Ch' 17 Additional Aspects of Aqueous Equilibria

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Ch. 17 Additional Aspects of Aqueous Equilibria

• The Common Ion Effect
• The dissociation, (or solubility), of a weak
  electrolyte is decreased by the addition of a
  strong electrolyte that has an ion in common with
  the weak electrolyte.
• Consider the equilibrium established when acetic
  acid, HC2H3O2, is added to water
• HC2H3O2(aq)?? H(aq) C2H3O2-(aq)
• What do you think would happen to the
  equilibrium if we were to dissolve NaC2H3O2 , (a
  strong electrolyte), into the solution?
• (Remember Le Châteliers principle!)

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Common Ion Effect

• As expected, the equilibrium would shift to the
  left.
• The H would decrease by reacting with the
  extra acetate ions to produce HC2H3O2 .
• The HC2H3O2 increases as the equilibrium is
  re-established.
• Since NaC2H3O2 is a strong electrolyte and
  HC2H3O2 is a weak electrolyte, the C2H3O2- is
  based on the amount of salt added to the solution
  not on the weak acids dissociation.
• Lets do a practice problem with numbers
• What is the pH of a 0.085 M HNO2 solution, (Ka
  4.5 x 10-4) when 0.1 M of KNO2 is added? (
  HNO2 ?? H NO2- )

0
0.1 M
0.085 M
x
x
x
0.085 x
0.10 x
x
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Common Ion Effect
0
0.10
0.085 M
x
x
x
0.085 x
0.10 x
x

• Ka 4.5 x 10-4 HNO2/HNO2 x0.10
  x/0.085 x
• Assuming x is going to be a very small answer
• 4.5 x 10-4 x0.10/0.085
• Solving for xx 0.00038 M
• This is equal to H, so
• pH -(log0.00038) 3.42
• F.Y.I.Without the salt, pH 2.2

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Common Ion Effect

• The previous example used the dissociation of a
  weak acid. But the ionization of a weak base can
  also be decreased by the addition of a strong
  electrolyte
• H2O NH3 ?? NH4 OH-
• If we were to add NH4Cl to the solution, the
  equilibrium would shift to the left, and the pH
  would decrease as OH- reacts with the additional
  NH4 in the solution.

5
Buffered Solutions (a.k.a. Buffers)

• The solutions we have been describing can resist
  drastic changes in pH as a strong acid, (H), or
  a strong base, (OH-), is added.
• These types of solutions are called buffers.
• A buffer consists of a mixture of a weak acid
  (HX) and its conjugate base (X ) HX(aq) ??
  H(aq) X(aq)
• So a buffer contains both
• An acidic species (to neutralize OH) and
• A basic species (to neutralize H).
• Buffers have many important applications in the
  lab and in medicineyour blood is a buffered
  solution that functions in the pH range of 7.35
  7.45 (See p.669 for more details!)

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Buffered Solutions (a.k.a. Buffers)

• When a small amount of OH is added to the
  buffer, the OH reacts with HX to produce X and
  water
• HX OH ? X H2O
• But the HX/ X ratio remains more or less
  constant, so the pH is not significantly changed.
• When a small amount of H is added to the
  buffer, X is consumed to produce HX
• X H ? HX
• Once again, the HX/ X ratio is more or
  less constant, so the pH does not change
  significantly.

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Buffers
8
Buffer Capacity

• Buffer capacity is the amount of acid or base
  that can be neutralized by the buffer before
  there is a significant change in pH.
• Buffer capacity depends on the concentrations of
  the components of the buffer.
• The greater the concentrations of the conjugate
  acid-base pair, the greater the buffer capacity.
• The pH of the buffer is related to Ka and to the
  relative concentrations of the acid and base

• This equation is known as the Henderson-Hasselbal
  ch equation.
• When base acid, the pH pKa.

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Buffer Calculations

• When a small amount of strong acid or strong
  base is added to a buffer, we assume that it is
  completely consumed by the reaction with the
  buffer.
• Buffer problems consist of 2 parts
  stoichiometry equilibrium
• Practice Problem
• A buffer is made by adding 0.300 moles of HOAc
  and 0.300 moles of NaOAc to enough water to make
  1.00 L of solution. The pH of the buffer equals
  4.74. What is the pH after 0.020 moles of NaOH
  is added?
• First, we use stoichiometry to determine the
  effect of the addition of OH- to the buffer
• OH- HOAc ? H2O OAc-
• We see that HOAc will decrease and OAc- will
  increase by the concentration of OH- added to the
  buffer.
• Lets tabulate this information

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Buffer Calculations

• OH- HOAc ? H2O OAc-
• Before reaction 0.020 mol 0.300 mol 0.300
  mol
• After reaction 0 0.280 mol
  0.320 mol
• Now we can use the Henderson-Hasselbalch equation
  to find pH
• pH pKa log ( base/acid )
• pH (log 1.8 x 105) log ( 0.320/0.280
  )
• pH 4.74 0.06 4.80
• A couple of things to note
• 1) You can use mole amounts in place of
  concentrations in the equation.
• 2) If we added 0.020 moles of a strong acid,
  then the pH would have decreased by 0.06 instead.

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Buffer Calculations
12
Acid-Base Titrations

• In an acid-base titration
• A solution of base of known concentration is
  added to an acid.
• Acid-base indicators or a pH meter are used to
  signal the equivalence pointthe point at which
  the moles of H moles of OH-.
• The end point in a titration is the point where
  the indicator changes color.
• The plot of pH versus volume during a titration
  is called a pH titration curve.
• Lets look at the titration curve for a strong
  acid-strong base
• ( HCl NaOH ? H2O NaCl )

13
Strong Acid-Strong Base Titration Curve

• Things of interest
• The initial pH is based on the HCl at the
  start.
• The pH rises slowly at first then jumps
  dramatically after the equivalence point.
• The equivalence point is at a pH of 7.0
• The final pH is based upon the NaOH in excess.
• We should choose an indicator that changes color
  in the portion of the graph where the pH rises
  rapidly.

14
Strong Acid-Strong Base Titration Curve

• Heres what the graph would look like if the
  strong acid were titrated into a strong
• basethe equivalence point is still at a pH of 7.0

15
Weak Acid-Strong Base Titrations

• Consider the titration of acetic acid, HC2H3O2
  with NaOH
• Again, we divide the titration into four general
  regions
• (1) Before any base is added the solution
  contains only weak acid. Therefore, pH is given
  by the equilibrium calculation.
• (2) Between the initial pH and the equivalence
  point the strong base consumes a stoichiometric
  quantity of weak acid
• HC2H3O2(aq) OH(aq) ? C2H3O2(aq) H2O(l)
• However, there is an excess of acetic acid.
  Therefore, we have a mixture of weak acid and its
  conjugate base.
• Thus the composition of the mixture is that of a
  buffer. The pH is given by the buffer
  calculation(Henderson-Hasselbalch equation.)
• moles base added moles of acid consumed moles
  of acetate ion formed

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Weak Acid-Strong Base Titrations
(3) 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 depends on the
C2H3O2 concentration. - The pH gt 7 at the
equivalence point since C2H3O2- is a weak
base(All weak acid-strong base titrations have
equivalence points at pHs greater than
7!) - The equivalence point is determined by the
Ka of the acid. (4) After the equivalence point
the pH is based on the excess strong base.
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Weak Acid-Strong Base Titration Curves

• More on choosing indicators
• 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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Weak Acid-Strong Base Titration Curves

• Things to notice
• For a weak acid-strong base titration, the
  initial pH rise is more steep than the strong
  acid-strong base case.
• Then there is a leveling off due to buffer
  effects.
• The equivalence point is higher for the weaker
  acids.
• The shape of the curves after equivalence point
  is the same because pH is determined by the
  strong base in excess.

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Polyprotic Acid Titration Curves

• Things to notice
• 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 H3PO3 with NaOH, the first
  proton dissociates to form H2PO3- then the
  second proton dissociates to form HPO3-2.

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Solubility Equilibria

• All dissolving is an equilibrium Solid ??
  Dissolved
• Example BaSO4 (s)?? Ba2(aq) SO4-2(aq)
• If there is not much solid, it will all
  dissolve.
• As more solid is added the solution will become
  saturated.
• At equilibrium, the solid will precipitate as
  fast as it dissolves.
• Ksp Ba2 SO4-2
• Ksp is the solubility product. It is the molar
  concentration of ions raised to their
  stoichiometric powers(BaSO4 is ignored because
  it is a pure solid, so its concentration is
  constant.)
• Solubility product is an equilibrium constant.
  It doesnt change except with temperature.

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Solubility Definition

• Solubility is not the same as solubility
  product.
• Solubility is the amount (grams) of substance
  that dissolves to form a saturated solution.
  Solubility is an equilibrium position for how
  much can dissolve.
• Molar solubility is the number of moles of
  solute dissolving to form a liter of saturated
  solution.
• A common ion can change the solubility of a
  substance.
• Relative Solubilities
• Ksp will only allow us to compare the solubility
  of solids that fall apart into the same number of
  ions.
• The bigger the Ksp, the more soluble the
  substance.
• If they fall apart into different number of
  pieces you have to do an equilibrium calculation
  to see which is more soluble.

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Converting Solubility into Ksp
Practice Problem The molar solubility of CaF2
is 1.24 x 10-3 M at 35º C. What is the
solubility product of CaF2 at 35º C? CaF2(s) ??
Ca2(aq) 2F-(aq) Ksp Ca2 F-2 Ca2
1.24 x 10-3 M F- 2 x (1.24 x 10-3 M) 2.48 x
10-3 M Ksp 1.24 x 10-3 2.48 x 10-32
7.63 x 10-9 Note In order to calculate Ksp ,
all concentrations must be in units of moles/L
(or molarity, M). Reminder grams/FormulaWeight
moles
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Solubility and the Common Ion Effect

• This is an application of Le Châteliers
  principle!
• For example
• If more F- is added, (lets say from the addition
  of a strong electrolyte such as NaF), the
  equilibrium shifts away from the increase.
• Therefore, CaF2(s) is formed and precipitation
  occurs.
• As NaF is added to the system, the solubility of
  CaF2 decreases. If the F- is removed, then the
  equilibrium shifts to the right, and more CaF2
  dissolves.
• F- can be removed by adding a strong acid
• This means that the pH affects the solubility of
  CaF2!

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Solubility and pH

• Another example of pH changing the solubility
• Mg(OH)2(s) ?? Mg2(aq) 2OH-(aq)
• What happens to the equilibrium if we were to
  add a strong acid?
• Again we apply Le Châteliers principle
• If OH is removed, then the equilibrium
  shifts toward the right and Mg(OH)2 dissolves
  OH(aq) H(aq) ? H2O(aq)
• As pH decreases, H increases and the
  solubility of Mg(OH)2 increases.
• The effect is most significant if one or both
  ions involved are at least somewhat acidic or
  basic.
• In general
• The solubility of slightly soluble salts
  containing basic ions increases as pH decreases.
  
• The more basic the anion, the greater the effect.

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Formation of Complex Ions

• Some metal ions, (cations), can form soluble
  complex ions when in solutions.

• Consider the formation of Ag(NH3)2
• The Ag(NH3)2 is called a complex ion.
• NH3 (the attached Lewis base) is called a ligand.
• The equilibrium constant for the reaction is
  called the formation constant, Kf Kf
  Ag(NH3)2/Ag NH32

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Formation of Complex Ions

• Consider the addition of ammonia to AgCl (white
  precipitate)
• The overall reaction is
• Effectively, the Ag(aq) has been removed from
  solution.
• By Le Châteliers principle, the forward reaction
  (the dissolving of AgCl) is favored.

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Formation of Complex Ions

• Finally, amphoteric oxides will dissolve in
  either a strong acid or a strong base by forming
  complex ions with several hydroxide ligands
  attached to the metal.
• For example
• We wont be dealing with these unique solubility
  situations much, but I wanted you to be aware
  that they exist.

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Solubility Guidelines for Common Ionic Compounds
in Water

• Our dividing line between soluble and insoluble
  will be 0.01 M at 25 C. (Some people use 0.1 M
  as a dividing line, so you may see some
  discrepancies between solubility guidelines.)
• Any substance that can form a solution with a
  concentration of 0.01 M or more is soluble. Any
  substance that fails to reach 0.01 M is defined
  to be insoluble.
• This value was picked with a purpose. VERY FEW
  substances have their maximum solubility near to
  0.01 M. Almost every substance of any importance
  in chemistry is either much MORE soluble or much
  LESS soluble.
• In the past, some teachers would have a third
  category slightly soluble. For the most part,
  that category has been cast aside.

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Solubility Guidelines for Common Ionic Compounds
in Water

• All alkali metal(Group 1A lithium, sodium,
  potassium, rubidium, and cesium) and ammonium
  compounds are soluble.
• (2) All acetate, perchlorate, chlorate, and
  nitrate compounds are soluble.
• (3) Silver, lead, and mercury (I) compounds are
  insoluble.
• (4) Chlorides, bromides, and iodides are soluble.
• (5) Carbonates, oxides, phosphates, chromates,
  fluorides, and silicates are insoluble.
• (6) Hydroxides and sulfides are insoluble (EXCEPT
  for Ca2, Ba2, and Sr2 .)
• (7) Sulfates are soluble (EXCEPT for Ca2, Ba2,
  and Sr2).
• - These rules are to be applied in the order
  given. For example, PbSO4 is insoluble because
  Rule 3 comes before Rule 7. In like manner, AgCl
  is insoluble because Rule 3 takes precedence
  over Rule 4.
• - Please be aware that these rules are
  guidelines only. For example, there are some
  alkali metal compounds that are insoluble.
  However, these are rather exotic compounds and
  can be safely ignored at an introductory level.

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Here are some of these guidelines shown in a
table format
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Precipitation of Ions

• We can use our knowledge of Ksp and equilibrium
  to determine if a precipitate will form when
  solutions are mixed.

• For example
• At any instant in time, Q Ba2SO42-.
• If Q gt Ksp, precipitation occurs until Q Ksp.
• If Q Ksp, equilibrium exists.
• If Q lt Ksp, solid dissolves until Q Ksp.
• (Well look at a practice problem in a little
  bit.)
• Based on our solubility guidelines, ions can be
  selectively removed from solutions.
• Removal of one metal ion from a solution is
  called selective precipitation.
• The textbook goes into detail about the
  procedure, but we wont.

32

• Qualitative analysis is designed to detect the
  presence of metal ions.
• Quantitative analysis is designed to determine
  how much metal ion is present.

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Forming Precipitates

• Practice Problem Will Ag2SO4 precipitate when
  100 mL of 0.050 M AgNO3 is mixed with a 10 mL
  solution of 0.050 M Na2SO4?
• Ag2SO4(s) ?? 2 Ag SO42
• Ksp 1.5 x 105 Ag2 SO42
• Before we can find the concentrations of the
  ions in the mixture, we need to determine the
  moles of each ion in the solutions before they
  were mixed together(Reminder moles Liters x
  Molarity)
• moles Ag 100 mL x (1 L /1000 mL) x ( 0.050
  moles/L) 0.005 moles
• moles SO4-2 10 mL x (1 L /1000 mL) x ( 0.050
  moles/L) 0.0005 moles
• After the solutions are mixed, the total volume
  is 110 mL or 0.11 L, so
• Ag 0.005 moles/0.11 L 0.045 M
• SO4-2 0.0005 moles/0.11 L 0.0045 M
• Q 0.0452 0.0045 9.1 x 10-6 lt Ksp , so NO
  PRECIPITATION will occur.