Applications of Aqueous Equilibria

1
Applications of Aqueous Equilibria
2
Solutions of Acids or Bases Containing a Common
Ion
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Solution containing weak acid HA and its salt NaA

• Salt dissolves in water and breaks up completely
  into its ions-it is a strong electrolyte
• NaA(s) ? Na(aq) A-(aq)

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Common Ion Effect

• When AgNO3 is added to a saturated solution of
  AgCl, it is often described as a source of a
  common ion, the Ag ion.
• By definition, a common ion is an ion that enters
  the solution from two different sources.
• Solutions to which both NaCl and AgCl have been
  added also contain a common ion in this case,
  the Cl- ion.
• There is an effect of common ions on solubility
  product equilibria.

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• The common-ion effect can be understood by
  considering the following question What happens
  to the solubility of AgCl when we dissolve this
  salt in a solution that is already 0.10 M NaCl?
• As a rule, we can assume that salts dissociate
  into their ions when they dissolve.
• A 0.10 M NaCl solution therefore contains 0.10
  moles of the Cl- ion per liter of solution.
• Because the Cl- ion is one of the products of the
  solubility equilibrium, LeChatelier's principle
  leads us to expect that AgCl will be even less
  soluble in an 0.10 M Cl- solution than it is in
  pure water.

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• Calculate the solubility of AgCl in 0.10 M NaCl.
• In pure waterCs 1.3 x 10-5 M
• In 0.10 M NaCl Cs 1.8 x 10-9 M
• These calculations show how the common-ion effect
  can be used to make an "insoluble" salt even less
  soluble in water.

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• The common ion effect can be applied to other
  equilibria, as well. Consider what happens when
  we add a generic acid (HA) to water. We now have
  two sources of a common ion ? the H3O ion.
• HA(aq) H2O(l) !H3O(aq) A-(aq)
• 2 H2O(l)! H3O(aq) OH-(aq)
• Thus, it isn't surprising that adding an acid to
  water decreases the concentration of the OH- ion
  in much the same way that adding another source
  of the Ag ion to a saturated solution of AgCl
  decreases the concentration of the Cl- ion.

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• The solubility of a solid is lowered if the
  solution already contains ions common to the
  solid
• Dissolving silver chloride in a solution
  containing silver ions
• Dissolving silver chloride in a solution
  containing chloride ions
• The common-ion effect can also be used to prevent
  a salt from precipitating from solution.
• Instead of adding a source of a common ion, we
  add a reagent that removes the common ion from
  solution.

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• Ksp (Solubility Product Constant, Solubility
  Product)
• Ksp Ca2F-2
• Experimentally determined solubility of an ionic
  solid can by used to calculate its Ksp value
• The solubility of an ionic solid can be
  calculated if its Ksp value is known
• Relative Solubilities
• IF the salts being compared produce the same
  number of ions in solution, Ksp can be used to
  directly compare solubility
• NaCl(s), KF(s)
• Ksp cationanion x2
• IF the salts being compared produce different
  numbers of ions, Ksp cannot be directly compared
• Ag2S(s)
• Ksp 2x2x
• Bi2S3(s) 2x23x3

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Common Ion-ion produced by both acid and its salt

• Ion provided in solution by an aqueous acid (or
  base) as well as a salt
• HF(aq) and NaF (F- in common)
• HF(aq) ? H(aq) F-(aq)
• NaF(s) (in water)? Na(aq) F-(aq)
• Excess F- added by NaF
• Equilibrium shifts away from added component.
• Fewer H ions present, making solution less
  acidic
• pH is higher than expected.
• NH4OH and NH4Cl (NH4 in common)
• NH3(aq) H2O(l) ? NH4(aq) OH-(aq)
• NH4Cl(s) (in water)? NH4(aq) Cl-(aq)
• Equilibrium shifts to the left.
• pH of solution decreases due to decrease in OH-
  conc.
• Common ion effect-shift in equilibrium position
  that occurs because of the addition of an ion
  already involved in the equilibrium reaction

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Equilibrium Calculations

• Consider initial concentration of ion from salt
  when calculating values for H and OH-
• Calculate the pH of a solution that contains 0.10
  M HC2H3O2 and 0.050 M NaC2H3O2. (The pH of 0.10
  M HC2H3O2 is 2.9. The addition of the common ion
  would shift the equilibrium to the left. As a
  result the H3O would decrease and the pH would
  rise.)

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• Ka H3OC2H3O2-/HC2H3O2
• 1.8 x 10-5 x(0.050 x)/0.10 x 0.050x/0.10
• x H3O 3.6 x 10-5 M
• pH -logH3O -log(3.6 x 10-5) 4.4
• as we predicted, the pH rose from 2.9 to 4.4.

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Buffered Solutions
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Until now, we have considered solutions
containing only pure weak acids, weak bases, or
their salts dissolved in distilled water

• A buffer is a solution with a very stable pH
• You can add acid or base to a buffer solution
  without greatly affecting the pH of the solution
• The pH of a buffer will also remain unchanged if
  the solution is diluted with water or if water is
  lost through evaporation
• A mixture that contains a conjugate acid-base
  pair is known as buffer solution because the pH
  changes by a relatively small amount if a strong
  acid or base is added to it
• Buffer systems are used to control pH in many
  biological and chemical reactions

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A buffer is created by placing a large amount of
a weak acid or base into a solution along with
its conjugate

• A weak acid and its conjugate base (B-) will
  remain in solution together without neutralizing
  each other
• A weak base and its conjugate acid (A) will do
  the same

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When both the acid and the conjugate base are
together in the solution, any hydrogen ions that
are added will be neutralized by the base while
any hydroxide ions that are added will be
neutralized by the acid without having much of an
effect on the solutions pH

• A buffer system is a chemical sponge for H3O
  or OH- ions that may be produced within a
  solution
• When additional H3O ions are added, they react
  with the conjugate base component of the buffer
  system to produce H2O and a weak acid that is
  only slightly ionized, so the pH drops only
  minimally
• H3O(aq) conjugate base-(aq) ? weak acid(aq)
  H2O(l)
• When additional OH- ions are added, the basic OH-
  ions react with the acid component of the buffer
  system to produce H2O and a weak base, so the pH
  rises only minimally
• OH-(aq) conjugate acid (aq) ? weak base-(aq)
  H2O(l)

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These solutions are governed by the same
equilibrium law as are weak acids and weak bases

• Buffered solutions-Equilibrium systems that
  resist changes in acidity and maintain constant
  pH when acids or bases are added to them
• Most effective pH range for any buffer is at or
  near the pH where the acid and salt
  concentrations are equal (pKa)
• H3O of weak acid/conjugate base buffer equals
  the Ka of the weak acid
• OH- of weak base/conjugate acid buffer equals
  the Kb of the weak base
• In general the most effect pH range of a buffer
  system is (optimum pH ? 1.0 pH unit

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• Since Ka for a weak acid is small, equilibrium
  concentrations of the weak acid and its conjugate
  base are very nearly their initial concentrations
• All weak acid-conjugate base buffer systems have
  general expression
• H3O Ka x weak acid0/conjugate base0
• Convert the buffer relationship into its
  logarithmic form
• pH for buffer-pH pKa logA-/HA pKa log
  conjugatebase0/weak acid0-obtained from
  equation for weak acid equilibrium (Ka
  HA-/HA)
• this relationship is the Henderson-Hasselbalch
  equation
• OH- Kb x weak base0/conjugate acid0
• pOH pKb log x conjugate acid0/weak base0
• Buffered solutions contain either
• A weak acid and its salt
• A weak base and its salt

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Preparing a buffer

• A buffer can be prepared by dissolving 0.200 mole
  of HF and 0.100 mole of NaF in water to make 1.00
  liter of solution
• Dissociation reaction of HF is
• HF(aq) ? F-(aq) H(aq)
• Equilibrium law is
• Ka F-H/HF 6.8 x 10-4
• Set up an equilibrium table using the given
  information

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• Substituting the information from the equilibrium
  line into the equilibrium law yields
• Ka (0.100 x)(x)/0.200 x 6.8 x 10-4
• Assuming that x is small compared to both 0.100
  and 0.200, we simplify the equation to
• Ka (0.100)(x)/0.200 6.8 x 10-4
• x 1.36 x 10-3
• This value of x satisfies the assumptions made,
  and the last line of the equilibrium table can be
  completed
• The last column gives us H 1.36 x 10-3 M
• The pH of the buffer solution is calculated as
  2.87
• For almost all buffer solutions the assumptions
  hold true, and this type of problem is solved by
  simply entering the given concentrations of the
  conjugate acid and conjugate base directly into
  the equilibrium law

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Calculate the pH of the following buffer solutions

• 0.250 M acetic acid and 0.150 M sodium acetate
• Ka C2H3O2-H/HC2H3O2
• 1.8 x 10-3 (0.150)H/0.250
• H 3.0 x 10-3, and pH 2.52
• A solution of 10.0 g each of formic acid and
  potassium formate dissolved in 1.00 L of H2O
• Ka CHO2-H/HCHO2
• 1.8 x 10-4 (0.417)H/0.217
• H 9.4 x 10-5, and pH 4.03

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• 0.0345 M ethylamine and 0.0965 M ethyl ammonium
  chloride
• Kb C2H5NH3OH-/C2H5NH2
• 4.3 x 10-4 (0.0965)OH-/0.0345
• OH- 1.5 x 10-4, pOH 3.82 and pH 10.18
• 0.125 M hydrazine and 0.321 M hydrazine
  hydrochloride
• Kb N2H5OH-/N2H4
• 9.6 x 10-7 (0.321)OH-/0.125
• OH- 3.7 x 10-7, pOH 6.43 and pH 7.57

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Shortcuts in buffer calculations

• Equilibrium law used for buffers involves a ratio
  of the concentrations of the conjugate acid and
  conjugate base
• We may use the moles of conjugate acids and moles
  of conjugate base instead of the acid and base
  molarities
• If the concentration of the conjugate acid and
  conjugate base are equal, their ratio is exactly
  1.00
• With equal molarities or equal numbers of moles
  of conjugate acid and conjugate base
• Ka H and pKa pH for a buffer made from a
  weak acid and its conjugate base
• Kb OH- and pKb pOH for a buffer made from a
  weak base and its conjugate acid

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pH changes in buffers

• Addition of a strong acid to a buffer will
  decrease the pH slightly
• Addition of a strong base to a buffer will
  increase the pH slightly
• To determine the amount that the pH changes when
  a strong acid or base is added to a buffer, we
  must calculate the change in concentration of the
  conjugate acid or base in the buffer

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• Steps
• Determine either the molarity or the number of
  moles of the conjugate acid and conjugate base in
  the original buffer
• Determine the amount of strong acid or base added
  in the same units as the conjugate acid and base
  in step 1
• If a strong acid is added to the buffer, add the
  value from step 2 to the conjugate acid and
  subtract it from the conjugate base
• If a strong base is added to the buffer, add the
  value from step 2 to the conjugate base and
  subtract it fro the conjugate acid
• Substitute the new values for the conjugate acid
  and conjugate base into the equilibrium law and
  calculate the pH as shown before

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• In an acetate buffer, acetic acid, HC2H3O2, is
  the conjugate acid, and the acetate ion, C2H3O2-,
  is the conjugate base
• Addition of a strong acid increases the
  concentration of acetic acid and decreases the
  concentration of acetate ions.
• Addition of a strong base to this buffer
  increases the acetate ion concentration and
  decreases the acetic acid concentration
• HC2H3O2 OH- ? C2H3O2-
• An acetate buffer is prepared with 0.250 M acetic
  acid and 0.100 M NaC2H3O2. If 0.002 mol of solid
  NaOH is added to 100 mL of this buffer, calculate
  the change in pH of the buffer due to the
  addition of the NaOH.
• To calculate the change in pH, the pH of the
  original buffer is needed, along with the final
  pH after the base is added
• Dissociation reaction for acetic acid
• HC2H3O2 ? H C2H5O2-
• Equilibrium law
• Ka H C2H5O2-/ HC2H3O2
• Value for Ka and the acetate and acetic acid
  concentrations are substituted into the
  equilibrium law
• 1.8 x 10-5 H)(0.100)/0.250

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• Hydrogen ion concentration is calculated as
• H 4.5 x 10-5 M, and the pH is 4.35
• To calculate the pH of the buffer after the NaOH
  is added, we must first convert either the
  molarities of the conjugate acids and bases to
  moles or the moles of NaOH to molarity so that
  all the units are the same
• Molarity NaOH 0.00200 mol NaOH/0.100 L 0.0200
  M
• Adding this value to the acetate concentration
  gives us 0.120 M C2H3O2-
• Subtracting it from the acetic acid concentration
  yields 0.230 M HC2H3O2
• Substituting these new values into the
  equilibrium law
• 1.8 x 10-5 H(0.120)/0.230
• H 3.45 x 10-5 M
• pH 4.46
• There is an increase of 0.11 pH units when the
  NaOH is added
• The answer is reasonable since the pH is expected
  to rise slightly when a base is added

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Calculations Involving Buffered Solutions
Containing Weak Acids

• "Buffered solutions are simply solutions of weak
  acids or bases containing a common ion. The pH
  calculations on buffered solutions require
  exactly the same procedures introduced in Chapter
  14. This is not a new type of problem."
• When a strong acid or base is added to a
  buffered solution, it is best to deal with the
  stoichiometry of the resulting reaction first.
  After the stoichiometric calculations are
  completed, then consider the equilibrium
  calculations. This procedure can be presented as
  follows

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• OH- HA ? A- H2O
• weak acid conjugate base
• OH- ions are not allowed to accumulate but are
  replaced by A- ions

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• Ka HA- / HA
• H Ka x HA / A-
• If the amounts of HA and A- originally present
  are very large compared with the amount of OH-
  added, the change in HA/A- will be small.
  Therefore the pH change will be small

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• Buffering also works for addition of protons
  instead of hydroxide ions
• H A- ? HA
• Conjugate base Weak acid
• Buffering with a Weak Base and Its Conjugate Acid
• Weak base B reacts with any H added
• B H ? BH
• Base Conjugate acid
• Conjugate acid BH reacts with any added OH
• BH OH- ? B H2O
• Conjugate acid Base

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Summary

• Buffered solutions contain relatively large
  concentrations of a weak acid and the
  corresponding weak base. They can involve a weak
  acid HA and the conjugate base A- or a weak base
  and the conjugate acid BH
• When H is added to a buffered solution, it
  reacts essentially to completion with the weak
  base present
• When OH- is added to a buffered solution, it
  reacts essentially to completion with the weak
  acid present
• The pH in the buffered solution is determined by
  the ratio of the concentrations of the weak acid
  and the weak base. The pH remains relatively
  unchanged as long as the concentrations of
  buffering materials are large compared with the
  amounts of H or OH- added

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Buffer Capacity
38
Buffering Capacity

• The amount of protons or hydroxide ions the
  buffer can absorb without a significant change in
  pH.
• The pH of a buffered solution is determined by
  the ratio A-/HA
• The capacity of a buffered solution is determined
  by the magnitudes of HA and A-

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Preparing a Buffer

• Preparation of a buffer starts with the selection
  of the desired pH
• Then a table of Ka and Kb values for weak acids
  and bases is consulted to find an appropriate
  conjugate acid-base pair to use
• These 2 steps determine the ratio of the salt to
  the weak acid or base
• Next, the moles of acid or base that need to be
  buffered are estimated
• The total number of moles of the conjugate
  acid-base pair should be at least 20 times the
  amount of the acid or base that needs to be
  buffered
• The volume of buffer solution needed is
  determined next

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• Method for preparing the buffer is decided on
• Conjugate acid and conjugate base are measured
  and dissolved to the desired volume
• Measuring the desired amount of conjugate acid
  and then adding the appropriate amount of a
  strong base to convert some of the conjugate acid
  into its conjugate base
• Conjugate base is measured and the necessary
  amount of strong acid is added to convert some of
  the conjugate base into its conjugate acid
• Optimal buffering occurs when HA is equal to
  A- ( A-/HA 1 )
• The pKa of the weak acid to be used should be as
  close as possible to the desired pH

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Titration

• Titration technique used for chemical analysis
  utilizing reactions of 2 solutions

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• One reactant solution is placed in a beaker and
  the other in a buret (long, graduated tube with
  stopcock)
• The stopcock (valve that allows chemist to add
  controlled amounts of solution from buret to
  beaker)
• An indicator is added to solution in beaker
  (substance which undergoes a color change in the
  pH interval of the equivalence point)

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• Chemist reads volume of solution in buret at
  start of experiment and again at point where
  indicator changes color
• Difference in these volumes represents the volume
  of reactant delivered from the buret
• Controlled addition of a solution of known
  concentration (titrant) in order to determine
  concentration of solution of unknown
  concentration
• The crucial point about the titration experiment
  is that the indicator is designed to change color
  when the amount of reactant delivered from the
  buret is exactly the amount needed to react with
  the solution in the beaker (Equivalence Point
  (Stoichiometric Point)-point in a titration at
  which the reaction between titrant and unknown
  has just been completed.)
• Classic chemical reaction-Fe2 and permanganate
  ion MnO4-
• 5Fe2 MnO4- 8H ? Mn2 5Fe3 4H2O
• The purple permanganate ion is the indicator of
  the point where the correct amount has been added
  to completely react all of the Fe2 ions in the
  sample

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Titration Curve (pH Curve)-plotting of the pH of
the solution as a function of the volume of
titrant added

• Monitors the pH of the solution as an acid-base
  titration proceeds from beginning to well beyond
  the equivalence point
• pH is plotted on the y-axis and volume of base
  (or acid) delivered is plotted on the x-axis
• Shape of titration curve makes it possible to
  identify the equivalence point and is an aid in
  selecting an appropriate indicator
• It can also be used to calculate the Ka or Kb of
  a weak acid or base

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• There are four major points of interest during a
  titration experiment
• Start of the titration, where the solution
  contains only one acid or base
• pH is calculated as previously described
• Region where titrant is added up to the end
  point, and the solution now contains a mixture of
  unreacted sample and products
• Mixture of a conjugate acid and its conjugate
  base
• If weak acid or base is involved, this is a
  buffer solution
• If only strong acids and bases are used, this
  region is unbuffered
• End point, where all the reactant has been
  converted into product
• Solution is the salt of the acid or base
• pH is calculated as previously described
• titration curve used to determined pH during
  titration
• Region after end point, where solution contains
  product and excess titrant
• pH depends on excess titrant used

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• Points on titration curve
• Midpoint of most vertical part of graph
  corresponds to exact endpoint.
• This will also correspond to the equivalence
  point, or the point at which the equivalents of
  acid equals the equivalents of base.
• In addition, the midpoint will also determine the
  pH of the salt that was formed during the
  titration.

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• Half-equivalence point is at the center of the
  buffer region
• pH increases more quickly at first, then levels
  out into buffer region
• At the half-equivalence point, enough base has
  been added to convert exactly half of the acid
  into conjugate base
• The concentration of the acid is equal to the
  concentration of the conjugate base
• The curve remains fairly flat until just before
  the equivalence point, when the pH increases
  sharply

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• Strong Acid-Strong Base Titration-titration curve
  can be divided into three segments
• Prior to the equivalence point (0-x-amount at
  equivalence point)
• The pH rises slowly at first because the solution
  contains a sizeable excess of H3O ions
  (buffering effect created by relatively high
  concentration of H in solution)
• H (thus the pH) can be calculated
• We begin to see significant rise in pH close to
  the equivalence point
• At the equivalence point (x)
• All excess H3O ions have been removed
• H3O is now 1.000 x 10-7 M, which corresponds
  to pH of 7.000
• Beyond the equivalence point (x total amount
  used)
• Once the endpoint has been passed, the rate of pH
  change diminishes again-resembles first part of
  graph except at higher pH
• The solution now contains an excess of OH- ions
  and as pH rises, it approaches that of the
  titrant
• OH- can be calculated from excess OH- and
  volume of solution
• H and pH can be calculated from OH- and Kw

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Strong Base with Strong Acid
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• Weak Acids-Strong Bases Titration-segments of
  titration curve
• Before titrant added
• pH of solution from given
• Prior to the equivalence point (0-x)
• As NaOH is added below the equivalence point
• HC2H3O2(aq) OH-(aq) ? C2H3O2-(aq) H2O(l)
• Solution is now a buffer system that contains
  both the weak acid (HC2H3O2) and its conjugate
  base (C2H3O2-)
• At the point of half-neutralization, HC2H3O2
  C2H3O2- and H3O Ka
• At the equivalence point (x)
• All of the HC2H3O2 has been exhausted and the
  total volume is now double the original volume of
  the base
• The equivalence point corresponds to the
  pH-always gt 7
• The stronger the basic anion, the higher the pH
  of the equivalence point
• Beyond the equivalence point (x-total)
• Solution now contains excess of OH- ions and as
  the pH rises, it approaches that of the titrant
• If the acid being titrated is weak, then the
  graph will not be nearly as vertical at the
  endpoint.
• Weaker acid is, the more the graph deviates from
  being vertical.

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• Weak Bases-Strong Acids Titration
• Before HCl has been added-pH of solution
  calculated from given
• Prior to the equivalence point (0-x)
• As HCl is added below the equivalence point
• NH3(aq) H3O(aq) ? NH4(aq) H2O(l)
• The solution is now a buffer system that contains
  both the weak base (NH3) and its conjugate acid
  (NH4)
• At the point of half-neutralization, NH3
  NH4 and OH- Kb
• At the equivalence point (x)
• All of the NH3 has been exhausted and the total
  volume is now double the original volume of the
  acid
• The equivalence point corresponds to pH (always lt
  7.00)
• The conjugate acid of the weak base lowers the pH
• Beyond the equivalence point (x- total)
• The solution now contains an excess of H3O ions
  and as the pH falls, it approaches that of the
  titrant

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• Polyprotic Acid-Strong Base Titration
• When titrating a polyprotic acid, the graph will
  show a steep rise, or endpoint, for each of the
  protons in the acid.
• An acid with 2 protons will have two endponts,
  one for each hydrogen.
• An acid with 3 protons will have three endpoints,
  one for each proton.
• With each successive ionization, the
  characteristic shape of the titration curve
  becomes less distinct
• First endpoint is fairly obvious, the second
  endpoint is not as well defined, the third
  endpoint is worse still, and so on.

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• When a monoprotic acid is completely neutralized
  by a monohydroxide base, equal numbers of moles
  of H3O and OH- ions react
• H3O(aq) OH-(aq) ? 2H2O(l)
• At neutralization, nH30 nOH-
• M n/V
• N MV
• (MV)acid (MV)base
• When 25.0 mL of HCl is neutralized by 0.100 M
  NaOH, 35.0 mL of the base are consumed.
  Calculate the molarity of the acid.
• MVacid MVbase
• M(25.0 mL) (0.100 M)(35.0 M)
• Macid 0.140 M

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• If the acid contains more than one ionizable
  proton, and/or the base contains more than one
  dissociable hydroxide ion, then the equation
  given above must be modified in order to include
  these numbers
• (M x V x H3O)acid (M x V x OH-)base where
  indicates the number of acidic hydrogens or
  dissociable hydroxide ions in chemical formula
• What volume of 0.200 M H2SO4 is needed to be
  completely neutralized by 300. mL of 0.450 M KOH?
• MVacid MVbase
• (0.200 M)V(2) (0.450)(300 mL)(1)
• Vacid 338 mL

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Acid-Base Indicators
63
Determination of Equivalence Point

• Use a pH meter, find midpoint of vertical line in
  the titration curve
• Use of indicators
• Indicators used in titrations must have two
  important properties
• The pH at the end point of the titration curve
  must change by at least 2 pH units very rapidly
• The pK of the indicator must be close to the
  end-point pH of the titration
• Titration curve shows us required large change in
  pH often occurs
• Proper selection of an indicator mandates that
  the pH at the end point and the pKa of the
  indicator be close to each other

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How Indicators Work

• Indicators are weak acids and weak bases whose
  respective conjugate bases and conjugate acids
  have different colors
• The reason is that the loss or gain of a proton
  changes the energy of the electrons within their
  structures
• This, in turn, changes the energy of light
  absorbed, which is observed as a change in color

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• They change colors within a specified pH range
• If conjugate acid has 1 color, its conjugate base
  has another
• The eye will see these colors clearly only if
  there is 10x more of one color than of the other
• If acid is yellow/base is blue, we see yellow if
  conjugate acid is 10x more concentrated than
  conjugate base, and blue if conjugate base is 10x
  more conc. than conjugate acid
• conjugate base/conjugate acid lt 0.1 (yellow
  observed)
• conjugate base/conjugate acid gt 10 (blue
  observed
• Between these 2 ratios various shades of green
  observed

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• Since they are weak acids/bases, they undergo
  acid-base equilibrium
• HIn(aq) H2O(l) ? H3O In-(aq)
• KIn H3OIn-/In
• The species HIn is known as the acid form of the
  indicator and In- is known as the basic form of
  the indicator
• Each form has its own distinct color
• We can rearrange the equilibrium constant
  expression to yield
• H3O KIn x HIn/In-
• When HIn gt In-, indicator has color
  associated with acid form
• When In- gt HIn, indicator has color
  associated with basic form
• When HIn In-, color of indicator changes
• At this point, H3O KIn
• Ideally, this condition should represent end
  point of titration

67

• Since indicators are weak acids/bases,
  significance of this color phenomenon best shown
  w/Henderson-Hasselbalch equation
• pH pKa log(A-/HA) pKa
  log(base/acid)
• HA molar conc. of undissociated weak acid (M)
• A- molar conc. of conjugate base (M)
• pOH pKb log x HB/B
• B molar conc. of weak base
• HB molar conc. of conjugate acid
• Substituting 0.1 into the log term yields
• pH pKa log(0.1) pKa 1.0
• pH pKa log(10) pKa 1.0
• For a particular buffering system (acid-conjugate
  base pair), all solutions that have same ratio
  A-/HA will have same pH
• In order to select a suitable indicator for a
  titration, it is best if the pKIn lies within one
  pH unit of the equivalence point
• pKIn pHequivalence point ? 1

68

• We have a buffer solution with concentrations of
  0.20 M HC2H3O2 and 0.50 M C2H3O2-. The acid
  dissociation constant for HC2H3O2 is 1.8 x 10-5.
  Find the pH of the solution.
• pH pKa log x C2H3O2-/ HC2H3O2
• pH -log(1.8 x 10-8) log x (0.50 M)/(0.20)
• pH -log(1.8 x 10-8) log(2.5)
• pH (4.7) (0.40) 5.1
• If both concentrations are 0.20M
• pH pKa log x C2H3O2-/ HC2H3O2
• pH -log(1.8 x 10-8) log x (0.20 M)/(0.20)
• pH -log(1.8 x 10-8) log(1)
• pH (4.7) (0) 4.7
• Notice that when the concentrations of acid and
  conjugative base in solution are the same, pH
  pKa (and pOH pKb)
• When you choose an acid for a buffer solution, it
  is best to pick an acid with a pKa that is close
  to the desired pH
• That way you can have almost equal amounts of
  acid and conjugate base in the solution, which
  will make the buffer as flexible as possible in
  neutralizing both added H and OH-.

69
Indicators

• Indicator color changes will be sharp, occurring
  with the addition of a single drop of titrant
• Suitable indicators must be selected based on the
  equivalence point
• Strong acid-strong base titrations may use
  indicators with end points as far apart as pH 5
  and pH 9
• Titration of weak acids or weak bases requires
  more careful selection of an indicator with
  appropriate transition interval

70
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71
Solubility Equilibria and the Solubility Product
72
Why Do Some Solids Dissolve in Water?

• Sugar is a molecular solid, in which the
  individual molecules are held together by
  relatively weak intermolecular forces.
• When sugar dissolves in water, the weak bonds
  between the individual sucrose molecules are
  broken, and these C12H22O11 molecules are
  released into solution.

73

• It takes energy to break bonds between C12H22O11
  molecules
• It also takes energy to break H bonds in water
  that must be disrupted to insert one of these
  sucrose molecules into solution
• Sugar dissolves in water because energy is given
  off when slightly polar sucrose molecules form
  intermolecular bonds with the polar water
  molecules.
• The weak bonds that form between the solute and
  the solvent compensate for the energy needed to
  disrupt the structure of both the pure solute and
  the solvent.
• In the case of sugar and water, this process
  works so well that up to 1800 grams of sucrose
  can dissolve in a liter of water.

74

• Ionic solids (or salts) contain ()and (-) ions,
  which are held together by the strong force of
  attraction between particles w/opposite charges.
• When one dissolves in water, ions that form solid
  are released into solution, where they become
  associated w/polar solvent molecules.
• H2O
• NaCl(s) ? Na(aq) Cl-(aq)

75

• We can generally assume that salts dissociate
  into their ions when they dissolve in water.
• Ionic compounds dissolve in water if the energy
  given off when the ions interact with water
  molecules compensates for the energy needed to
  break the ionic bonds in the solid and the energy
  required to separate the water molecules so that
  the ions can be inserted into solution.

76
Solubility Equilibria-based on following
assumption

• When solids dissolve in water, they dissociate to
  give the elementary particles from which they are
  formed.
• Molecular solids dissociate to give individual
  molecules
• H2O
• C12H22O11(s) ? C12H22O11(aq)
• Ionic solids dissociate into solutions of ()/(-)
  ions they contain
• H2O
• NaCl(s) ? Na(aq) Cl-(aq)

77

• When the salt is first added, it dissolves and
  dissociates rapidly.
• The conductivity of solution therefore increases
  rapidly at first.
• Dissolve
• NaCl(s) ? Na(aq) Cl-(aq)
• Dissociate

78

• Concentrations of these ions soon become large
  enough that the reverse reaction starts to
  compete with forward reaction, which leads to a
  decrease in rate at which Na and Cl- ions enter
  solution.
• Associate
• Na(aq)Cl-(aq) ? NaCl(s)
• Precipitate
• Eventually, the Na and Cl- ion concentrations
  become large enough that the rate at which
  precipitation occurs exactly balances the rate at
  which NaCl dissolves.
• Once that happens, there is no change in the
  concentration of these ions with time and the
  reaction is at equilibrium.
• AT equilibrium, solution is saturated, because it
  contains the maximum conc. of ions that can exist
  in equilibrium w/solid salt.
• The amount of salt that must be added to a given
  volume of solvent to form a saturated solution is
  called solubility of salt.

79
Solubility Rules-patterns obtained from measuring
solubility of different salts and are based on
the following definitions of the terms

• A salt is soluble if it dissolves in water to
  give a solution with a concentration of at least
  0.1 moles per liter at room temperature.
• A salt is insoluble if the concentration of an
  aqueous solution is less than 0.001 M at room
  temperature.
• Slightly soluble salts give solutions that fall
  between these extremes.

80
The Solubility Product Expression

• Silver chloride is so insoluble in water (.0.002
  g/L) that a saturated solution contains only
  about 1.3 x 10-5 moles of AgCl per liter of
  water.
• H2O   
• AgCl(s) ? Ag(aq) Cl-(aq)
• Equilibrium constant expressions for this
  reaction gives following result.
• Water isn't included because it is neither
  consumed nor produced in this reaction, even
  though it is a vital component of the system.

81

• Ag and Cl- terms represent concentrations of
  the Ag and Cl- ions in moles per liter when this
  solution is at equilibrium.
• AgCl is more ambiguous.
• It doesn't represent conc. of AgCl dissolved in
  water- assume that AgCl dissociates into Ag/Cl-
  ions when it dissolves in water
• It can't represent amount of solid AgCl in
  system-equilibrium is not affected by the amount
  of excess solid added to the system.
• The AgCl term has to be translated quite
  literally as the number of moles of AgCl in a
  liter of solid AgCl.

82

• Concentration of solid AgCl calculated from
  density molar mass of AgCl.
• This quantity is a constant, however.
• The number of moles per liter in solid AgCl is
  the same at the start of the reaction as it is
  when the reaction reaches equilibrium.

83

• Since the AgCl term is a constant, which has no
  effect on the equilibrium, it is built into the
  equilibrium constant for the reaction.
• AgCl- Kc x AgCl
• This equation suggests that product of the
  equilibrium concentrations of Ag and Cl- ions in
  this solution is equal to a constant.
• Since this constant is proportional to solubility
  of the salt, it is called the solubility product
  equilibrium constant for reaction, or Ksp.
• Ksp AgCl-
• Ksp expression for a salt is product of the
  concentrations of the ions, with each
  concentration raised to power equal to
  coefficient of that ion in the balanced equation
  for the solubility equilibrium.

84
The Relationship Between Ksp And the Solubility
of a Salt

• Ksp is called the solubility product because it
  is literally the product of the solubilities of
  the ions in moles per liter.
• Solubility product of salt can be calculated from
  its solubility, or vs.

85

• Photographic films are based on the sensitivity
  of AgBr to light. When light hits a crystal of
  AgBr, a small fraction of the Ag ions are
  reduced to silver metal. The rest of the Ag ions
  in these crystals are reduced to silver metal
  when the film is developed. AgBr crystals that do
  not absorb light are then removed from the film
  to "fix" the image.
• Example Let's calculate the solubility of AgBr
  in water in grams per liter, to see whether AgBr
  can be removed by simply washing the film.
• We start with the balanced equation for the
  equilibrium.
• H2O   
• AgBr(s) ? Ag(aq) Br-(aq)
• We then write the solubility product expression
  for this reaction.
• Ksp AgBr- 5.0 x 10-13
• One equation can't be solved for two
  unknowns-Ag/ Br- ion conc.
• We can generate a second equation, however, by
  noting that one Ag ion is released for every Br-
  ion.
• Because there is no other source of either ion in
  this solution, concentrations of these ions at
  equilibrium must be the same.
• Ag Br-

86

• Substituting this equation into Ksp expression
  gives following result.
• Ag2 5.0 x 10-13
• Taking the square root of both sides of this
  equation gives the equilibrium concentrations of
  the Ag and Br- ions.
• Ag Br- 7.1 x 10-7M
• Once we know how many moles of AgBr dissolve in a
  liter of water, we can calculate the solubility
  in grams per liter.
• The solubility of AgBr in water is only 0.00013
  gram per liter. It therefore isn't practical to
  try to wash the unexposed AgBr off photographic
  film with water.

87

• Solubility product calculations with 11 salts
  such as AgBr are relatively easy to perform.
• In order to extend such calculations to compounds
  with more complex formulas we need to understand
  the relationship between the solubility of a salt
  and concentrations of its ions at equilibrium.
• Symbol Cs describes the amount of a salt that
  dissolves in water.

88
The Role of the Ion Product (Qsp) In Solubility
Calculations
89
Consider a saturated solution of AgCl in water.

•   H2O   
• AgCl(s) ? Ag(aq)Cl-(aq)
• Because AgCl is a 11 salt, the concentrations of
  the Ag and Cl- ions in this solution are equal.
• Saturated solution of AgCl in water
• Ag Cl-

90

• Imagine what happens when a few crystals of solid
  AgNO3 are added to this saturated solution of
  AgCl in water.
• According to the solubility rules, silver nitrate
  is a soluble salt.
• It therefore dissolves and dissociates into Ag
  and NO3- ions.
• As a result, there are two sources of the Ag ion
  in this solution.
• AgNO3(s) ?Ag(aq)NO3-(aq)     
•   H2O    
• AgCl(s) ? Ag(aq) Cl-(aq)
• Adding AgNO3 to a saturated AgCl solution
  therefore increases the Ag ion concentrations.
• When this happens, the solution is no longer at
  equilibrium because product of the concentrations
  of the Ag and Cl- ions is too large.
• In more formal terms, we can argue that the ion
  product (Qsp) for the solution is larger than the
  solubility product (Ksp) for AgCl.
• Qsp (Ag)(Cl-) gt Ksp

91
The ion product is literally the product of the
concentrations of the ions at any moment in time.

• When it is equal to the solubility product for
  the salt, the system is at equilibrium.
• The reaction eventually comes back to equilibrium
  after the excess ions precipitate from solution
  as solid AgCl.
• When equilibrium is reestablished, however, the
  concentrations of the Ag and Cl- ions won't be
  the same.
• Because there are two sources of the Ag ion in
  this solution, there will be more Ag ion at
  equilibrium than Cl- ions
• Saturated solution of AgCl to which AgNO3 has
  been added
• Ag gt Cl-

92
Now imagine what happens when a few crystals of
NaCl are added to a saturated solution of AgCl in
water.

• There are two sources of the chloride ion in this
  solution.
•   H2O   
• NaCl(s) ? Na(aq) Cl-(aq)   
•   
•   H2O   
• AgCl(s) ? Ag(aq)Cl-(aq)
• Once again, the ion product is larger than the
  solubility product.
• Qsp (Ag)(Cl-) gt Ksp
• This time, when the reaction comes back to
  equilibrium, there will be more Cl- ion in the
  solution than Ag ion.
• Saturated solution of AgCl to which NaCl has been
  added
• Ag lt Cl-

93
The figure below shows a small portion of the
possible combinations of the Ag and Cl- ion
concentrations in an aqueous solution.

• Any point along the curved line in this graph
  corresponds to a system at equilibrium, because
  the product of the Ag and Cl- ion concentrations
  for these solutions is equal to Ksp for AgCl.

94

• Point A represents a solution at equilibrium that
  could be produced by dissolving two sources of
  the Ag ion such as AgNO3 and AgCl in water.
• Point B represents a saturated solution of AgCl
  in pure water, in which the Ag and Cl- terms
  are equal.
• Point C describes a solution at equilibrium that
  was prepared by dissolving two sources of the Cl-
  ion in water, such as NaCl and AgCl.

95

• Any point that is not along the solid line in the
  above figure represents a solution that is not at
  equilibrium.
• Any point below the solid line (such as Point D)
  represents solution for which the ion product is
  smaller than the solubility product.
• Point D Qsp lt Ksp
• If more AgCl were added to solution at Point D,
  it would dissolve
• If Qsp lt Ksp  AgCl(s) Ag(aq) Cl-(aq)
• Points above the solid line (such as Point E)
  represent solutions for which the ion product is
  larger than the solubility product.
• Point E  Qsp gt Ksp
• The solution described by Point E will eventually
  come to equilibrium after enough solid AgCl has
  precipitated.
• If Qsp gt Ksp Ag(aq) Cl-(aq) AgCl(s)

96
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