Applications of Aqueous Equilibria

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Applications of Aqueous Equilibria

• Buffers, Acid-Base Titrations and Precipitation
  Reactions

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Buffers

• Buffers are solutions of a weak acid and its
  conjugate base.
• Buffers resist changes in pH when small amounts
  of strong acid or base are added. This is
  because there is both an acid and its conjugate
  base present initially. Buffers typically have
  the acid and its conjugate base (or salt) present
  in roughly equal concentrations.

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Buffers

• Consider a buffer of weak acid HA and NaA in
  equimolar concentrations. The predominant
  reaction is
• HA(aq) H2O(l) ? H3O(aq) A-(aq)
• If a small amount of strong acid is added,
  there is enough A- available to become
  protonated. The equilibrium shifts to the left,
  and the H3O and pH do not change much.

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Buffers

• HA(aq) H2O(l) ? H3O(aq) A-(aq)
• Likewise, if a small amount of strong base is
  added, some of the hydronium will react with it.
• H3O(aq) OH-(aq) ? 2 H2O(l)
• As hydronium ion reacts, more HA will
  dissociate so as to replenish the H3O. As a
  result, the pH remains fairly constant.

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Buffers
Because buffers contain both a weak acid and its
conjugate base, they can react with strong acids
or bases and maintain their pH.

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Buffers
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Problem Buffers

• Determine the pH of a solution that contains
  0.50M HClO and 0.60M NaClO. (Ka for HClO 3.5 x
  10-8)
• - Write the major reactions
• NaClO(aq) ? Na(aq) ClO-(aq)
• The NaClO serves as a source of ClO- ions. The
  sodium ion is a spectator.

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Problem Buffers

• Determine the pH of a solution that contains
  0.50M HClO and 0.60M NaClO.
• Even though both HClO and ClO- are present
  initially, its important to realize that they
  are in equilibrium with each other, and go on
  opposite sides of the equation.
• HClO(aq) H2O(l) ? H3O(aq) ClO-(aq)

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Problem Buffers

• Determine the pH of a solution that contains
  0.50M HClO and 0.60M NaClO. (Ka for HClO 3.5 x
  10-8)
• HClO(aq) H2O(l) ? H3O(aq) ClO-(aq)
• Ka H3OClO- 3.5 x 10-8
• HClO

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Problem Buffers

• Determine the pH of a solution that contains
  0.50M HClO and 0.60M NaClO. (Ka for HClO 3.5 x
  10-8)
• HClO(aq) H2O(l) ? H3O(aq) ClO-(aq)

HClO H3O ClO-
initial 0.50 0 0.60
change -x x x
equil. .50-x x .60x
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The Henderson-Hasselbalch Equation

• For buffers, regardless of the value of Ka, you
  can always assume that x will be small compared
  to the concentrations of the acid or its
  conjugate base.
• This is because the reaction will proceed only
  slightly to the right due to the presence of the
  conjugate base initially.
• HA(aq) H2O(l) ? H3O(aq) A-(aq)

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The Henderson-Hasselbalch Equation

• As a result, the Ka expression can be written
  in logarithmic form.
• Ka H3OA-
• HA
• H3O Ka HA
• A-
• pH pKa log A-
• HA

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The Henderson-Hasselbalch Equation

• In more general terms, the equation is written
  as
• pH pKa log
• This equation is for buffers only, and is a
  quick way to calculate pH.

base
acid
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Problem Adding Acid to a Buffer

• Calculate the change in pH if 5.00 mL of 0.010M
  HNO3 is added to 50.0 mL of the HClO/NaClO
  buffer.
• The nitric acid will protonate some of the ClO-
  to form additional HClO.
• HNO3(aq) H2O(l) ? H3O(aq) NO3-(aq)
• H3O(aq) ClO-(aq)? HClO (aq) H2O(l)

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Problem Adding Acid to a Buffer

• HNO3(aq) H2O(l) ? H3O(aq) NO3-(aq)
• H3O(aq) ClO-(aq)? HClO (aq) H2O(l)
• Before you can use the Henderson-Hasselbalch
  equation, you need to calculate the new
  concentration of HClO and ClO- that results from
  adding the nitric acid.

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Problem Adding Acid to a Buffer

• HNO3(aq) H2O(l) ? H3O(aq) NO3-(aq)
• H3O(aq) ClO-(aq)? HClO (aq) H2O(l)
• For each mole of nitric acid added, you form a
  mole of HClO and react a mole of ClO-.

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Problem Adding Acid to a Buffer

• 1. You will need to calculate the moles of HNO3
  added (5.00 mL of 0.010M HNO3).
• 2. For each mole of nitric acid added, you will
  produce a mole of HClO and lose a mole of ClO-.
• 3. Since you need concentrations, remember that
  the new volume of the solution has changed to
  55.0 mL to 50.0 mL.

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Problem Compare to Water

• Calculate the change in pH when 5.00 mLs of
  0.0100M HNO3 is added to 50.0 mL of water.

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Preparation of Buffer Solutions

• Scientists often need to make a solution that
  is buffered to a specific pH. They may need to
  mimic biological conditions, test the
  corrosiveness of metal parts at a specific pH,
  etc.
• Conjugate acid/base pairs have a pH range at
  which they can effectively serve as buffers.

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Choosing the Acid and Salt

• For optimum buffering, choose an acid with a Ka
  value close to the desired H3O of the buffer,
  or with a pKa value near the desired pH.
• You can then use the Henderson-Hasselbalch
  equation (or Ka expression) to determine the
  relative concentration of base to acid needed to
  prepare the buffer.

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Preparing a Buffer

• Choose an appropriate acid and base to make a
  buffer with a pH of 6.50. Calculate the relative
  concentration of acid and base needed.

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Acid-Base Titrations

• Titration is a laboratory technique in which
  the amount and concentration of one reactant is
  known, and the concentration of the other
  reactant is determined.
• Titrations can be applied to redox reactions,
  and precipitation reactions, but they are most
  commonly used for analyzing solutions of acids or
  bases.

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Acid-Base Titrations

• Acid-base titrations are based on
  neutralization reactions, in which an acid is
  completely neutralized by a base. The progress
  of the reaction may be monitored using a pH
  meter, or pH indicators.

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Titrations

• In either method, base is added to the acid (or
  vice versa) until the equivalence point is
  reached.
• The equivalence point has been reached when the
  moles of acid exactly equals the moles of base.
  It is sometime called the stoichiometric point.

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Strong Acid/Strong Base Titration
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Titrations

• Not all titrations reach the equivalence point
  at a pH of 7. The nature of the acid and base
  (weak or strong) will influence the pH at the
  equivalence point. As a result, indicators much
  be carefully chosen to change color at the
  desired pH.

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Strong Acid Strong Base

• When a strong acid is titrated with a strong
  base, a neutral salt and water result. Since
  the conjugate base of a strong acid has no
  tendency to accept protons, and the conjugate
  acid of a strong base has no tendency to donate
  protons, the pH at the equivalence point will be
  7.

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Strong Acid Strong Base

• If the pH is monitored during the titration
  using a pH meter, a graphical presentation of the
  pH versus volume of titrant can be obtained.

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Strong Acid Strong Base

• The graph has certain features characteristic
  of a strong acid-strong base titration.

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Strong Acid Strong Base

• Note the equivalence point of 7, and the nearly
  vertical rise in pH from 4-10.

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Strong Acid Strong Base

• Because the rise in pH is vertical over such a
  broad range, there are several indicators that
  may be used with accurate results.

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Strong Acid Strong Base

• Phenolphthalein, which changes color at a pH of
  around 9, and methyl red, which changes color at
  a pH of approximately 5 will both give highly
  accurate results.

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Strong Acid Strong Base

• Phenolphthalein is usually used because it goes
  from colorless to pink, and our eyes are better
  at detecting the presence of color than a change
  in color.

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Weak Acid Strong Base

• When a weak acid is titrated with a strong
  base, the salt that is produced will contain the
  conjugate base of the weak acid. As a result,
  the pH will be greater than 7 at the equivalence
  point.

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Weak Acid Strong Base

• When a weak acid is titrated with a strong
  base, the salt that is produced will contain the
  conjugate base of the weak acid. As a result,
  the pH will be greater than 7 at the equivalence
  point.

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Weak Acid Strong Base

• The titration curve for acetic acid with NaOH
  shows an equivalence point at a pH of 9. The
  lack of a steep vertical rise in pH means that
  selection of the proper indicator is crucial.

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Weak Acid Strong Base

• Another feature of the curve is the flattening
  out or leveling off of pH near the half-way
  point. This is not observed in strong
  acid-strong base titrations.

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Weak Acid Strong Base

• Another feature of the curve is the flattening
  out or leveling off of pH near the half-way
  point. This is not observed in strong
  acid-strong base titrations.

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Weak Acid Strong Base

• The pH levels off due to the formation of a
  buffer mid-way through the titration. Half of
  the acid has been de-protonated to form its
  conjugate base.

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Titration of Benzoic Acid

• 25.0 mL of 0.10M benzoic acid (Ka 6.4 x 10-5)
  is titrated with 0.10M NaOH. Calculate the pH
• - Initially
• - After 12.5 mL of NaOH have been added
• - At the equivalence point
• Choose an appropriate indicator for the
  titration.

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Titration of Benzoic Acid

• 25.0 mL of 0.10M benzoic acid (Ka 6.4 x 10-5)
  is titrated with 0.10M NaOH.
• We will need to know how much NaOH is needed
  for complete neutralization. In this case, since
  the concentrations of acid and base are the same,
  25.0 mL of 0.10M acid will require an equal
  volume of 0.10M base.

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Calculating Volume for Neutralization

• In less obvious cases, the following
  relationship can be used. We are assuming
  complete neutralization of a monoprotic acid by a
  monobasic base.
• moles acid moles base
• MacidVacid MbaseVbase

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Titration of Benzoic Acid

• 25.0 mL of 0.10M benzoic acid (Ka 6.4 x 10-5)
  is titrated with 0.10M NaOH. Calculate the pH
• - Initially
• Major Reaction
• HA(aq) H2O(l) ? H3O(aq) A-(aq)
• init 0.10 0 0
• chg -x x x
• equ. .10-x x x

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Titration of Benzoic Acid

• 25.0 mL of 0.10M benzoic acid (Ka 6.4 x 10-5)
  is titrated with 0.10M NaOH. Calculate the pH
• - After 12.5 mL of NaOH have been added
• Titration Reaction
• HA(aq) OH-(aq) ? H2O(l) A-(aq)
• For each mole of NaOH added, a mole of HA has
  been converted to A-.

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Titration of Benzoic Acid

• 25.0 mL of 0.10M benzoic acid (Ka 6.4 x 10-5)
  is titrated with 0.10M NaOH. Calculate the pH
• - After 12.5 mL of NaOH have been added
• This problem can be greatly simplified if you
  realize that this is the mid-point of the
  titration. Since complete neutralization will
  require 25.0 mLs of NaOH, at 12.5 mL base, half
  of the acid has been converted to its conjugate
  base.

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Titration of Benzoic Acid

• 25.0 mL of 0.10M benzoic acid (Ka 6.4 x 10-5)
  is titrated with 0.10M NaOH. Calculate the pH
• - After 12.5 mL of NaOH have been added
• In essence, a buffer has been formed, with HA
  A-. (This is the place in the titration
  curve where the pH begins to level off.)

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Titration of Benzoic Acid

• 25.0 mL of 0.10M benzoic acid (Ka 6.4 x 10-5)
  is titrated with 0.10M NaOH. Calculate the pH
• - After 12.5 mL of NaOH have been added
• We can use the Henderson-Hasselbalch equation,
  because a buffer is present. Since HA A-,
  we need not calculate the actual concentration,
  as they will cancel out.

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Titration of Benzoic Acid

• 25.0 mL of 0.10M benzoic acid (Ka 6.4 x 10-5)
  is titrated with 0.10M NaOH. Calculate the pH
• - After 12.5 mL of NaOH have been added
• pH pKa log (A-/HA)
• pH -log (6.4 x 10-5) log (1.0)
• pH 4.19

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Titration of Benzoic Acid

• Calculate the pH at the equivalence point.
• At the equivalence point, enough NaOH has been
  added (25.0 mL) to completely neutralize the
  benzoic acid.
• There are two reactions to consider.
• 1. The neutralization reaction.
• 2. The subsequent reaction of the benzoate ion
  with water.

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Titration of Benzoic Acid

• At the equivalence point, benzoate ion will react
  with water, since it is the conjugate base of a
  weak acid.
• Major reactions
• HA(aq) OH-(aq) ? H2O(l) A-(aq)
• A-(aq) H2O(l) ? OH-(aq) HA(aq)

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Titration of Benzoic Acid

• - At the equivalence point, the solution will be
  basic, due to the high concentration of benzoate
  ion.
• Major Reaction
• A-(aq) H2O(l) ? OH-(aq) HA(aq)
• The A- equals the number of moles of HA
  initially present divided by the total volume of
  solution at the equivalence point.

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Titration of Benzoic Acid

• Major Reaction
• A-(aq) H2O(l) ? OH-(aq) HA(aq)
• The A- equals the number of moles of HA
  initially present divided by the total volume of
  solution at the equivalence point.
• A- 25.0 mL(0.10M)/50.0 mL 0.050M

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Titration of Benzoic Acid

• Major Reaction
• A-(aq) H2O(l) ? OH-(aq) HA(aq)
• init. 0.050 0 0
• chg. -x x x
• equil .050-x x x
• Since the reaction involves a conjugate base
  reacting with water, we need to write the Kb
  expression, and calculate the value of Kb.

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Titration of Benzoic Acid

• Major Reaction
• A-(aq) H2O(l) ? OH-(aq) HA(aq)
• Since the reaction involves a conjugate base
  reacting with water, we need to write the Kb
  expression, and calculate the value of Kb.
• Kb OH-HA Kw
• A- Ka
• Kb 1.0x 10-14/ 6.4 x 10-5 1.6 x 10-10

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Titration of Benzoic Acid

• Major Reaction
• A-(aq) H2O(l) ? OH-(aq) HA(aq)
• init. 0.050 0 0
• chg. -x x x
• equil .050-x x x
• 1.6 x 10-10 (x)(x)/(.050-x)

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Titration of Benzoic Acid

• The pH at the equivalence point will be 8.45.
• Selection of a indicator must be made
  carefully, as there isnt a steep vertical rise
  in pH near the end point when titrating a weak
  acid.

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Acetic Acid/NaOH Titration
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Titration of Weak Acids

• As the acid being titrated gets weaker, the
  titration curve flattens out, and the end point
  becomes less obvious.

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Titration of Weak Acids

• Very weak acids cannot be analyzed using
  titration.

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Titration of a Weak Base

• NH3(aq) HCl(aq) ? NH4(aq) Cl-(aq)
• The titration of ammonia with a strong acid
  will produce a solution of ammonium ion. Since
  ammonium ion is the conjugate acid of a weak
  base, the pH at the equivalence point will be
  less than 7.

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Titration of a Weak Base

• Again, selection of the proper indicator is
  crucial.

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Ammonia/HCl Titration