
Solubility Products, KSP
KSP = 1.1 x 10-12 = [ Ag+ ]2 [ CrO42- ]
[CrO42-]
[ Ag+ ]
= 0.010 M
= ( KSP / [ CrO42- ] ) 1/2
= (1.1 x 10-12 / 0.010 M ) 1/2
= 1.1 x 10 -5 M
[ Ag+ ] << [ CrO42- ]
so our assumption was valid.
Relative Solubility
• Be careful when comparing solubility
products (Ksp)
• Compare
– AgI Ksp = 1.5 x 10-16
– CuI Ksp = 5.0 x 10-12
• Which is more soluble?
Copper (I) iodide is more soluble. Each compound
produces the same number ions
Relative Solubility
• CuS
• Ag2S
• Bi2S3
Ksp = 8.5 x 10-45
Ksp = 1.6 x 10-49
Ksp = 1.1 x 10-73
Which is more soluble?
Each has a different number of ions,
so calculate the solubility of each.
Estimate (guess)
three groups and calculate the solubility
Factors influencing solubility
Common ion and salt effects.
As with other equilibria we’ve discussed, adding
a ‘common’ ion will result in a shift of a
solubility equilibrium.
AgCl (s)
Ag+ (aq) + Cl- (aq)
KSP = [Ag+] [Cl-]
Adding either Ag+ or Cl- to our equilibrium
system will result in driving it to the left.
Factors That Affect Solubility
Common-Ion Effect
Factors influencing solubility
Complex ion formation.
The solubility of slightly soluble salts can be
increased by complex ion formation.
Example. Addition of excess ClAgCl(s)
Ag+(aq) + Cl-(aq)
A large excess of
+
chloride results in
the formation of the
2Cl-(aq)
complex.
AgCl2-(aq)
More AgCl will
dissolve as a result.
Factors That Affect Solubility
Formation of Complex Ions
Factors That Affect Solubility
Formation of Complex Ions
• Consider the addition of ammonia to AgCl (white
precipitate):
AgCl(s)
+
Ag+(aq) + Cl-(aq)
Ag (aq) + 2NH3(aq)
• The overall reaction is
AgCl(s) + 2NH3(aq)
+
Ag(NH3)2(aq)
+
Ag(NH3)2(aq) + Cl-(aq)
• Effectively, the Ag+(aq) has been removed from solution.
• By Le Châtelier’s principle, the forward reaction (the
dissolving of AgCl) is favored.
• Knet = Ksp•Kf = (1.8 x 10-10)(1.7 x 107) = 3.1 x 10-3 at 25ºC
Factors That Affect Solubility
Amphoterism
• Amphoteric oxides will dissolve in either a strong acid
or a strong base.
• Examples: hydroxides and oxides of Al3+, Cr3+, Zn2+,
and Sn2+.
• The hydroxides generally form complex ions with four
hydroxide ligands attached to the metal:
Al(OH3)(s) + OH-(aq)
Al(OH)4-(aq)
• Hydrated metal ions act as weak acids. Thus, the
amphoterism is interrupted:
Factors That Affect Solubility
Amphoterism
• Hydrated metal ions act as weak acids. Thus, the
amphoterism is interrupted:
Al(H2O)63+(aq) + OH-(aq)
Al(H2O)5(OH)2+(aq) + OH-(aq)
Al(H2O)4(OH)+(aq) + OH-(aq)
Al(H2O)3(OH)3(s) + OH-(aq)
Al(H2O)5(OH)2+(aq) + H2O(l)
Al(H2O)4(OH)2+(aq) + H2O(l)
Al(H2O)3(OH)3(s) + H2O(l)
Al(H2O)2(OH)4-(aq) + H2O(l)
Factors influencing solubility
Hydrolysis.
If the anion of a weak acid, or cation of a weak
base is part of a KSP, solubilities are greater than
expected.
This competing
+
AgCN(s)
Ag (aq) + CN (aq)
equilibrium causes
the CN- to be lower
+
than expected.
H2O(l)
More AgCN will
dissolve as a result.
HCN(aq) + OH-(aq)
Factors That Affect Solubility
Solubility and pH
• Again we apply Le Châtelier’s principle:
CaF2(s)
Ca2+(aq) + 2F-(aq)
– If the F- is removed, then the equilibrium shifts towards the
decrease and CaF2 dissolves.
– F- can be removed by adding a strong acid:
+
F (aq) + H (aq)
HF(aq)
– As pH decreases, [H+] increases and solubility increases.
• The effect of pH on solubility is dramatic.
Factors That Affect Solubility
Solubility and pH
Solubility Equilibria
Solubility-Product Constant, Ksp
Consider
BaSO4(s)
Ba2+(aq) + SO42-(aq)
for which
2
2
K sp  [Ba ][SO4 ]
Ksp is the solubility product. (BaSO4 is ignored because
it is a pure solid so its concentration is constant.)
Solubility Equilibria
Solubility-Product Constant, Ksp
• In general: the solubility product is the molar
concentration of ions raised to their stoichiometric
powers.
• Solubility is the amount (grams) of substance that
dissolves to form a saturated solution.
• Molar solubility is the number of moles of solute
dissolving to form a liter of saturated solution.
Solubility Equilibria
Solubility and Ksp
To convert solubility to Ksp
• solubility needs to be converted into molar solubility
(via molar mass);
• molar solubility is converted into the molar
concentration of ions at equilibrium (equilibrium
calculation),
• Ksp is the product of equilibrium concentration of
ions.
Solubility Equilibria
Solubility and Ksp
Precipitation Reactions
• We will use “Q” and Ksp
• If Q is bigger than Ksp, what will happen?
– Smaller
• The Q in a solubilty problem is called the
Ion Product
A solution prepared by adding 750 mL of 4.00 x 103 M Ce(NO ) to 300.0 mL of 2.00 x 10-2 M KIO .
3 3
3
Will Ce(IO3)3 precipitate? (Ksp = 1.9 x 10-10)
• Step one: Calculate the concentration of Ce+3 and IO3-1
before a reaction occurs
[Ce+3]0 = (750.0 mL)(4.00x10-3 M) = 2.86 x 10-3 M
(750. + 300.mL)
[IO3-1]0 = (300.0 mL)(2.00x10-3 M) = 5.71 x 10-3 M
(750. + 300.mL)
Q = (2.86 x 10-3 M) x (5.71 x 10-3 M)3 = 5.32 x 10-10
Q > Ksp therefore the precipitation reaction will occur
Precipitation and Separation of Ions
BaSO4(s)
Ba2+(aq) + SO42-(aq)
• At any instant in time, Q = [Ba2+][SO42-].
– If Q < Ksp, precipitation occurs until Q = Ksp.
– If Q = Ksp, equilibrium exists.
– If Q > Ksp, solid dissolves until Q = Ksp.
• Based on solubilities, ions can be selectively
removed from solutions.
• Consider a mixture of Zn2+(aq) and Cu2+(aq). CuS
(Ksp= 610-37) is less soluble than ZnS(Ksp=210-25),
CuS will be removed from solution before ZnS.
Precipitation and Separation of Ions
• As H2S is added to the green solution, black CuS
forms in a colorless solution of Zn2+(aq).
• When more H2S is added, a second precipitate of
white ZnS forms.
Selective Precipitation of Ions
• Ions can be separated from each other based on their
salt solubilities.
• Example: if HCl is added to a solution containing Ag+
and Cu2+, the silver precipitates(Ksp for AgCl is
1.810-10) while the Cu2+ remains in solution, since
CuCl2.
• Removal of one metal ion from a solution is called
selective precipitation.
Qualitative Analysis for Metallic Elements
• Qualitative analysis is
designed to detect the
presence of metal ions.
• Quantitative analysis is
designed to determine
how much metal ion is
present.
Qualitative Analysis for Metallic Elements
• We can separate a complicated mixture of ions into
five groups:
– Add 6 M HCl to precipitate insoluble chlorides (AgCl,
Hg2Cl2, and PbCl2).
– To the remaining mix of cations, add H2S in 0.2 M HCl to
remove acid insoluble sulfides (e.g. CuS, Bi2S3, CdS, PbS,
HgS, etc.).
– To the remaining mix, add (NH4)2S at pH 8 to remove base
insoluble sulfides and hydroxides (e.g. Al(OH)3, Fe(OH)3,
ZnS, NiS, CoS, etc.).
– To the remaining mixture add (NH4)2HPO4 to remove
insoluble phosphates (Ba3(PO4)2, Ca3(PO4)2, MgNH4PO4).
– The final mixture contains alkali metal ions and NH4+.