Chapter 15
Applying equilibrium
15.1 The Common Ion Effect
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When the salt with the anion of a weak acid is
added to that acid, it reverses the dissociation
of the acid.
NaF(s)
Na+(aq) + F-(aq)
adding


NaCl( aq ) shifts equilibrium position
Lowers the percent dissociation of the acid.
The same principle applies to salts with the
cation of a weak base.
The calculations are the same as last chapter.
Calculate [H+] and % dissociation of HF in a
solution containing 1.0 M HF (Ka=7.2x10-4)
and 1.0 M NaF Ex.15.1
15.2 Buffered Solutions
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A solution that resists change in its pH when
either H+ or OH- are added.
Either a weak acid and its salt or a weak base
and its salt.
We can make a buffer of any pH by varying the
concentrations of these solutions.
Same calculations as before.This is not a new
type of problem.
Calculate the pH of a solution that is 0.50 M
HAc and 0.50 M NaAc (Ka = 1.8 x 10-5) Ex.15.2
Practice
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A buffer is made by dissolving NH4Cl and
NH3 in water. Write equations to show how
this buffer neutralizes added H+ and OH-. #22
Calculate the pH of each of the following
solutions. #24
A) 0.100 M HONH2 (Kb = 1.1 x 10-8)
B) 0.100 M HONH3Cl
C) Pure H2O
D) A mixture of 0.100 M HONH2 and 0.100 M
HONH3Cl
Compare % dissociation of A with D #26
Adding a strong acid or base
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Do the stoichiometry first.
A strong base will grab protons from the
weak acid reducing [HA]0
A strong acid will add its proton to the anion
of the salt reducing [A-]0
Then do the equilibrium problem.
What is the pH of 1.0 L of a solution(0.50M
HC2H3O2 & 0.50M NaC2H3O2 ) when 0.010
mol of solid NaOH is added? Ex.15.3
What is the pH when .010 mol of solid NaOH
is added to plain water? Ex.15.3
General equation to HH
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Ka = [H+] [A-]
[HA]
so [H+] = Ka [HA]
[A-]
The [H+] depends on the ratio [HA]/[A-]
taking the negative log of both sides
pH = -log(Ka [HA]/[A-])
pH = -log(Ka)-log([HA]/[A-])
pH = pKa + log([A-]/[HA])
This is called the HendersonHasselbach equation
pH = pKa + log([A-]/[HA])
 pH = pKa + log(base/acid)
 Calculate the pH of the following
mixtures:
 0.75 M lactic acid (HC3H5O3) and 0.25 M
sodium lactate (Ka = 1.4 x 10-4) Ex.15.4
 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x
10-5) Ex.15.5
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Prove they’re buffers
0.75 M lactic acid (HC3H5O3) and 0.25
M sodium lactate (Ka = 1.4 x 10-4)
 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8
x 10-5)
 What would the pH be if 0.10 mol of
HCl is added to 1.0 L of both of the
preceding solutions. Ex.15.6
 What would the pH be if 0.050 mol of
solid NaOH is added to each of the
proceeding.
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Buffer Summary
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Buffers contain relatively large amounts of
weak acid and corresponding base.
Added H+ reacts to completion with the
weak base.
Added OH- reacts to completion with the
weak acid. (YDVD)
The pH is determined by
the ratio of the
concentrations of the weak
acid and weak base.
QuickTime™ and a
Cinepak Codec by Radius decompressor
are needed to see this picture.
15.3 Buffer Capacity
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Represents the amount of H+ or OH- the
buffer can absorb without a significant
change in pH.
The pH of a buffered solution is determined
by the ratio [A-]/[HA].
As long as this doesn’t change much the pH
won’t change much.
The more concentrated these two are the
more H+ and OH- the solution will be able to
absorb.
Larger concentrations bigger buffer capacity.
Buffer Capacity
Calculate the change in pH that occurs
when 0.010 mol of HCl(g) is added to
1.0L of each of the following: Ex.15.7
 5.00 M HAc and 5.00 M NaAc
 0.050 M HAc and 0.050 M NaAc
 Ka= 1.8x10-5
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Buffer Capacity
The best buffers have a ratio [A-]/[HA]
=1
 This is most resistant to change.
 True when [A-] = [HA]
 Make pH = pKa (since log 1 = 0)
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Practice
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A chemist needs a solution buffered at a
pH of 4.30 and can choose from the
following acids (and their sodium salts):
Ex.15.8
Chloroacetic acid (Ka = 1.35 x 10-3)
 Propanioc acid
(Ka = 1.3 x 10-5)
 Benzoic acid
(Ka = 6.4 x 10-5)
 Hypochlorous acid (Ka = 3.5 x 10-8)
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Practice
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A solution is composed of 0.60 M HF
and 1.00 M KF. #36 Ka=7.2 x10-4
– Calculate the pH after 0.10 mol of
NaOH is added to 1.00L of this
solution.
– Calculate the pH after 0.20 mol of
HCl is added to 1.00L of this solution.
Practice
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A buffered solution is made by adding 50.0 g
NH4Cl to 1.00L of a 0.75 M solution of NH3.
Calculate the pH of the final solution. #38 Kb =1.8
x 10-5
An aqueous solution contains dissolved
C6H5NH3Cl and C6H5NH2. The conc. of
C6H5NH2 is 0.50 M and pH 4.20. Kb =3.8 x 10-10
Calculate the conc. of C6H5NH3+ in this buffer
solution. #40
Calculate the pH after 4.0 g of NaOH(s) is added
to 1.0 L of this solution
15.4 Titrations
Millimole (mmol) = 1/1000 mol
 Molarity = mmol/mL = mol/L
 Makes calculations easier because we
will rarely add Liters of solution.
 Adding a solution of known
concentration until the substance being
tested is consumed.
 This is called the equivalence point.
 Graph of pH vs. mL is a titration curve.
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Strong Acid with Strong Base
Do the stoichiometry.
 There is no equilibrium .
 They both dissociate completely.
 The titration of 50.0 mL of 0.200 M HNO3
with 200 ml of 0.100 M NaOH (Case Study)
 Analyze the pH
 0 ml, 10 ml, 20 ml, 100 ml, 150 ml, 200 ml
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*Look at GA SASB
Weak acid with Strong base
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There is an equilibrium.
Do stoichiometry. The reaction is assumed to
run to completion - then determine remaining
species.
Then do equilibrium. Determine the position
of weak acid equilibrium and calculate pH.
Titrate 50.0 mL of 0.10 M Hac (Ka = 1.8 x 10-5)
with 0.10 M NaOH (Case Study)
0 ml, 10 ml, 25 ml, 40 ml, 50 ml, 60 ml, 75 ml
*Look at GA WASB
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The pH at equivalence point of a titration of a
weak acid with a strong base is always
greater than 7. Why?
Weak acid/Stong base titration curve.Where are the following?
•(a) Stoichiometric (equivalence) point.
•(b)Region of maximum buffering.
•c. pH=pKa
•(d)pH depends only on [HA]
•(e)pH depends only on [A-]
•(f) pH depends only on the amount of excess base added.
Weak Base/Strong Acid
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100.0 ml of 0.100 M H2NNH2 (Kb = 3.0 x 10-6)
was titrated by 0.200 M HNO3. Calculate the
pH of the resulting solution after the
following volumes of HNO3 have been
added: 0.0 ml, 20.0 ml, 25.0 ml, 40.0 ml, 50.0
ml, 100.0 ml #56
Titrate 100.0 mL of 0.050 M NH3 with 0.10 M
HCl. (Kb = 1.8 x 10-5) case study
What is the pH at the following points: 0 ml,
at the half way point, at equivalence point,
beyond equivalence point. (60ml)
Weak Base/Strong Acid
In the titration of 50.0 ml of 1.0 M
CH3NH2 (Kb = 4.4 x 10-4), with 0.50 M
HCl, calculate the pH under the
following conditions. #62
 A) after 50.0 ml of 0.50 M has been
added
 B) at the equivalence point
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Titration Curves
Strong acid with strong Base
 Equivalence at pH 7
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pH
7
mL of Base added
Weak acid with strong Base
 Equivalence at pH >7
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pH
>7
mL of Base added
Strong base with strong acid
 Equivalence at pH 7
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pH
7
mL of Base added
Weak base with strong acid
 Equivalence at pH <7
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pH
<7
mL of Base added
Summary
Strong acid and base - just stoichiometry.
 Determine Ka, use for 0 mL base
 Weak acid before equivalence point
–Stoichiometry first
–Then Henderson-Hasselbach
 Weak acid at equivalence point Kb
 Weak base after equivalence - leftover
strong base.
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Summary
Determine Ka, use for 0 mL acid.
 Weak base before equivalence point.
–Stoichiometry first
–Then Henderson-Hasselbach
 Weak base at equivalence point Ka.
 Weak base after equivalence - leftover
strong acid.
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15.5 Indicators
Weak acids that change color when they
become bases.
 Weak acid written HIn
 Weak base
 HIn
H+ + Inclear
red
 Equilibrium is controlled by pH
 End point - when the indicator changes
color.
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Indicators
Since it is an equilibrium the color change
is gradual.
 It is noticeable when the ratio of
[In-]/[HI] or [HI]/[In-] is 1/10
 Since the Indicator is a weak acid, it has a
Ka.
 The pH the indicator changes at is:
 pH=pKa +log([In ]/[HI]) = pKa +log(1/10)
 pH=pKa - 1 on the way up
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Indicators
pH=pKa + log([HI]/[In-]) = pKa +
log(10)
 pH=pKa+1 on the way down
 Choose the indicator with a pKa 1 less
than the pH at equivalence point if you
are titrating with base.
 Choose the indicator with a pKa 1
greater than the pH at equivalence
point if you are titrating with acid.
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Practice
Two drops of indicator HIn (Ka=1.0 x 10-9),
where HIn is yellow and In- is blue, are
placed in 100.0 ml of 0.10 M HCl. # 65
 What color is the solution initially?
 The solution is titrated with 0.10 M NaOH.
At what pH will the color change occur?
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Practice
Using Fig. 15.8 (next page), which
indicators could be used for the following
titrations with pH at equivalence point
given. #68,70
 7.00
 4.82
 8.79
 3.27
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15.6 Solubility Equilibria
Will it all dissolve, and if not, how
much?
All dissolving is an equilibrium.
 If there is not much solid it will all
dissolve.
 As more solid is added the solution will
become saturated.
 Solid
dissolved
 The solid will precipitate as fast as it
dissolves .
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(BDVD) Supersaturated, A-16,A-17
Solubility Product
For solids dissolving to form aqueous
solutions.
 Bi2S3(s)
2Bi3+(aq) + 3S2-(aq)
 Ksp = solubility product constant, therefore:
 Ksp = [Bi3+]2[S2-]3
 Write a balanced equation and the solubility
expression for the following:
 Ag2CO3, Ce(IO3)3, BaF2
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Watch out
Solubility is not the same as solubility
product.
 Solubility product is an equilibrium
constant.
 It doesn’t change except with
temperature.
 Solubility is an equilibrium position for
how much can dissolve.
 A common ion can change this.
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Calculating Ksp
Find the Ksp of the following:
 The solubility of copper(I) bromide is
2.0 x 10-4 mol/L Ex.15.12
 The solubility of Bi2S3 is 1.0 x 10-15
mol/L Ex.15.13
Vizzini: You fell victim to one of the classic blunders! The
first is never get involved in a land war in Asia. The
second, only slightly less well known, is this: never go up
against a Sicilian when death is on the line! Ha ha ha ha ha
ha -Richardson: The third: always pay attention to mole ratios
when dealing with concentration.
Calculating Ksp
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The concentration of Ag+ in a solution
saturated with Ag2C2O4(s) is 2.2 x 10-4 M.
Calculate the Ksp for Ag2C2O4. #78
Calculating Solubility
The solubility is determined by
equilibrium.
 Its an equilibrium problem.
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Calculate the solubility of Cu(IO3)2,
with a Ksp of 1.4 x 10-7. Ex.15.14
 Calculate the solubility of Sr3(PO4)2,
with a Ksp of 1.0 x 10-31 #80c
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Relative Solubilities
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Ksp will only allow us to compare the
solubility of solids that fall apart into the
same number of ions.
The bigger the Ksp of those the more soluble.
AgI
Ksp = 1.5 x 10-16
CuI
Ksp = 5.0 x 10-12
CaSO4
Ksp = 6.1x10-5
If they fall apart into different number of
pieces you have to do the math.
Common Ion Effect
If we try to dissolve the solid in a
solution with either the cation or anion
already present less will dissolve.
 Calculate the solubility of CaF2, with a
Ksp of 4.0 x 10-11 in a solution of 0.025 M
NaF. Ex.15.15
 Calculate the solubility of SrSO4, with a
Ksp of 3.2 x 10-7 in a solution of 0.010 M
Na2SO4. Find M and g/L.
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pH and solubility
OH- can be a common ion.
 More soluble in acid.
 For other anions if they come from a
weak acid they are more soluble in a
acidic solution than in water.
 CaC2O4
Ca+2 + C2O4-2
 H+ + C2O4-2
HC2O4 Reduces C2O4-2 in acidic solution.
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(BDVD) Ag
pH and solubility
For which salt of each group will the
solubility depend on pH?
 AgF, AgCl, AgBr
 Pb(OH)2, PbCl2
 Sr(NO3)2, Sr(NO2)2
 Ni(NO3)3, Ni(CN)2
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15.7 Precipitation
Ion Product, Q =[M+]a[Nm-]b
 If Q> Ksp a precipitate forms.
 If Q< Ksp No precipitate.
 If Q = Ksp equilibrium.
-3
 A solution of 750.0 mL of 4.00 x 10 M
Ce(NO3)3 is added to 300.0 mL of 2.00 x
10-2 M KIO3. Will Ce(IO3)3 (Ksp = 1.9 x
10-10 M) precipitate from the mixed
solutions? If so, what is the
concentration of the ions? Ex.15.16
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Precipitation
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Will a precipitate form when 100 ml of
4.0 x 10-4 M Mg(NO3)2 is added to 100
ml of 2.0 x 10-4 M NaOH? (Ksp=8.9 x 10-12)
#92
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A solution is prepared by mixing 150.0
ml of 1.00 x 10-2 M Mg(NO3)2 and 250.0
ml of 1.00 x 10-1 M NaF. Calculate the
concentrations of Mg2+ and F- at
equilibrium with solid MgF2 (Ksp = 6.4 x
10-9) Ex. 15.17
#96 if time
Selective Precipitations
Used to separate mixtures of metal ions
in solutions.
 Add anions that will only precipitate
certain metals at a time.
 Used to purify mixtures.
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Often use H2S because in acidic
solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4
will precipitate.
Selective Precipitation
In basic solutions, the H+is removed
from H2S making more S-2. More
soluble sulfides will precipitate.
 Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3,
Al(OH)3
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Selective precipitation
Follow the steps first with insoluble
chlorides (Ag, Pb, Ba).
 Then sulfides in Acid.
 Then sulfides in base.
 Then insoluble carbonate (Ca, Ba, Mg)
 Alkali metals and NH4+ remain in
solution.
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Selective precipitation
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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 I- necessary to begin
precipitation of each salt. Ex. 15.18
Complex ion Equilibria
A charged ion surrounded by ligands.
 Ligands are Lewis bases using their lone
pair to stabilize the charged metal ions.
 Common ligands are NH3, H2O, Cl-, CN Coordination number is the number of
attached ligands. Most common are 4 and
6. (If guessing, 2 times the charge on the
cation).
 Cu(NH3)4+2 has a coordination # of 4
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