Chapter 15 - Applications of Aqueous
Equilibrium
AP Chemistry
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Common Ion Effect
COMMON ION EFFECT „ Calculate [H+] for a 1.0 M HF solution (Ka = 7.2 x
10-4). Then calculate % ionization for this solution.
If a solution of NaF is added to the HF solution,
which way will the system shift? _____ What
____
would happen to
[H+]?
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Common Ion Effect
„
Calculate [H+] and percent dissociation of HF in a
solution containing 1.0 M HF and 1.5 M NaF.
Q - What happened to the [H+] when a common ion
was added? ____
Q - What happened to the % dissociation? _____
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Buffers
Buffer solutions are solutions which are resistant to changes in
pH when either an acid or base is added.
„ These are made by having both a weak acid and its conjugate
base present.
Ex. 1) A buffered soln contains 0.40 M acetic acid (Ka = 1.8 x 10-5)
and 0.80 M sodium acetate. Calculate the pH of this soln.
„
1.
2.
3.
Write an ICE chart for the ionization of acetic acid (using BOTH initial concentrations).
Write Ka expression and plug in concentrations in terms of x.
Solve for X = [H+].
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Buffers
Additional base (OH-) will be neutralized by
the weak acid (HA).
„ Additional Acid (H+) will be neutralized by the
conjugate base (A-).
„
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Buffers
Write net ionic eqns for when the following solns are mixed.
1)
Hydrochloric acid is added to a buffer made with sodium
cyanide and cyanic acid, HCN.
2)
Sodium hydroxide is added to a buffer made with sodium
cyanide and cyanic acid.
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Buffers
3)
Hydrochloric acid is added to a buffer made with
methylamine and the salt of its conjugate acid, CH3NH3Cl.
4) Sodium hydroxide is added to a buffer made with sodium
methylamine and the salt of its conjugate acid, CH3NH3Cl.
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STEPS TO SOLVE FOR pH WHEN
ADDING STRONG ACID/BASE
Addition of Strong Acid to Buffer
1) Write an ICE chart for the H+ of
the strong acid reacting with the
weak base. (Be sure to plug in the
initial concentrations for BOTH the
weak acid and the weak base.)
Addition of Strong Base to Buffer
1) Write an ICE chart for the OH- of
the strong acid reacting with the
weak acid. (Be sure to plug in the
initial concentrations for BOTH the
weak acid and the weak base.)
2) Write an ICE Chart for the dissociation for the
weak acid and use Ka to solve for [H+] to calculate
pH.
OR
Write an ICE Chart for the dissociation for the
weak base and use Kb to solve for [OH-] to
calculate pH.
(Be sure to plug in the initial concentrations for
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the weak
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REVIEW - Dissociation
Equations
Weak Acid
Dissociation
Corresponding Weak
Base Dissociation
BH+ Ù H+ + B
OR
HA Ù A- + H+
B + H2OÙ BH+ + OHOR
A- + H2O Ù HA + OH-
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Buffer Example Problems
Ex.) A buffer is made by mixing 0.125 moles NaCH3COO in
0.250 L of 0.500 M CH3COOH. (Ka CH3COOH = 1.8 x 10-5)
A) Use net ionic equations to show how this solution resists
pH change when either HCl or NaOH is added.
Addition of HCl
Addition of NaOH
B) Find [H+] and pH for the above buffer.
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EFFECTS OF ADDITION OF
ACID OR BASE:
(A)
Find [H+] & pH for the above buffer soln after
0.010 mol HCl is added.
HENDERSON-HASSELBACH EQUATION: (next slide)
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EFFECTS OF ADDITION OF
ACID OR BASE:
HENDERSON-HASSELBACH EQUATION:
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EFFECTS OF ADDITION OF
ACID OR BASE:
(B) Find [H+] & pH for the original buffer soln
after 0.050 mol NaOH is added.
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EFFECTS OF ADDITION OF
WATER:
Ex.) Find [H+] & pH for the above buffer solution
after 250 mL H2O is added.
NOTE: dilution of a buffer changes/does NOT change the pH.
This is different thanJodidiluting
anHigh
acid
or base.
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Buffer Summary
Summary of
Buffer Effects
starting buffer
after 2x dilution
after 0.010 mol H+
after 0.050 mol OH-
pH
4.74
4.74
4.73
5.26
pH
If pure water
7.00
after 0.010 mol H+ 1.40
after 0.050 mol OH- 13.30
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How a Buffer Works
Ka = [H+][A-]
[HA]
0.50 = 1
0.50
=
[H+] =
Ka
[HA] =
[A-]
[HA] =
[A-]
The change in the ratio [HA]/[A-] is very small,
thus the [H+] and pH remain essentially
constant.
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Buffer Capacity
Buffers have the greatest capacity when ratio
of acid/conj. base is: _____
Another way to look at it:
pH = pKa + log [A-]
[HA]
„ Usual range for [A-] ratio is 10 --> 1
[HA]
1
10
Thus pH of buffer ±
pH unit(s) from pKa
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Buffer Capacity
Q – What’s a better buffer?
____ 1.
a. 0.20 M NaCH3COO OR b. 0.30 M NaCH3COO
0.40 M CH3COOH
0.30 M CH3COOH
____ 2.
a. 0.50 M NaCH3COO OR
0.50 M CH3COOH
b. 0.10 M NaCH3COO
0.10 M CH3COOH
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Preparing a Buffer Solution
(Choosing solutions)
Acid
Ka
HNO2
4.0 x 10-4
HC3H5O2 1.3 x 10-5
HCN
6.2 x 10-10
pKa
3.40
4.89
9.21
Henderson-Hasselbach is a useful tool!
pH = pKa + log [A-]
[HA]
NOTICE: If the ratio of [A-]/[HA] = 1, then pH = ___
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Preparing a Buffer Solution
(Choosing solutions)
Acid
Ka
HNO2
4.0 x 10-4
HC3H5O2 1.3 x 10-5
HCN
6.2 x 10-10
pH of the
Buffer
needed
3.40
3.75
9.21
9.10
What soln’s
could you
use?
pKa
3.40
4.89
9.21
What if you
only have the
weak acid?
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What if you
have a 50:50
mixture of
WA/WB?
Preparing a Buffer Solution
(Choosing solutions)
SUMMARY:
pH of Buffer Needed
pH= pKa
Amts & Soln Used
pH < pKa
pH > pKa
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Preparing a Buffer Solution
(Choosing solutions)
NOTE: Buffers can also be done with a weak
base and its CA.
Base
Kb
pKb
NH3
1.8 x 10-5 4.75
C2H5NH2 4.38 x 10-4 3.36
Acid
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pKa
Titration of Strong Acid with
Strong Base
Reaction:
H+(aq) + OH- (aq) Æ H2O (aq)
Krxn =
EXAMPLE - 50.00 mL of 1.000 M HCl titrated
w/ 1.000 M NaOH.
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EX - 50.00 mL of 1.000 M HCl
titrated w/ 1.000 M NaOH.
Important Concepts:
1) At the Start, before
Titration
Calculations:
1) Find pH at Start,
before Titration
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EX - 50.00 mL of 1.000 M HCl
titrated w/ 1.000 M NaOH.
2) On the way to the
equivalence point
2) Calculate the pH one
drop before equiv pt.
(1 drop ~ 0.05 mL)
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EX - 50.00 mL of 1.000 M HCl
titrated w/ 1.000 M NaOH.
3) At the equivalence
point
3) At the equivalence pt.
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EX - 50.00 mL of 1.000 M HCl
titrated w/ 1.000 M NaOH.
4) Past the equivalence
point
4) Calculate the pH one
drop after equivalence
pt. (1 drop ~ 0.05 mL)
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Strong Acid/Base Summary
NOTE: 2 drops causes a [H+] change of 5 x 10-4 M
Æ 2 x 10-11 M
When mixing strong acids and bases, the net
ionic reaction is always:
Write an ice chart for H+ + OH- Æ H2O.
„ The ion is present in excess determines the pH.
„ If there is no excess, the pH = ____.
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Titration of a Weak AcidStrong Base
Reaction:
HB + OH- Æ B- + H2O
Thus Krxn =
Example - 50.00 mL of 1.000 M Formic Acid,
HCOOH, is titrated with 1.00 M NaOH. (Ka
HCOOH = 1.9 x 10-4)
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50.00 mL of 1.000 M Formic Acid,
HCOOH, is titrated with 1.00 M NaOH.
(Ka HCOOH = 1.9 x 10-4)
Important Concepts:
1) At the Start, before
Titration
Calculations:
1) Find pH at Start,
before Titration
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50.00 mL of 1.000 M Formic Acid,
HCOOH, is titrated with 1.00 M NaOH.
(Ka HCOOH = 1.9 x 10-4)
2) On the way to the
equivalence point
2) a. Calculate pH
when15.00 mL 1.000 M
NaOH added.
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50.00 mL of 1.000 M Formic Acid,
HCOOH, is titrated with 1.00 M NaOH.
(Ka HCOOH = 1.9 x 10-4)
special: at the halfequivalence point
2) b. Calculate pH when
25.00 mL 1.000 M
NaOH added.
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50.00 mL of 1.000 M Formic Acid,
HCOOH, is titrated with 1.00 M NaOH.
(Ka HCOOH = 1.9 x 10-4)
3) At the equivalence
point
3) Calculate pH when
50.00 mL 1.000 M
NaOH added.
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50.00 mL of 1.000 M Formic Acid,
HCOOH, is titrated with 1.00 M NaOH.
(Ka HCOOH = 1.9 x 10-4)
4) Past the equivalence
point
4) Calculate the pH when
60.00 mL 1.000 M
NaOH added.
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Titration of a Weak BaseStrong Acid
Reaction: B- + H+ Æ HB
OR
B + H+ --> HB+
Thus Krxn =
Example - 50.00 mL of 1.000 M Methylamine,
CH3NH2, is titrated with 1.00 M HCl. (Ka
CH3NH3+ = 6.2 x 10-10)
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Ex- 50.00 mL of 1.000 M Methylamine,
CH3NH2, is titrated with 1.00 M HCl.
(Ka CH3NH3+ = 6.2 x 10-10)
Important Concepts:
1) At the Start, before
Titration
Calculations:
1) Find pH at Start,
before Titration
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Ex- 50.00 mL of 1.000 M Methylamine,
CH3NH2, is titrated with 1.00 M HCl.
(Ka CH3NH3+ = 6.2 x 10-10)
2) On the way to the
equivalence point
2) Calculate pH when
20.00 mL 1.00 M HCl
added
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Ex- 50.00 mL of 1.000 M Methylamine,
CH3NH2, is titrated with 1.00 M HCl.
(Ka CH3NH3+ = 6.2 x 10-10)
special: at the halfequivalence point
2) b. Calculate pH when
25.00 mL 1.00 M HCl
added.
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Ex- 50.00 mL of 1.000 M Methylamine,
CH3NH2, is titrated with 1.00 M HCl.
(Ka CH3NH3+ = 6.2 x 10-10)
3) At the equivalence
point
3) Calculate pH when
50.00 mL 1.00 M HCl
added.
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Ex- 50.00 mL of 1.000 M Methylamine,
CH3NH2, is titrated with 1.00 M HCl.
(Ka CH3NH3+ = 6.2 x 10-10)
4) Past the equivalence
point
4) Calculate the pH when
55.00 mL 1.00 M HCl
added.
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Titration Summary
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Indicators
Indicators - are typically weak acids where
HB and B- are different colors.
Ka =
HB Ù H+ + BIf Add H+ shifts _____; color = _______
If Add OH- shifts _____; color = _______
If the intermediate color is present [HB] [B-]
[H+] = ___
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Indicators
equivalence point –
end point –
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Indicators
Ex: Methyl Red
HMr Æ H+ + Mr- Ka ~ 1 x 10-5
Ka = [H+][Mr-] => Δ’s color @ pH = 5.00
[HMr]
„ Calculate the ratio [Mr-]/[HMr] at pH = 6.00
„
Calculate the ratio [Mr-]/[HMr] at pH = 4.00
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Strong Acid/Strong Base 1
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Strong Acid/Strong Base 2
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Strong Acid/Strong Base 3
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Weak Acid/Strong Base 2
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Weak Acid/Strong Base 3
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Weak Acid/Strong Base 3
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Weak Acid Curves
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Indicators
Ex: Methyl Red
HMr Æ H+ + Mr- Ka ~ 1 x 10-5
Ka = [H+][Mr-] => Δ’s color @ pH = 5.00
[HMr]
„ Calculate the ratio [Mr-]/[HMr] at pH = 6.00
„
Calculate the ratio [Mr-]/[HMr] at pH = 4.00
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Indicators
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Most indicators
change color over a 2
pH range.
„ The intermediate
color is used as the
end point.
„ In titration we
select an indicator
that changes color
very near the
equivalence point of
the reaction.
„
Solubility Equilibrium, Ksp
{Solubility Product}
Soluble –
Solubility –
„ A Saturated solution exists at solubility
equilibrium given the following dissolving eqn.
Ag2CrO4 (s) Ù 2 Ag+ (aq) + CrO42- (aq)
„ Write the equilibrium expression {called Ksp}
Ksp =
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Solubility Equilibrium, Ksp
{Solubility Product}
Ag2CrO4 (s) Ù 2 Ag+ (aq) + CrO42- (aq)
Ksp = [Ag+]2 [CrO42-]
Given - When equilibrium is established 0.208 g of Ag2CrO4
dissolves in 10.0 L of water at 25°C
Find - the value for Ksp of Ag2CrO4 (330 g/mol)
• Any time equilibrium exists the value of the Ksp
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Solubility Equilibrium, Ksp
{Solubility Product}
Ex. A precipitate of Ag2CrO4 has formed after mixing
solutions of AgNO3 and K2CrO4
(A) Find [CrO42-] if K2CrO4 is added to a solution in
which [Ag+] = 3 x 10-3 M
(B)
Find [Ag+] if Ag2CrO4 is added to a solution in
which [CrO42-] = 4 x 10-2
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Solubility Equilibrium, Ksp
{Solubility Product}
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Ksp and Water solubility
Ksp and Water solubility are related but NOT
equal to each other.
„ Find solubility in distilled water for:
„
(A) AgCl Ksp = 1.8 x 10-10
AgCl (s) Ù Ag+ + ClKsp =
(B) Ag2CrO4 Ksp = 1 x 10-12
Ag2CrO4 (s) Ù 2 Ag+ + CrO42Ksp =
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Ksp and Water solubility
Which compound in each column is the most
soluble?
1) BaF2 Ksp = 2.4 x 10-5
MgF2 Ksp = 6.4 x 10-9
PbF2 Ksp = 4 x 10-8
SrF2
Ksp = 7.9 x 10-10
„
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Ksp and Water solubility
Which compound in each column is the most
soluble?
2) AgBr
Ksp = 5.0 x 10-13
AgCl
Ksp = 1.6 x 10-16
SrSO4
Ksp = 3 x 10-7
ZnS
Ksp = 2.5 x 10-22
„
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Ksp and Water solubility
Which compound in each column is the most
soluble?
Ksp = 1.4 x 10-8
3) PbI2
CaF2
Ksp = 4.0 x 10-11
CaSO4 Ksp = 6.1 x 10-5
Ag2SO4 Ksp = 1.2 x 10-5
„
Consider both value AND unit power of Ksp.
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Solubility & Common Ions
COMMON ION EFFECT:
Find solubility of AgCl in 0.020 M NaCl.
(Here Cl- is in both AgCl and NaCl and called a
common ion.)
An ionic solid is always more/less soluble in a
solution containing a common ion than it is in pure
water.
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Formation of Precipitates
„
„
„
„
A precipitate will form if you try to dissolve more
than what will dissolve.
The Ksp value relates to
The value of the trial ion product (Q)
When Q > Ksp
When Q < Ksp
When Q = Ksp
The trial ion product (Q) is the same math
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Formation of Precipitates
PROBLEM: If 10.0 mL of 5 x 10-4 M AgNO3 is mixed with 40.0
mL of 1 x 10-4 M K2CrO4 will a ppt of Ag2CrO4 form? ( Ksp
= 1x 10-12)
1. Find possible ion concentration
2. Substitute & find value of Q using expression used for Ksp
3. Then make your decision
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Calculating Equilibrium
Concentrations After
Precipitation has occurred:
Ex. Calculate the concentration of Pb2+ and I- in solution after 100.0 mL of
0.0500 M Pb(NO3)2 and 200.0 mL of 0.100 M NaI have been mixed.
1) Calculate Trial Ion Product:
2) If the equilibrium goes virtually to completion, do stoichiometric
calculations before doing the equilibrium calculations.
• In other words, let the system go completely in the direction toward
which it tends (formation of a precipitate), then adjust back to
equilibrium.
Ksp = 1.4 x 10-8
Pb2+ (aq) + 2 I- (aq) --> PbI2 (s)
Krxn =
PbI2 (s) <==> Pb2+ (aq) + 2 I- (aq)
Ksp = 1.4 x 10-8
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Selective Precipitation
„
Selective Precipitation: a method of separating metal ions
from an aqueous mixture by using a reagent show anion
forms a precipitate with only one or a few of the ions in the
mixture.
„
Ex. A solution contains 1.0 x 10-4 M Cu+ and 2.0 x 10-3 M Pb2+. If a
source of I- is gradually added to this solution, will PbI2 (Ksp = 1.4 x 10-8)
or CuI (Ksp = 5.3 x 10-12) precipitate first? Specify the concentration of Inecessary to begin precipitation of each salt.
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Selective Precipitation and
Qualitative Analysis
„
Since metal sulfides differ greatly in their solubilities, the
sulfide ion is often used to separate solutions of metal ions.
Important feature: Solubility of S2- ion can be controlled by pH
changes
H2S <==> H+ + HSKa1 = 1.0 x 10-7
HS- <==> H+ + S2Ka2 ~ 10-19
NOTE: From the small Ka2 value that S2- ions have a _____
affinity for H+
==> In an acidic solution (large H+), [S2-] will be ________.
==> In a basic solution, [S2-] will be ___________.
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Selective Precipitation and
Qualitative Analysis
Ex. A solution of Fe2+ and Mn2+
Ksp FeS = 3.7 x 10-19 MnS = 2.3 x 10-13
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Complex Ion Equilibria
Complex Ion - a charges species consisting of
a metal ion surrounded by ligands
Ligand - a lewis base
Coordination Number - the number of ligands
attached to a metal ion
„ metal ions add ligands one at a time in steps
like the dissociation of polyprotic acids.
Each equilibrium has its own formation
constant (stability
constant).
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Complex Ion Equilibria
Complex Ion - a charges species consisting of
a metal ion surrounded by ligands
Ligand - a lewis base
Coordination Number - the number of ligands
attached to a metal ion
„ metal ions add ligands one at a time in steps
like the dissociation of polyprotic acids.
Each equilibrium has its own formation
constant (stability
constant).
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Complex Ion Equilibria
Complex Ion
Ag(NH3)2+
Cation
Ligands
Cu2+
H2O
Fe3+
CN-
Al3+
OH-
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Coordination #
Complex Ions and Solubility
The formation of a complex ion decreases the
concentration of the free metal ion, thus ligands
can be added to increase the solubility of an ionic
compound.
Ex. AgCl Ù Ag+ + ClKsp = 1.6 x 10-10
Ag+ + NH3 Ù Ag(NH3)+
K1 = 2.1 x 103
Ag(NH3)+ + NH3 Ù Ag(NH3)2+ K2 = 8.2 x 103
„
Æ AgCl solubility can be ___________ by adding
excess NH3 to the
solution.
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TWO STRATEGIES FOR
DISSOLVING IONIC CPDS:
„
1.
2.
„
KEY - the key to increasing an ionic compound’s solubility
is to lower the concentration of one of the ions it breaks up
into, thus shifting the equilibrium right and causing more of
the compound to dissolve.
If the anion is a good base, the solubility is greatly
increased by acidifying the solution.
In cases where the anion is not sufficiently basic, the ionic
solid often can be dissolved in a solution containing a
ligand that forms stable complex ions with its cation.
Sometimes heating is sufficient to dissolve ionic solids (the
dissolving of most ionic solids is an _____thermic
process).
Jodi Grack; Wayzata High School;
Images from Zumdahl Chemistry 7th
edition
Qualitative Analysis
Group I - insoluble chlorides
add dilute HCl
-->
AgCl, PbCl2, Hg2Cl2
rest soluble
Group II - sulfides insoluble in acid solution
add H2S
-->
Hg2+, Cd2+, Bi3+, Cu2+, Sn4+
acidic so only most insoluble sulfides ppt.
rest still dissolved
Group III - sulfides insoluble in basic solution
soln made basic
-->
Co2+, Zn2+, Mn2+, Ni2+, Fe2+, (Cr3+, Al3+)
more H2S added
higher [S2-], more ppt’s
Group IV - insoluble carbonates (all except group 1A and 2A @ this point)
add CO32-->
Ba2+, Ca2+, Mg2+
group 2A cations form ppt’s
Group V - alkali metal and ammonium ions
flame test
(Li - crimson, Na - yel/orange, K - pinkish violet)
Jodi Grack; Wayzata High School;
Images from Zumdahl Chemistry 7th
edition