Updates
• Assignment 06 is due today (in class)
• Midterm 2 is THIS Thurs., March 15 and will
cover Chapters 16 & 17
– Huggins 10, 7-8pm
– For conflicts: ELL 221, 6-7pm (must arrange at
least one week in advance)
Acid-Base Equilibria and
Solubility Equilibria
Chapter 17
Precipitation and separation of ions
• Predicting what precipitate might form from a
mixture of ions (solubility rules, pg. 97)
• Using quantitative means to decide whether a
precipitate will form
• Predicting selective precipitation of an ion
from a mixture of ions
Will a Precipitate Form?
Remember Q is the reaction quotient, which is obtained by
substituting the initial concentrations into the equilibrium
expression.
• In a solution,
– If Q = Ksp, the system is at equilibrium and the
solution is saturated.
– If Q < Ksp, more solid will dissolve until Q = Ksp.
– If Q > Ksp, the salt will precipitate until Q = Ksp.
If 2.00 mL of 0.200 M NaOH are added to 1.00 L of
0.100 M CaCl2, will a precipitate form?
1. Note what ions present in solution: Na+, OH-, Ca2+, Cl-.
2. Note that the only possible precipitate is Ca(OH)2 (solubility
rules).
3. Is Q > Ksp for Ca(OH)2?
[Ca2+]0 = 0.100 M
[OH-]0 = 4.0 x 10-4 M
Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8
Ksp = [Ca2+][OH-]2 = 8.0 x 10-6
Q < Ksp
No precipitate will form
17.6
What concentration of Ag is required to precipitate ONLY
AgBr in a solution that contains both Br- and Cl- at a
concentration of 0.02 M?
1. Use Ksp for AgBr to figure out the solubility of Ag+ when [Br] is 0.02 M
2. Use Ksp for AgCl to figure out the solubility of Ag+ when [Cl-]
is 0.02 M
3. Is it possible to choose a concentration between the two
solubility values where only AgBr precipitates? This will
only work if the Ksp for AgCl is greater than that for AgBr
since the concentrations of the counterions are identical in
the problem.
17.7
What concentration of Ag is required to precipitate ONLY
AgBr in a solution that contains both Br- and Cl- at a
concentration of 0.02 M?
AgBr (s)
Ag+ (aq) + Br- (aq)
Ksp = 7.7 x 10-13
AgCl (s)
Ag+ (aq) + Cl- (aq)
Ksp = 1.6 x 10-10
AgBr is less soluble than AgCl so a selective precipitation should be possible
-13
K
7.7
x
10
sp
-11 M
=
=
3.9
x
10
[Ag+] =
0.020
[Br-]
[Ag+]
Ksp
=
[Cl-]
1.6 x 10-10
=
= 8.0 x 10-9 M
0.020
When [Ag+] is greater than 3.9 x 10-11 M, AgBr will precipitate; when [Ag+] is
greater than 8.0 x 10-9, AgCl will precipitate.
3.9 x 10-11 M < [Ag+] < 8.0 x 10-9 M
17.7
Factors that affect solubility
• We have considered the solubility of ionic
compounds in pure water; we noted that
temperature and ionic strength has an effect,
but we did not discuss these influences
further
• We will now examine three factors that affect
the solubility of ionic compounds in water
– Presence of common ions
– pH of solution
– Presence of complexing agents
The common ion effect is the shift in equilibrium caused by the
addition of a compound having an ion in common with the
dissolved substance.
The presence of a common ion suppresses
the ionization of a weak acid or a weak base.
Consider mixture of CH3COONa (strong electrolyte) and
CH3COOH (weak acid).
CH3COONa (s)
Na+ (aq) + CH3COO- (aq)
CH3COOH (aq)
H+ (aq) + CH3COO- (aq)
common
ion
17.2
What is the pH of a solution containing 0.30 M HCOOH
and 0.52 M HCOOK?
Mixture of weak acid and conjugate base!
HCOOH (aq)
Initial (M)
Change (M)
Equilibrium (M)
Common ion effect
0.30 – x  0.30
0.52 + x  0.52
H+ (aq) + HCOO- (aq)
0.30
0.00
0.52
-x
+x
+x
0.30 - x
x
0.52 + x
[HCOO-]
pH = pKa + log
[HCOOH]
[0.52]
= 4.01
pH = 3.77 + log
[0.30]
HCOOH pKa = 3.77
17.2
Common ions affect solubility
• CaF2
Ca2+ + F-
• The presence of either Ca2+
or F- (from another source)
reduces the solubility of
CaF2, shifting the solubility
equilibrium of CaF2 to the left
The Common Ion Effect and Solubility
The presence of a common ion decreases
the solubility of the salt.
What is the molar solubility of AgBr in (a) pure water
and (b) 0.0010 M NaBr?
AgBr (s)
Ag+ (aq) + Br- (aq)
Ksp = 7.7 x 10-13
The solubility of Ag and Br ions equals
1/2 only when AgBr is the only source of
(K
)
2
sp
s = Ksp
Ag and Br ions (a)!
s = 8.8 x 10-7
17.8
The Common Ion Effect and Solubility
The presence of a common ion decreases
the solubility of the salt.
What is the molar solubility of AgBr in (a) pure water
and (b) 0.0010 M NaBr?
Na+ (aq) + Br- (aq)
AgBr (s)
Ag+ (aq) + Br- (aq) NaBr (s)
[Br-] = 0.0010 M
Ksp = 7.7 x 10-13
AgBr (s)
Ag+ (aq) + Br- (aq)
s2 = Ksp
[Ag+] = s
s = 8.8 x 10-7
[Br-] = 0.0010 + s  0.0010
Ksp = 0.0010 x s
s = 7.7 x 10-10
17.8
pH and Solubility
•
•
•
The presence of a common ion decreases the solubility.
Insoluble bases dissolve in acidic solutions
Insoluble acids dissolve in basic solutions
remove
add
Mg(OH)2 (s)
Mg2+ (aq) + 2OH- (aq)
Ksp = [Mg2+][OH-]2 = 1.2 x 10-11
Ksp = (s)(2s)2 = 4s3
4s3 = 1.2 x 10-11
s = 1.4 x 10-4 M
[OH-] = 2s = 2.8 x 10-4 M
pOH = 3.55 pH = 10.45
17.9
At pH less than 10.45
Lower [OH-]
OH- (aq) + H+ (aq)
H2O (l)
Increase solubility of Mg(OH)2
At pH greater than 10.45
Raise [OH-]
Decrease solubility of Mg(OH)2
Complex Ions Affect Solubility
• Complex Ions
– The formation
of these
complex ions
increases the
solubility of
these salts.
Complex Ion Equilibria and Solubility
A complex ion is an ion containing a central metal cation
bonded to one or more molecules or ions.
CoCl42- (aq)
Co2+ (aq) + 4Cl- (aq)
The formation constant or stability constant (Kf) is the
equilibrium constant for the complex ion formation.
Co(H2O)2+
6
CoCl24
Kf =
[CoCl42- ]
[Co2+][Cl-]4
Kf
stability of
complex
17.10
17.10
17.11
Selective Precipitation of Ions
• Common cations can be
divided into five groups
• Insoluble chlorides
• Acid-insoluble sulfides
• Base-insoluble sulfides and
hydroxides
• Insoluble phosphates
• Alkali metal ions and NH4+
remain in solution; each ion
can be tested for
individually using a flame
test
Qualitative
Analysis of
Cations
17.11
Flame Test for Cations
lithium
sodium
potassium
copper
17.11
Chemistry In Action: How an Eggshell is Formed
Ca2+ (aq) + CO32- (aq)
CaCO3 (s)
carbonic
CO2 (g) + H2O (l)
H2CO3 (aq)
anhydrase
H2CO3 (aq)
H+ (aq) + HCO3- (aq)
HCO3- (aq)
H+ (aq) + CO32- (aq)
…explain why “as the polarity of H-X
bonding increases, the acid strength
increases”? …especially the section
involving oxoacids
Binary acids (HX, H2X, H3X, H4X)
•
Bond strength determines acidity within the same group (column), size
•
Bond polarity determines acidity within the same period (row), electronegativity
Oxyacids
Central atoms derived from same group (same oxidation state)
•
More electronegative central atom polarizes the OH bond more,
facilitating ionization (effect is weakening the O-H bond)
•
More electronegative central atom better able to stablize resulting
negative charge following ionization, making a happier (more stable)
conjugate base
16.41) Calculate the concentrations of all
the species (HCN, H+, CN- and OH-) in
a 0.15 M HCN solution.
2
[H  ][CN  ]
x
Ka 
 4.9 x10 10 
HCN
0.15
x  8.6 x10 6 M  [H  ]  [CN  ]
If [H+] is 8.6 x 10-6, then [OH-]:
1x10 14
9

1
.
2
x
10
M
6
8.6 x10
If [H+] = [CN-] = 8.6 x 10-6, then we have lost this amount
of HCN, so:
[HCN] = 0.15 – (8.6 x 10-6) = 0.15 M
16.97) Henry’s law constant for CO2 at 38oC is
2.28 x 10-3 mol/L.atm. Calculate the pH of a
solution of CO2 at 38oC in equilibrium with the
gas at a partial pressure of 3.20 atm.
Remember that Henry’s law describes the effect of pressure
on the solubility of gases. The solubility of CO2 can be
calculated from Henry’s law:
2.28 x 10-3 mol/L.atm x 3.20 atm = 7.30 x 10-3 mol/L.
Remember that CO2 dissolves in water to form H2CO3.
Therefore, the pH will depend on the extent of ionization of
H2CO3, which can be found from Ka (4.2 x 10-7):
4.2 x 10-7 = x2/(7.30 x 10-3 M) = 5.54 x 10-5 M;
pH = -log x, pH = 4.26.
17.47) insert