Chapter 17 Additional Aspects of Aqueous Equilibria

1
Chapter 17Additional Aspects of Aqueous
Equilibria
Chemistry, The Central Science, 10th
edition Theodore L. Brown H. Eugene LeMay, Jr.
and Bruce E. Bursten
John D. Bookstaver St. Charles Community
College St. Peters, MO ? 2006, Prentice Hall, Inc.
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The Common-Ion Effect

• Consider a solution of acetic acid
• If acetate ion is added to the solution, Le
  Châtelier says the equilibrium will shift to the
  left.

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The Common-Ion Effect

• The extent of ionization of a weak electrolyte
  is decreased by adding to the solution a strong
  electrolyte that has an ion in common with the
  weak electrolyte.

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The Common-Ion Effect

• Calculate the fluoride ion concentration and pH
  of a solution that is 0.20 M in HF and 0.10 M in
  HCl.
• Ka for HF is 6.8 ? 10-4.

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The Common-Ion Effect
Because HCl, a strong acid, is also present, the
initial H3O is not 0, but rather 0.10 M.
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The Common-Ion Effect

• x
• 1.4 ? 10-3 x

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The Common-Ion Effect

• Therefore, F- x 1.4 ? 10-3
• H3O 0.10 x 1.01 1.4 ? 10-3 0.10 M
• So, pH -log (0.10)
• pH 1.00

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Solution   Analyze We are asked to determine the
pH of a solution of a weak electrolyte (HC2H3O2)
and a strong electrolyte (NaC2H3O2) that share a
common ion, C2H3O2. Plan In any problem in
which we must determine the pH of a solution
containing a mixture of solutes, it is helpful to
proceed by a series of logical steps 1.
Identify the major species in solution, and
consider their acidity or basicity. 2. Identify
the important equilibrium that is the source of
H and therefore determines pH. 3. Tabulate the
concentrations of ions involved in the
equilibrium. 4. Use the equilibrium-constant
expression to calculate H and then pH.
(We have written the equilibrium using H(aq)
rather than H3O(aq), but both representations of
the hydrated hydrogen ion are equally
valid.) Third, we tabulate the initial and
equilibrium concentrations much as we did in
solving other equilibrium problems in Chapters 15
and 16
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Comment In Section 16.6 we calculated that a
0.30 M solution of HC2H3O2 has a pH of 2.64,
corresponding to H 2.3 ? 103 M.Thus, the
addition of NaC2H3O2 has substantially decreased
H, as we would expect from Le Châteliers
principle.
PRACTICE EXERCISE Calculate the pH of a solution
containing 0.085 M nitrous acid (HNO2 Ka 4.5 ?
104) and 0.10 M potassium nitrite (KNO2).
Answer 3.42
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Comment Notice that for all practical purposes,
H is due entirely to the HCl the HF makes a
negligible contribution by comparison.
PRACTICE EXERCISE Calculate the formate ion
concentration and pH of a solution that is 0.050
M in formic acid (HCHO2 Ka 1.8 ? 104) and
0.10 M in HNO3.
Answer CHO2 9.0 ? 105 pH 1.00 
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Buffers

• Solutions of a weak conjugate acid-base pair.
• They are particularly resistant to pH changes,
  even when strong acid or base is added.

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Buffers

• If a small amount of hydroxide is added to an
  equimolar solution of HF in NaF, for example, the
  HF reacts with the OH- to make F- and water.

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Buffers

• If acid is added, the F- reacts to form HF and
  water.

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Buffer Calculations

• Consider the equilibrium constant expression for
  the dissociation of a generic acid, HA

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Buffer Calculations

• Rearranging slightly, this becomes

Taking the negative log of both side, we get
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Buffer Calculations

• So

• Rearranging, this becomes

• This is the HendersonHasselbalch equation.

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HendersonHasselbalch Equation

• What is the pH of a buffer that is 0.12 M in
  lactic acid, HC3H5O3, and 0.10 M in sodium
  lactate? Ka for lactic acid is
• 1.4 ? 10-4.

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HendersonHasselbalch Equation
pH 3.85 (-0.08) pH 3.77
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pH Range

• The pH range is the range of pH values over which
  a buffer system works effectively.
• It is best to choose an acid with a pKa close to
  the desired pH.

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When Strong Acids or Bases Are Added to a Buffer

• it is safe to assume that all of the strong acid
  or base is consumed in the reaction.

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Addition of Strong Acid or Base to a Buffer

• Determine how the neutralization reaction affects
  the amounts of the weak acid and its conjugate
  base in solution.
• Use the HendersonHasselbalch equation to
  determine the new pH of the solution.

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Calculating pH Changes in Buffers

• A buffer is made by adding 0.300 mol HC2H3O2 and
  0.300 mol NaC2H3O2 to enough water to make 1.00 L
  of solution. The pH of the buffer is 4.74.
  Calculate the pH of this solution after 0.020 mol
  of NaOH is added.

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Calculating pH Changes in Buffers

• Before the reaction, since
• mol HC2H3O2 mol C2H3O2-
• pH pKa -log (1.8 ? 10-5) 4.74

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Calculating pH Changes in Buffers
The 0.020 mol NaOH will react with 0.020 mol of
the acetic acid HC2H3O2(aq) OH-(aq) ???
C2H3O2-(aq) H2O(l)
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Calculating pH Changes in Buffers
Now use the HendersonHasselbalch equation to
calculate the new pH
pH 4.74 0.06 pH 4.80
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PRACTICE EXERCISE Calculate the pH of a buffer
composed of 0.12 M benzoic acid and 0.20 M sodium
benzoate. (Refer to Appendix D.)
Answer 4.42
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Comment Because NH4 and NH3 are a conjugate
acid-base pair, we could use the
HendersonHasselbalch equation (Equation 17.9) to
solve this problem. To do so requires first using
Equation 16.41 to calculate pKa for NH4 from the
value of pKb for NH3. We suggest you try this
approach to convince yourself that you can use
the HendersonHasselbalch equation for buffers
for which you are given Kb for the conjugate base
rather than Ka for the conjugate acid.
PRACTICE EXERCISE Calculate the concentration of
sodium benzoate that must be present in a 0.20 M
solution of benzoic acid (HC7H5O2) to produce a
pH of 4.00.
Answer 0.13 M
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SAMPLE EXERCISE 17.5 Calculating pH Changes in
Buffers
A buffer is made by adding 0.300 mol HC2H3O2 and
0.300 mol NaC2H3O2 to enough water to make 1.00 L
of solution. The pH of the buffer is 4.74 (Sample
Exercise 17.1). (a) Calculate the pH of this
solution after 0.020 mol of NaOH is added. (b)
For comparison, calculate the pH that would
result if 0.020 mol of NaOH was added to 1.00 L
of pure water (neglect any volume changes).
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Comment Notice that we could have used mole
amounts in place of concentrations in the
HendersonHasselbalch equation and gotten the
same result. The volumes of the acid and base are
equal and cancel. If 0.020 mol of H was added
to the buffer, we would proceed in a similar way
to calculate the resulting pH of the buffer. In
this case the pH decreases by 0.06 units, giving
pH 4.68, as shown in the figure on the next
slide.
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SAMPLE EXERCISE 17.5 continued
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Note that although the small amount of NaOH is
enough to change the pH of water significantly,
the pH of the buffer changes very little.
PRACTICE EXERCISE Determine (a) the pH of the
original buffer described in Sample Exercise 17.5
after the addition of 0.020 mol HCl, and (b) the
pH of the solution that would result from the
addition of 0.020 mol HCl to 1.00 L of pure water.
Answers (a) 4.68, (b) 1.70
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Titration

• A known concentration of base (or acid) is
  slowly added to a solution of acid (or base).

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Titration

• A pH meter or indicators are used to determine
  when the solution has reached the equivalence
  point, at which the stoichiometric amount of acid
  equals that of base.

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Titration of a Strong Acid with a Strong Base

• From the start of the titration to near the
  equivalence point, the pH goes up slowly.

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Titration of a Strong Acid with a Strong Base

• Just before and after the equivalence point, the
  pH increases rapidly.

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Titration of a Strong Acid with a Strong Base

• At the equivalence point, moles acid moles
  base, and the solution contains only water and
  the salt from the cation of the base and the
  anion of the acid.

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Titration of a Strong Acid with a Strong Base

• As more base is added, the increase in pH again
  levels off.

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SAMPLE EXERCISE 17.6 Calculating pH for a Strong
AcidStrong Base Titration
Calculate the pH when the following quantities of
0.100 M NaOH solution have been added to 50.0 mL
of 0.100 M HCl solution (a) 49.0 mL, (b) 51.0 mL.
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PRACTICE EXERCISE Calculate the pH when the
following quantities of 0.100 M HNO3 have been
added to 25.0 mL of 0.100 M KOH solution (a)
24.9 mL, (b) 25.1 mL.
Answers (a) 10.30, (b) 3.70
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Titration of a Weak Acid with a Strong Base

• Unlike in the previous case, the conjugate base
  of the acid affects the pH when it is formed.
• The pH at the equivalence point will be gt7.
• Phenolphthalein is commonly used as an indicator
  in these titrations.

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Titration of a Weak Acid with a Strong Base

• At each point below the equivalence point, the
  pH of the solution during titration is determined
  from the amounts of the acid and its conjugate
  base present at that particular time.

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Titration of a Weak Acid with a Strong Base

• With weaker acids, the initial pH is higher and
  pH changes near the equivalence point are more
  subtle.

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Titration of a Weak Base with a Strong Acid

• The pH at the equivalence point in these
  titrations is lt 7.
• Methyl red is the indicator of choice.

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Titrations of Polyprotic Acids

• In these cases there is an equivalence point for
  each dissociation.

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Comment We could have solved for pH equally well
using the HendersonHasselbalch equation.
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SAMPLE EXERCISE 17.7 continued
PRACTICE EXERCISE (a) Calculate the pH in the
solution formed by adding 10.0 mL of 0.050 M NaOH
to 40.0 mL of 0.0250 M benzoic acid (HC7H5O2, Ka
6.3 ? 105). (b) Calculate the pH in the
solution formed by adding 10.0 mL of 0.100 M HCl
to 20.0 mL of 0.100 M NH3.
Answers (a) 4.20, (b) 9.26
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Solution Analyze We are asked to determine the
pH at the equivalence point of the titration of a
weak acid with a strong base. Because the
neutralization of a weak acid produces its anion,
which is a weak base, we expect the pH at the
equivalence point to be greater than 7. Plan We
should first determine how many moles of acetic
acid there are initially. This will tell us how
many moles of acetate ion there will be at the
equivalence point. We then must determine the
volume of the solution at the equivalence point
and the resultant concentration of acetate ion.
Because the acetate ion is a weak base, we can
calculate the pH using Kb and the concentration
of acetate as we did for other weak bases in
Section 16.7.
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PRACTICE EXERCISE Calculate the pH at the
equivalence point when (a) 40.0 mL of 0.025 M
benzoic acid (HC7H5O2, Ka 6.3 ? 105) is
titrated with 0.050 M NaOH (b) 40.0 mL of 0.100
M NH3 is titrated with 0.100 M HCl.
Answers (a) 8.21, (b) 5.28
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Solubility Products

• Consider the equilibrium that exists in a
  saturated solution of BaSO4 in water

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Solubility Products

• The equilibrium constant expression for this
  equilibrium is
• Ksp Ba2 SO42-
• where the equilibrium constant, Ksp, is called
  the solubility product.

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Solubility Products

• Ksp is not the same as solubility.
• Solubility is generally expressed as the mass of
  solute dissolved in 1 L (g/L) or 100 mL (g/mL) of
  solution, or in mol/L (M).

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In Appendix D we see that this Ksp has a value of
3.9 ? 1011.
PRACTICE EXERCISE Give the solubility-product-cons
tant expressions and the values of the
solubility-product constants (from Appendix D)
for the following compounds (a) barium
carbonate, (b) silver sulfate.
Answers (a) Ksp Ba2CO32 5.0 ? 109
(b) Ksp Ag2SO42 1.5 ? 105
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Check We obtain a small value, as expected for a
slightly soluble salt. Furthermore, the
calculated value agrees well with the one given
in Appendix D, 1.2 ? 1012.
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SAMPLE EXERCISE 17.10 continued
PRACTICE EXERCISE A saturated solution of Mg(OH)2
in contact with undissolved solid is prepared at
25C. The pH of the solution is found to be
10.17. Assuming that Mg(OH)2 dissociates
completely in water and that there are no other
simultaneous equilibria involving the Mg2 or OH
ions in the solution, calculate Ksp for this
compound.
Answer 1.6 ? 1012
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Solution Analyze We are given Ksp for CaF2 and
are asked to determine solubility. Recall that
the solubility of a substance is the quantity
that can dissolve in solvent, whereas the
solubility-product constant, Ksp, is an
equilibrium constant. Plan We can approach this
problem by using our standard techniques for
solving equilibrium problems. We write the
chemical equation for the dissolution process and
set up a table of the initial and equilibrium
concentrations. We then use the
equilibrium-constant expression. In this case we
know Ksp, and so we solve for the concentrations
of the ions in solution.
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Check We expect a small number for the
solubility of a slightly soluble salt. If we
reverse the calculation, we should be able to
recalculate Ksp Ksp (2.1 ? 104)(4.2 ? 104)2
3.7 ? 1011, close to the starting value for
Ksp, 3.9 ? 1011 . Comment Because F is the
anion of a weak acid, you might expect that the
hydrolysis of the ion would affect the solubility
of CaF2. The basicity of F is so small (Kb 1.5
? 1011), however, that the hydrolysis occurs to
only a slight extent and does not significantly
influence the solubility. The reported solubility
is 0.017 g/L at 25C, in good agreement with our
calculation.
PRACTICE EXERCISE The Ksp for LaF3 is 2 ? 1019 .
What is the solubility of LaF3 in water in moles
per liter?
Answer  9.28 ? 106 mol/L
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Factors Affecting Solubility

• The Common-Ion Effect
• If one of the ions in a solution equilibrium is
  already dissolved in the solution, the
  equilibrium will shift to the left and the
  solubility of the salt will decrease.

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The value of Ksp is unchanged by the presence of
additional solutes. Because of the common-ion
effect, however, the solubility of the salt will
decrease in the presence of common ions. We can
again use our standard equilibrium techniques of
starting with the equation for CaF2 dissolution,
setting up a table of initial and equilibrium
concentrations, and using the Ksp expression to
determine the concentration of the ion that comes
only from CaF2.
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SAMPLE EXERCISE 17.12 continued
Comment The molar solubility of CaF2 in pure
water is 2.1 ? 104 M (Sample Exercise 17.11). By
comparison, our calculations on the previous
slide show that the solubility of CaF2 in the
presence of 0.010 M Ca2 is 3.1 ? 105 M, and in
the presence of 0.010 M F ion it is 3.9 ? 107
M. Thus, the addition of either Ca2 or F to a
solution of CaF2 decreases the solubility.
However, the effect of F on the solubility is
more pronounced than that of Ca2 because F
appears to the second power in the Ksp expression
for CaF2, whereas Ca2 appears to the first power.
PRACTICE EXERCISE The value for Ksp for
manganese(II) hydroxide, Mn(OH)2 , is 1.6 ?
1013. Calculate the molar solubility of Mn(OH)2
in a solution that contains 0.020 M NaOH.
Answer  4.0 ? 1010 M
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Factors Affecting Solubility

• pH
• If a substance has a basic anion, it will be more
  soluble in an acidic solution.
• Substances with acidic cations are more soluble
  in basic solutions.

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SAMPLE EXERCISE 17.13 Predicting the Effect of
Acid on Solubility
Which of the following substances will be more
soluble in acidic solution than in basic
solution (a) Ni(OH)2(s), (b) CaCO3(s), (c)
BaF2(s), (d) AgCl(s)?
Solution Analyze The problem lists four
sparingly soluble salts, and we are asked to
determine which will be more soluble at low pH
than at high pH. Plan Ionic compounds that
dissociate to produce a basic anion will be more
soluble in acid solution.
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(d) The solubility of AgCl is unaffected by
changes in pH because Cl is the anion of a
strong acid and therefore has negligible basicity.
PRACTICE EXERCISE Write the net ionic equation
for the reaction of the following copper(II)
compounds with acid (a) CuS, (b) Cu(N3)2.
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Factors Affecting Solubility

• Complex Ions
• Metal ions can act as Lewis acids and form
  complex ions with Lewis bases in the solvent.

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Factors Affecting Solubility

• Complex Ions
• The formation of these complex ions increases the
  solubility of these salts.

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SAMPLE EXERCISE 17.14 Evaluating an Equilibrium
Involving a Complex Ion
Calculate the concentration of Ag present in
solution at equilibrium when concentrated ammonia
is added to a 0.010 M solution of AgNO3 to give
an equilibrium concentration of NH3 0.20 M.
Neglect the small volume change that occurs when
NH3 is added.
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SAMPLE EXERCISE 17.14 continued
Solving for x, we obtain x 1.5 ? 108 M
Ag. Thus, formation of the Ag(NH3)2 complex
drastically reduces the concentration of free Ag
ion in solution.
PRACTICE EXERCISE Calculate Cr3 in equilibrium
with Cr(OH)4 when 0.010 mol of Cr(NO3)3 is
dissolved in a liter of solution buffered at pH
10.0.
Answer 1 ? 1016 M
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Factors Affecting Solubility

• Amphoterism
• Amphoteric metal oxides and hydroxides are
  soluble in strong acid or base, because they can
  act either as acids or bases.
• Examples of such cations are Al3, Zn2, and Sn2.

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Will a Precipitate Form?

• In a solution,
• If Q Ksp, the system is at equilibrium and the
  solution is saturated.
• If Q lt Ksp, more solid will dissolve until Q
  Ksp.
• If Q gt Ksp, the salt will precipitate until Q
  Ksp.

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Solution Analyze The problem asks us to
determine whether a precipitate will form when
two salt solutions are combined. Plan We should
determine the concentrations of all ions
immediately upon mixing of the solutions and
compare the value of the reaction quotient, Q, to
the solubility-product constant, Ksp, for any
potentially insoluble product. The possible
metathesis products are PbSO4 and NaNO3. Sodium
salts are quite soluble PbSO4 has a Ksp of 6.3 ?
107 (Appendix D), however, and will precipitate
if the Pb2 and SO42 ion concentrations are high
enough for Q to exceed Ksp for the salt.
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PRACTICE EXERCISE Will a precipitate form when
0.050 L of 2.0 ? 102 M NaF is mixed with 0.010 L
of 1.0 ? 102 M Ca(NO3)2?
Answer yes, CaF2 precipitates because Q 4.6 ?
108 is larger than Ksp 3.9 ? 1011
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Solution Analyze We are asked to determine the
concentration of Cl necessary to begin the
precipitation from a solution containing Ag and
Pb2 ions, and to predict which metal chloride
will begin to precipitate first.  Plan We are
given Ksp values for the two possible
precipitates. Using these and the metal ion
concentrations, we can calculate what
concentration of Cl ion would be necessary to
begin precipitation of each. The salt requiring
the lower Cl ion concentration will precipitate
first.
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PRACTICE EXERCISE A solution consists of 0.050 M
Mg2 and 0.020 M Cu2. Which ion will precipitate
first as OH is added to the solution? What
concentration of OH is necessary to begin the
precipitation of each cation? Ksp 1.8 ? 1011
for Mg(OH)2, and Ksp 2.2 ? 1020 for Cu(OH)2.
Answer  Cu(OH)2 precipitates first. Cu(OH)2
begins to precipitate when OH exceeds 1.0 ?
109 M Mg(OH)2 begins to precipitate when OH
exceeds 1.9 ? 105 M.
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Selective Precipitation of Ions

• One can use differences in solubilities of salts
  to separate ions in a mixture.

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SAMPLE INTEGRATIVE EXERCISE Putting Concepts
Together
A sample of 1.25 L of HCl gas at 21C and 0.950
atm is bubbled through 0.500 L of 0.150 M NH3
solution. Assuming that all the HCl dissolves and
that the volume of the solution remains 0.500 L,
calculate the pH of the resulting solution.