Chapter 18: Equilibria in Solutions of Weak Acids and Bases

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Chapter 18 Equilibria in Solutions of Weak Acids
and Bases

• All weak acids behave the same way in aqueous
  solution they partially ionize
• In terms of the general weak acid HA, this can
  be written as
• Following the procedures in Chapter 16

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• Ka is called the acid ionization constant
• These are often reported as the pKa
• Table 18.1 and Appendix C list the Ka and pKa for
  a number of acids
• A large pKa,means a small value of Ka and
  only a small fraction of the acid molecules
  ionize
• A small pKa,means a large value of Ka and a
  large fraction of the acid molecules ionize

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• Weak bases behave in a similar manner in water
• For the general base B

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• Values of Kb and pKb for a number of weak bases
  are listed in Table 18.2 and in Appendix C
• Where, like for acids
• There is an interesting relations ship between
  the acid and base ionizations constants for a
  conjugate acid-base pair
• Using the general weak acid HA

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• Thus, for any conjugate acid-base pair
• Most tables of ionization constants only give
  values for the molecular member of the conjugate
  acid-base pair
• The ionization constant of the ion member of the
  conjugate acid-base pair is then calculated as
  needed

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Relative strengths of conjugate acid-base pairs.
The stronger the acid is, the weaker the
conjugate base. The weaker the acid, the stronger
the conjugate base. Very strong acids ionize 100
and their conjugate bases do not react to any
measurable extent.

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• The primary goal is usually to determine the
  equilibrium concentration for all species in the
  mass action expression
• The percentage ionization of the acid or base is
  defined as
• This, and the pH, are often used or requested in
  equilibrium calculations

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• Example Morphine is very effective at relieving
  intense pain and is a weak base. What is the Kb,
  pKb, and percentage ionization of morphine if a
  0.010 M solution has a pH of 10.10?
• ANALYSIS The reaction can be represented as
• At equilibrium, OH- x 10-pOH

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• SOLUTION Use pOH 14.00 pH, substituting

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Determining how to proceed in acid-base
equilibrium problems. The nature of the solute
species determines how the problem is approached.
This flowchart can help get you started in the
right direction.
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• As in Chapter 16, if the extent of reaction (x)
  is small it is possible to simplify the mass
  action expression
• For the case of a weak acid or base added to pure
  water
• If the initial solute concentration is at least
  400 times larger than the ionization constant,
  the initial concentration of the solute can be
  used as though they were the equilibrium
  concentration
• If the solute concentration is too small, or the
  equilibrium constant too large, then the
  quadratic equation must be used

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• Ions can also be acids or bases
• For example, NH4 is a weak acid and NO2- a weak
  base
• In salts both anions and cations are present,
  either of which could affect the pH
• These ions can arise from a number of sources
• Properties of their aqueous solutions can be
  summarized

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• Aqueous cations
• Cations that are conjugate acids of weak
  molecular bases are weak acids
• Metal cations with a high charge density (like
  Al3, Fe3, and Cr3) yield aqueous solutions
  that are acidic
• Aqueous anions
• The anion of a strong acid is too weak a base to
  influence the pH of a solution
• Anions of weak acids tend to make solution basic

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• There are four possibilities when a salt is added
  to pure water
• Neither the anion nor cation affects the pH and
  the solution remains neutral. For example NaCl
• Only the cation is acidic, so the solution
  becomes acidic. For example NH4Cl
• Only the anion is basic, so the solution becomes
  basic. For example NaNO2
• The anion is basic and the cation is acidic, the
  pH of the solution will be determined by the
  relative strengths of the acid and base. For
  example NH4NO2 produces an acidic solution and
  NH4OCl produces a basic solution

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• The quadratic equation can be used to solve
  equilibrium problems when simplifying assumptions
  are invalid
• Simplifying assumptions fail, for example, when
  the initial concentration of a weak acid or base
  in pure water is less that 400 times the
  ionization constant
• Solving the problem using the quadratic equation
  is more time consuming, so it is worth checking
  before it is used

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• Example Calculate the pH of a 0.0010 M solution
  of dimethylamine for which Kb9.6x10-4.
• ANALYSIS 400 Kbgt0.0010 M, so use of the
  quadratic equation is indicated.
• SOLUTION Set the problem up

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• Put in standard form
• Solve for x and the equilibrium concentrations

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• When a small amount of strong acid or base is
  added to certain solutions, only a small change
  in pH is observed
• These solutions are called buffers
• Buffer solution usually contains two solutes, one
  providing a weak acid and the other a weak base
• If the weak acid is molecular, then the conjugate
  base can be supplied as a soluble salt of the acid

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• Buffers work because the weak acid can react
  with added base and the weak base can react with
  added acid
• Consider the general buffer made so that both HA
  and A- are present in solution
• When base (OH-) is added
• When acid (H) is added
• Net result small changes in pH

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• Because of these reactions, calculations
  involving buffer solutions can be greatly
  simplified
• For buffer solutions, the initial concentration
  of both the weak acid and its conjugate base can
  be used as though they were equilibrium values
  (more complicated buffer systems where this
  assumption is not valid will not be encountered
  in this text)
• For buffer solutions only, either molar
  concentrations or moles can be used in the Ka (or
  Kb) expression to express the amounts of the
  members of the conjugate acid-base pair (the same
  units must be used for both members of the pair)

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• Example What is the pH of a buffer made by
  adding 0.10 mol NH3 and 0.11 mol NH4Cl to 2.0 L
  of solution? The Kb for ammonia is 1.8x10-5
• ANALYSIS This is a buffer, initial
  concentrations can be used as equilibrium values

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• SOLUTION Solve for OH- and use this to
  calculate the pH
• There are two important factors that determine
  the pH of a buffer solution

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• For the general weak acid HA
• Thus both the value of Ka and the ratio of the
  molarities (or the ratio of moles) affect the pH
• These last two relations are often expressed in
  logarithmic form

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• The first is called the Henderson-Hasselbalch
  equation, and is frequently encountered in
  biology courses
• When preparing a buffer, the concentration ratio
  is usually near 1, so the pH is mostly determined
  by the pKa of the acid

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• Typically, the weak acid is selected so the the
  desired pH is within one unit of the pKa
• A buffers capacity is determined by the
  magnitudes of the molarities of its components
• Generally, the pH change in an experiment must be
  limited to about

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• Example A buffer made from 0.10 mol HA
  (pKa7.20) and 0.15 mol NaA in 2.0 L has 0.02 mol
  of HCl added to it with no volume change. What is
  the pH change?
• ANALYSIS This buffer problem is best solved in
  terms of moles. HCl is a strong acid. The H it
  contributes to solution increases the amount of
  HA present at the expense of A-.
• SOLUTION The pH before addition of HCl was

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• After the HCl ionizes and reacts
• The pH change was greater than 0.1 unit. The
  buffer effectively resisted the pH change,
  however, because if the HCl had been added to
  pure water, the pH change would have been much
  larger

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• Acids that can donate more than one H to
  solution are called polyprotic acids
• Table 18.3 (and Appendix C) list the ionization
  constants of a number of polyprotic acids
• The ionization constants for these acids are
  numbered to keep tract of the degree of
  ionization
• Note that, for a given polyprotic acid, the
  magnitudes of the ionization constants are
  always Ka1 gt Ka2 (gt Ka3, if applicable)

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• Simplifying assumptions like those made for the
  acid ionization can be made
• The overall procedure for an acid-base titration
  was discussed in Section 5.13
• The titration stops at the end point, as
  indicated by a color change of an acid-base
  indicator

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• If the indicator was properly selected, the end
  point is very near the equivalence point (when
  stoichiometric amounts of acid and bases have
  been added) for the system
• If the pH of the solution is plotted versus
  volume of titrant added, a titration curve is
  obtained
• The pH of the solution can easily be obtained
  using a pH meter
• Some of the more important types of titrations
  will be considered

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• Titration of a strong acid by a strong base

Titration curve for the titration of 25.00 mL of
0.2000 M HCl (a strong acid) with the 0.2000 M
NaOH (a strong base). The equivalence point
occurs at 25.00 mL added base with a pH of 7.0
(data from Table 18.4).
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• Titration of a weak acid by a strong base
• This can be divided into four regions
• Before the titration begins this is simple a
  solution of weak acid
• During the titration, but before the equivalence
  point the solution is a buffer
• At the equivalence point the solution contains a
  salt of the weak acid, and hydrolysis can occur
• Past the equivalence point the excess added OH-
  is used to determine the pH of the solution
• Data for the titration of acetic acid with sodium
  hydroxide is tabulated in Table 18.5

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The titration curve for the titration of 25.00 mL
of 0.200 M acetic acid with 0.200 M sodium
hydroxide. Due to hydrolysis, the pH at the
equivalence point higher than 7.00 (data from
Table 18.5).
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• Titration of a weak base by a strong acid
• This is similar to the titration of a weak acid
  by strong base
• Again dividing into four regions
• Before the titration begins this is a solution
  of a weak base in water
• During the titration, but before the equivalence
  point the solution is a buffer
• At the equivalence point the solution contains
  the salt of the weak base, and hydrolysis can
  occur
• Past the equivalence point excess added H
  determines the pH of the solution

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Titration curve for the titration of 25.00 mL of
0.200 M NH3 with 0.200 M HCl. The pH at the
equivalence point is below 7.00 because of the
hydrolysis of NH4.
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• Titration curves for diprotic acids
• The features are similar to those for monoprotic
  acids, but two equivalence points are reached

The titration of the diprotic acid H2A by a
strong base. As each equivalence point is
reached, the pH rises sharply.
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• A few general comments about indicators can be
  made
• Most dyes that are acid-base indicators are weak
  acids, which can be represented as HIn
• The color change can be represented as

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• The color change will appear to the human eye
  near the equivalence point of the indicator
• At the equivalence point, the concentration of
  the acid and base form are equal, so that
• The best indicators have intense color(s) so only
  a small amount will produce an intense color
  change that is easy to see and wont consume
  too much of the titrant