Chapter 17: Applications of Aqueous Equilibria

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Chapter 17 Applications of Aqueous Equilibria

• By Ms. Buroker

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The Common Ion Effect

• Suppose we had a solution that contained the
  weak acid HF (Ka 7.2 x 10-4) and its salt NaF.
• What would be the major species in solution?
• HF, Na, F-, H2O
• The common ion is this solution is F- because it
  comes from both the HF and the NaF!!!

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So what effect does the presence of the NaF
have on the dissociation of HF?

• Lets compare
• 1.0M HF with (1.0M HF 1.0M NaF)
• According to LeChateliers principle we would
  expect the dissociation equilibrium for HF to be
• HF(aq) ? H(aq) F-(aq)

Added F- ions from NaF
Equilibrium shifts away from the added component.
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Buffered Solutions

• A solution that resists a change in pH when
  either hydroxide ions or protons are added.
• Buffered solutions contain either
• A weak acid and its salt
• A weak base and its salt

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Acid/Salt Buffering Pairs

• The salt will contain the anion of the acid, and
  the cation of a strong base (NaOH, KOH)

Weak Acid Acid Formula Example of salt of the weak acid
Hydrofluoric acid HF KF
Hydrocyanic acid HCN NaCN
Acetic Acid HC2H3O2 Mg(C2H3O2)2
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Base/Salt Buffering Pairs

• The salt will contain the cation of the base,
  and the anion of a strong acid (HCl, HNO3)

Weak Base Base Formula Example of salt of the weak base
Ammonia NH3 NH4Cl
Methylamine CH3NH2 CH3NH2Cl
Ethylamine C2H5NH2 C2H5NH3NO3
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Example 1

• What is the pH of a buffer solution composed of
  .100M propanic acid (HC3H5O2 Ka1.3x10-5) and
  .100M sodium propanate?

To decide pH, we need to figure out how the
presence of the salt affects the acids
equilibrium
HC3H5O2 ltgt H C3H5O2-
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Example 1 (continued)
HC3H5O2 ltgt H C3H5O2-
ICE

.100M 0 0.100M
-x x x
.100-x x 0.100 x
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Example 1 (continued)

we will assume that the effect of x can be
ignored (we will double check later)
K
x 1.3 x 10-5
pH -logx 4.88
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How do buffers work?

• The presence of the common ion from the conjugate
  helps moderate the equilibrium
• More of the common ion shifts the equilibrium
  toward the non-dissociated side of the
  equilibrium
• Adding more acid or base has a lesser effect on
  pH because of this

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Example 2
How does the addition of 0.010 moles of solid
NaOH affect the pH of 1 L of the buffer from the
last example?
HC3H5O2 OH- ?C3H5O2- H2O

• Stoichiometry Point of View
• 0.010 moles of the base will react with the same
  amount of the acid, and produce the same amount
  of the conjugate
• Therefore we have
• 0.100 - .010 .090 moles of the acid remaining
• 0.100 .010 .110 moles of the conjugate present

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Example 2 (continued)

HC3H5O2 ? H C3H5O2-
ICE
.090M 0 .110M
-x x x
.090-x x .110 x
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Example 2 (continued)

we will assume that the effect of x can be
ignored (we will double check later)
x 1.06 x 10-5
pH -logx 4.97
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Another Way to Solve

• The stability of a buffer can be explained
  another way

can be re-arranged to
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Another WayHenderson- Hasselbalch

Taking negative log of both sides gives
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Example 2 (again)

pH 4.97
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Example 3

• A buffered solution contains .50M NH3
  (Kb1.8x10-5)and .30M NH4Cl. What is its pH?

To solve H-H
Kw (Ka)(Kb)
1x10-14 (Ka)(1.8x10-5)
5.6x10-10 (Ka)
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Example 3 (continued)

pH 9.26 (-.22) 9.04
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Acid- Base Titrations

• Titration Quantitatively determining the
  concentration of a solution by reacting it with
  another solution of precisely known
  concentration.
• Terminology
• Titrant The solution of known concentration used
  to titrate a volume of the solution of unknown
  concentration.
• Standardized (standard) solution Solution of
  precisely known concentration
• Equivalence point The point at which
  stoichiometrically equivalent amounts of
  reactants have reacted
• Endpoint The point where an indicator (such as
  pH) changes color.

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Titration Curve
The pH at the equivalence point of a strong
acid-strong base titration is 7. The pH of the
equivalence point is taken as the mid-point in
the vertical portion of the pH verses volume of
titrant curve.
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Example

• What is the pH after 25.0mL of 0.100M NaOH has
  been added to 50.0mL of 0.100M HCl? What is the
  pH after 50.50mL of NaOH has been added?
• 1.) The first thing you need to do is find out
  how much of the acid will be left once the base
  has been added you know all 25mL of the base
  will react.

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Answer to First Part

• mol acid left (0.0500L)(0.100M)
  (.0250L)(0.100M)
• .0025mol
• New H3O .0025mol/.075L .033M
• pH -log(.033) 1.48

Remember now that the base has been added to the
acid the volume has increased!! Draw a picture
it helps to keep things straight.
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Now back to the problem

• What is the pH after 25.0mL of 0.100M NaOH has
  been added to 50.0mL of 0.100M HCl? What is the
  pH after 50.50mL of NaOH has been added?
• 2.) Now you will notice that all of the acid has
  reacted and you have excess base remaining. So
  now you just have to find the hydroxide ion
  concentration and go from there.

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Answer to Second Part
Remember now that the base has been added to the
acid the volume has increased!! Draw a picture
it helps to keep things straight.

• At 50.50mL of NaOH all of the HCl is neutralized
  and an extra 0.50mL of .100M NaOH exists in
  solution.
• OH- (0.0005L)(0.100M) / .10050L 4.975x10-4
• pOH -log(4.975x10-4) 3.303
• pH 14.0 3.303 10.7

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Weak Acid- Strong Base Titrations

• The pH before the titration begins is given by
  the equilibrium expression using Ka of the weak
  acid.
• The pH at the equivalence point is dominated by
  the conjugate base (in the salt that forms) from
  the weak acid that was titrated.
• The pH at the halfway point (half-equivalence
  point) is given by
• pH pKa log(A-/HA)
• at the halfway point A- HA and the
  log(1) 0 Therefore pH pKa

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Titration Curve
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Example

• What is the pH of the solution when 35.0mL of
  0.100M NaOH has been added to 100.0mL of 0.100M
  acetic acid?
• Answer
• HC2H3O2 NaOH ?NaC2H3O2 H2O
• Moles NaOH (.0350L)(0.100M) .0035 mol
• Moles HC2H3O2 (.1000L)(0.100M) .0100 mol
• pH pKa log (C2H3O2-/HC2H3O2)
• pH 4.74 log (.0259M / 0.0481M) 4.47

Moles of acid minus the moles of base it reacted
with divided by liters of solution.
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Example

• Calculate the pH after 75.0mL of 0.100M HCl has
  been added to 100.0mL of 0.100M NH3 (Kb of NH3
  1.8x10-5)
• Answer
• mol HCl (.0750L)(0.100M) .00750mol HCl
• initial moles NH3 (.100L)(0.100M) .0100mol
  NH3
• Final
• mol NH4 .00750mol
• mol NH3 .0100 - .00750 .0025mol
• pOH pKb log(NH4/NH3)
• pOH 4.74 log(.00750/.0025) 5.22
• pH 14.00 5.22 8.78

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Solubility of Salts

• For a given slightly soluble salt of the form
• AxBy(s) ?xAy(aq) yBx-(aq)
• Ksp AyxBx-y
• Where Ksp is the solubility product constant
  (equilibrium constant) for the salt (in aqueous
  solution)
• Can be used to determine solubility (in mol/L or
  g/100mL H2O, etc.)

Molar solubility (mol/L) is the number of moles
of solute dissolved in 1 L of a saturated
solution.
Solubility (g/L) is the number of grams of solute
dissolved in 1 L of a saturated solution.
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Solubility Equilibria
Ksp AgCl-
Ksp is the solubility product constant
Ksp Mg2F-2
Ksp Ag2CO32-
Ksp Ca23PO33-2
Dissolution of an ionic solid in aqueous solution
No precipitate
Q lt Ksp
Unsaturated solution
Q Ksp
Saturated solution
Precipitate will form
Q gt Ksp
Supersaturated solution
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What is the solubility of silver chloride in g/L ?
Ksp 1.6 x 10-10
0.00
0.00
Ksp AgCl-
Ksp x2
x
x
x
x
x 1.3 x 10-5
Ag 1.3 x 10-5 M
Cl- 1.3 x 10-5 M
1.9 x 10-3 g/L
Solubility of AgCl
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Example

• Knowing that the Ksp value for MgF2 is 5.2x10-11,
  calculate the solubility of the salt in
• moles per liter and
• grams per liter
• Answer
• 5.2x10-11 X2X2
• X 2.35x10-4 2.4x10-4mol/L
• 2.4x10-4mol/L(62.3g/mol) 0.015gMgF2/L

Reaction MgF2 ? Mg2(aq) 2F-(aq) Ksp
Mg2F-2
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If 2.00 mL of 0.200 M NaOH are added to 1.00 L of
0.100 M CaCl2, will a precipitate form?
The ions present in solution are Na, OH-, Ca2,
Cl-.
Only possible precipitate is Ca(OH)2 (solubility
rules).
Is Q gt Ksp for Ca(OH)2?
Ca20 0.100 M
OH-0 4.0 x 10-4 M
0.10 x (4.0 x 10-4)2 1.6 x 10-8
Ksp Ca2OH-2 8.0 x 10-6
Q lt Ksp
No precipitate will form
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Solubility and the Common Ion Effect

• The effect of having an ion already in solution
  that is common to one of the ions in a substance
  being added to the solution (with a given Ksp) is
  to reduce the solubility of that substance.
• Example
• Calculate the solubility of BaSO4 a) in pure
  water and b) in the presence of 0.010M Ba(NO3)2.
  Ksp for BaSO4 is 1.1x10-10.

Think LeChatelier!!!
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Answer

• a.) Ksp Ba2SO42-
• so 1.1x10-10 (x)(x)
• solving for x gives x 1.0x10-5mol/L
• b.) 1.1x10-10 (x .010)(x)
• solving for x gives x 1.1x10-8mol/L

This comes from the common ion that was added to
the solution of BaSO4.
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Complex Ion Equilibria and Solubility
A complex ion is an ion containing a central
metal cation bonded to one or more molecules or
ions.
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16.10
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Complex Ion Formation

• These are usually formed from a transition metal
  surrounded by ligands (polar molecules or
  negative ions).
• As a "rule of thumb" you place twice the number
  of ligands around an ion as the charge on the
  ion... example the dark blue Cu(NH3)42 (ammonia
  is used as a test for Cu2 ions), and Ag(NH3)2.
• Memorize the common ligands.

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Common Ligands
Ligands Names used in the ion
H2O aqua
NH3 ammine
OH- hydroxy
Cl- chloro
Br- bromo
CN- cyano
SCN- thiocyanato (bonded through sulfer) isothiocyanato (bonded through nitrogen)
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Names

• Names ligand first, then cation
• Examples
• tetraamminecopper(II) ion Cu(NH3)42
• diamminesilver(I) ion Ag(NH3)2.
• tetrahydroxyzinc(II) ion Zn(OH)4 2-
• The charge is the sum of the parts (2) 4(-1)
  -2.

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When Complexes Form

• Aluminum also forms complex ions as do some post
  transitions metals. Ex Al(H2O)63
• Transitional metals, such as Iron, Zinc and
  Chromium, can form complex ions.
• The odd complex ion, FeSCN2, shows up once in a
  while
• Acid-base reactions may change NH3 into NH4 (or
  vice versa) which will alter its ability to act
  as a ligand.
• Visually, a precipitate may go back into solution
  as a complex ion is formed. For example, Cu2 a
  little NH4OH will form the light blue
  precipitate, Cu(OH)2. With excess ammonia, the
  complex, Cu(NH3)42, forms.
• Keywords such as "excess" and "concentrated" of
  any solution may indicate complex ions. AgNO3
  HCl forms the white precipitate, AgCl. With
  excess, concentrated HCl, the complex ion,
  AgCl2-, forms and the solution clears.

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Coordination Number

• Total number of bonds from the ligands to the
  metal atom.
• Coordination numbers generally range between 2
  and 12, with 4 (tetracoordinate) and 6
  (hexacoordinate) being the most common.

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Some Coordination Complexes
molecular formula Lewis base/ligand Lewis acid donor atom coordination number
Ag(NH3)2 NH3 Ag N 2
Zn(CN)42- CN- Zn2 C 4
Ni(CN)42- CN- Ni2 C 4
PtCl6 2- Cl- Pt4 Cl 6
Ni(NH3)62 NH3 Ni2 N 6