Chapter 18 Solubility and Complex-Ion Equilibria

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Chapter 18 Solubility and Complex-Ion Equilibria

• Dr. Peter Warburton
• peterw_at_mun.ca
• http//www.chem.mun.ca/zcourses/1051.php

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Solubility equilibria

• MmXx (s) ? m Mn (aq) x Xy- (aq)
• The equilibrium is established when we have a
  saturated solution of ions forming the solid and
  solid is dissociating to form the ions in
  solution. The rates of these processes must be
  equal. (eqm definition)

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Solubility equilibria

• For a dissolution process, we give the
  equilibrium constant expression the name
  solubility product (constant) Ksp. For
• MmXx (s) ? m Mn (aq) x Xy- (aq)
• Ksp Mnm Xy-x

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Ksp is an equilibrium constant

• Since Ksp is an equilibrium constant we MUST
• refer to a specific balanced equation
• (by definition this balanced equation is one mole
  of solid becoming aqueous ions)
• at a specific temperature.

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Problem

• Write the expressions of Ksp of
• a) AgCl
• b) PbI2
• c) Ca3(PO4)2
• d) Cr(OH)3

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Problem

• If a saturated solution of BaSO4 is prepared by
  dissolving solid BaSO4 in water, and Ba2
  1.05 x 10-5 mol?L-1, what is the Ksp for BaSO4?

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Molar solubility

• If we know the Ksp value for a solid, we can
  calculate the molar solubility, which is the
  number of moles of the solid that can dissolve in
  a given amount of solvent before the solution
  becomes saturated.
• The molar solubility leads to the solubility (by
  using the molar mass) which is the mass of the
  solid that can dissolve in a given amount of
  solvent before the solution becomes saturated.

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Molar solubility

• Alternatively, if we know the
• molar solubility
• OR
• the molar mass and the solubility
• we can calculate
• the Ksp for the solid.

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Molar solubility

• A good (but NOT physically correct) way to think
  about molar solubility is to treat our solid
  dissolution AS IF there are two separate
  processes
• The molar solubility has the same value as the
  concentration of aqueous MmXx after the first
  step. We can figure out this concentration based
  on concentrations of ions in the second step.

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Problem

• A handbook lists the aqueous solubility of AgOCN
  as 7 mg per 100 mL at 20 ?C. What is the Ksp of
  AgOCN at 20 ?C? The molar mass of AgOCN is
  149.885 g?mol-1.

Answer Ksp 2 x 10-7
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Problem

• A handbook lists the aqueous solubility of
  lithium phosphate (Li3PO4) as 0.034 g per 100 mL
  at 18 ?C. What is the Ksp of lithium phosphate
  at 18 ?C? The molar mass of Li3PO4 is 115.794
  g?mol-1.

Answer Ksp 2.0 x 10-9
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Problem

• Which has the greater molar solubility
• AgCl with Ksp 1.8 x 10-10
• or
• Ag2CrO4 with Ksp 1.1 x 10-12?

Answer The molar solubility of AgCl is 1.3 x
10-5 M while the molar solubility of Ag2CrO4 is
6.5 x 10-5 M. Silver chromate has a higher molar
solubility.
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Problem

• How many milligrams of BaSO4 (molar mass is
  233.391 g?mol-1) are dissolved in a 225 mL sample
  of saturated aqueous barium sulphate? Ksp 1.1
  x 10-10 at 25 ?C.

Answer mass 0.55 mg
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The common-ion effect

• MmXx (s) ? m Mn (aq) x Xy- (aq)
• If we have dissolved a solid in pure water and we
  add to this solution another solution containing
  one of the common ions, then Le Chataliers
  Principle tells us what will happen
• The presence of the common-ion in the added
  solution will force the dissolution reaction to
  the left, meaning more solid will form!

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The common-ion effect
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Figure
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The common-ion effect

• MmXx (s) ? m Mn (aq) x Xy- (aq)
• If instead of dissolving a solid in pure water we
  try and dissolve it into a solution that already
  contains one of the common ions, then Le
  Chataliers Principle tells us what will happen
• The presence of the common-ion already in
  solution will force the dissolution reaction to
  the left, meaning less solid will dissolve than
  would dissolve in pure water!

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Problem

• Calculate the molar solubility of MgF2 (Ksp 7.4
  x 10-11) in pure water and in 0.10 mol?L-1 MgCl2
  at 25 C.

Answer The molar solubility is 2.6 x 10-4 M in
pure water and 1.4 x 10-5 M in 0.10 M magnesium
chloride.
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Problem

• What is the the molar solubility of Fe(OH)3 (Ksp
  4 x 10-38) in a buffered solution with pH
  8.20 at 25 C.

Answer The molar solubility is 1 x 10-20 M in
the buffered solution.
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Limitations of Ksp

• If our solid is more than slightly soluble then
  we really should use the activities of our ions
  in solution rather than concentrations.
• These two measures are nearly the same for very
  dilute ion concentrations, but can become quite
  different at higher concentrations!

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The diverse (uncommon) ion effect

• The activities of ions tend to be LESS than the
  concentration value as the total ionic
  concentration increases.

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The diverse (uncommon) ion effect

• Adding a salt that does NOT feature a common ion
  to the solution will tend to decrease the
  activity (the effective concentration) of the
  ions in solution.
• The ion concentrations appear smaller than they
  should be at equilibrium so more solid dissolves
  to reach the appropriate equilibrium
  concentrations.

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Ion-pair formation and Ksp

• We assume in Ksp calculations that the solid
  dissociates completely into ions in solution.
• If this is not true, then the ionic
  concentrations we measure do not include
  dissolved but undissociated molecules or ion
  pairs which come from solid dissolution.

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Ion-pair formation and Ksp

• Positive ions and negative ions are attracted to
  each other and so they can form an ion pair that
  has a chemical identity different from each of
  the individual ions!

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Ion-pair formation and Ksp

• Say we measure a molar solubility for magnesium
  fluoride to be 4 x 10-3 M and assume that
• Mg2 4 x 10-3 M and
• F- 8 x 10-3 M
• to give a
• calculated Ksp of 3 x 10-7

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Ion-pair formation and Ksp

• IN REALITY the presence of undissociated MgF2
  (aq)
• and MgF- ion pairs
• means that not all of the solid that has
  dissolved is found as free ions, so our ionic
  concentrations are LOWER than we assumed, and so
  Ksp is actually smaller than we calculated.

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Simultaneous equilibria

• Weve seen in reference to Le Chataliers
  Principle that if more than one reaction can take
  place in a container, then the reactions might
  not be able to be treated independently. Other
  equilibrium processes may affect the solubility
  of the solid and lead to miscalculated Ksp values.

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Simultaneous equilibria

• For example, weve seen that AgI becomes more
  soluble when we add ammonia because of the
  formation of a complex of silver ions and
  ammonia.
• Well look more closely at complex formation a
  little later.

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Assessing the limitations of Ksp

• Most tabulated Ksp values are actually based on
  activities, and not concentrations.
• We use concentrations in our examples, so our
  calculations represent an ideal, and not reality.
  
• Generally we could be in error by over 100 times!

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Criteria for precipitation and its completeness

• Can we predict if a solid will form if we mix two
  solutions of different ions?
• Consider the mixing of two different solutions,
  one with Ca2 ions and one with F- ions. A
  formation of solid is the dissolution reaction in
  reverse, so we can express the reaction using the
  dissolution equation
• CaF2 (s) ? Ca2 (aq) 2 F- (aq) Ksp Ca2
  F-2

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Criteria for precipitation and its completeness

• When we mix the solutions
• (BE CAREFUL mixing ALWAYS changes the
  concentrations of both our ions!)
• the system is most likely not at equilibrium.
• Like in other equilibrum problems, we can use a
  reaction quotient Qsp (often called the ion
  product) to tell us in which direction the system
  must go to reach equilibrium
• Qsp Ca2 F-2

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Criteria for precipitation and its completeness
If Qsp gt Ksp, the solution is supersaturated, so
the system is not at equilibrium. The
concentration of the ions is greater than it
would be at equilibrium, and so the reaction
wants to shift from ions towards the solid. We
expect precipitation to occur! If Qsp Ksp, the
solution is saturated, and the system is at
equilibrium. No precipitation occurs!
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Criteria for precipitation and its completeness
If Qsp lt Ksp, the solution is unsaturated, so the
system is not at equilibrium. The concentration
of the ions is less than it would be at
equilibrium, and so the reaction wants to shift
from solid towards the ions. No precipitation
can occur!
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Mixing and equilibrium take time!
We must wait until dilution is completed and
equilibrium is established BEFORE we say
precipitation occurred!
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Problem

• Will a precipitate form when 0.150 L of 0.10
  mol?L-1 Pb(NO3)2 and 0.100 L of 0.20 mol?L-1 NaCl
  are mixed?
• Ksp of PbCl2 is 1.2 x 10-5

Answer Qsp 3.8 x 10-4 gt Ksp so precipitation
should occur.
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Problem

• How many drops (1 drop 0.05 mL) of 0.20 M KI
  must we add to 100.0 mL of 0.010 M Pb(NO3)2 to
  get precipitation of lead iodide to start?
• Ksp of PbI2 is 7.1 x 10-9

Answer We require at least 9 drops.
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Complete precipitation

• Generally we treat precipitation as complete if
  99.9 of the original ion concentration has been
  lost to the precipitate.
• For example, if our initial Pb2 is 0.10 M,
  then precipitation by adding I- is complete when
  our solution contains a Pb2 less than 1 x 10-4
  M.

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Problem

• A typical Ca2 concentration in seawater is 0.010
  M. Will the precipitation of Ca(OH)2 be complete
  from a seawater sample in which OH- is
  maintained at 0.040 M?
• Ksp of Ca(OH)2 is 5.5 x 10-6

Answer Since the final Ca2 is 3.4 x 10-3 M,
which is 34 of 0.010 M, the precipitation is
not complete.
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Problem

• What OH- should be maintained in a solution if,
  after precipitation of Mg2 as solid magnesium
  hydroxide, the remaining Mg2 is to be at a
  level of 1?g?L-1?
• Molar mass Mg is 24.305 g?mol-1
• Ksp of Mg(OH)2 is 1.8 x 10-11

Answer OH- needed is 1.6 x 10-2 M.
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Fractional precipitation

• If we have a solution with
• both CrO42- ions and Br- ions
• and add a large amount of Ag ions at once,
• then both Ag2CrO4 and AgBr
• will precipitate
• in our container at the same time.

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Fractional precipitation

• If we slowly add the Ag solution instead the
  solid with the significantly lower molar
  solubility (AgBr in this case do the
  calculations to check this for yourself)
• will precipitate first
• and consume the added Ag preferentially.

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Fractional precipitation

• In other words,
• the concentration of Ag
• CAN NOT become large enough
• to precipitate Ag2CrO4
• until the AgBr
• precipitation is complete.

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Fractional precipitation
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Problem

• AgNO3 is slowly added to a solution with Cl-
  0.115 M and Br- 0.264 M. What percent of the
  Br- remains unprecipitated at the point at which
  AgCl (s) begins to precipitate?
• Ksp values
• AgCl 1.8 x 10-10 AgBr 5.0 x 10-13

Answer 0.12 of Br- remains.
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Solubility and pH

• If a solid dissolves to give a basic anion in
  solution, addition of strong acid will increase
  the solubility of the solid.
• CaCO3 (s) ? Ca2 (aq) CO32- (aq) Ksp 2.8 x
  10-9
• Carbonate, CO32-, is a basic anion that will
  react with a proton to give HCO3-
• CO32- (aq) H3O (aq) ? HCO3- (aq) H2O (l)

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Solubility and pH

• CO32- (aq) H3O (aq) ? HCO3- (aq) H2O (l)
• This reaction will shift to the right (products)
  as the pH becomes more acidic which means CO32-
  (aq) decreases. However, in our first
  equilibrium IT ALSO MUST DECREASE so the first
  equilibrium will also shift to the right to
  compensate.
• More solid will dissolve!

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Adding equilibria

• A better way to state the effect of pH on
  solubility comes when we add dissolution and weak
  base strong acid reactions together
• CaCO3 (s) ? Ca2 (aq) CO32- (aq)
• Ksp 2.8 x 10-9
• CO32- (aq) H3O (aq) ? HCO3- (aq) H2O (l)
• K Kb x 1/Kw (2.1 x 10-4) x (1.0 x 1014)
• K 2.1 x 1010

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Adding equilibria

• CaCO3 (s) H3O (aq) ?
• Ca2 (aq) HCO3- (aq) H2O (l)
• K K x Ksp (2.1 x 1010) x (2.8 x 10-9)
• K 59
• The solubility of the solid will increase in the
  presence of H3O!

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Problem

• Will a precipitate of Fe(OH)3 form from a
  solution that is 0.013 M Fe3 in a buffer
  solution that is 0.150 M acetic acid 0.250 M
  acetate?
• Ksp Fe(OH)3 4 x 10-38 Ka 1.8 x 10-5

Answer Since Qsp is 1 x 10-29, then
precipitation will occur since QspgtKsp.
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Problem

• What minimum NH4 must be present to prevent
  precipitation of Mn(OH)2 (s) from a solution that
  is 0.0050 M MnCl2 and 0.025 M NH3? For Mn(OH)2
  Ksp 1.9 x 10-13 and Kb for NH3 is 1.8 x 10-5.

Answer NH4 gt 0.073 M
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Formation of complex ions

• Solubility of a solid increases if there is the
  ability to form a complex ion.
• An example of a complex ion is Ag(NH3)2.
• Such complexes affect solubility by reducing the
  concentration of the cation so that the
  dissolution reaction must shift to the products
  to replace the cation concentration to
  re-establish equilibrium.

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Formation of complex ions

• Ag (aq) 2 NH3 (aq) ? Ag(NH3)2 (aq) Kf 1.7
  x 107
• AgCl (s) ? Ag (aq) Cl- (aq) Ksp 1.8 x
  10-10
• In the presence of ammonia, the dissolution of
  AgCl can be expressed by the sum of these two
  reactions
• 2 NH3 (aq) AgCl (s) ? Ag(NH3)2 (aq) Cl- (aq)
• K Kf x Ksp 1.7 x 107 x 1.8 x 10-10 3.1 x
  10-3

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Formation of complex ions

• We see the dissolution of AgCl occurs to a
  greater level of completion
• in the presence of ammonia
• (K 3.1 x 10-3)
• than it does in pure water
• (Ksp 1.8 x 10-10).

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Qualitative cation analysis

• Qualitative analysis is concerned with what do
  we have? and NOT how much do we have?
• If we want to identify what cations we have in a
  solution, we can use a series of precipitation
  reactions in a certain order to tell us.

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Groups of precipitated ions

• 1) Chloride group
• Pb2, Ag, Hg22
• 2) Hydrogen Sulphide group-
• Pb2, Hg2, Bi3, Cu2, Cd2, As3, Sn2, Sb3
• 3) Ammonium sulphide group
• Mn2, Fe2, Fe3, Ni2, Co2, Al3, Zn2, Cr3
• 4) Carbonate group
• Mg2, Ca2, Sr2, Ba2
• 5) Soluble group
• Na, K, NH4

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Reactions with hydrogen chloride

• Most metal ions form soluble salts with chloride
  EXCEPT Pb2, Hg22, and Ag. Adding aqueous HCl
  to our unknown solution will let us now if we
  have one or more of these ions because we will
  get white precipitate(s). If we want to know if
  we have more than one of these ions, we do
  further tests on the precipitated solids

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Further tests for insoluble chlorides

• In a) we have a mixture of AgCl, Hg2Cl2, and
  PbCl2.
• If we add ammonia, any AgCl should dissolve
  because of complex ion formation
• AgCl (s) 2 NH3 (aq) ?
• Ag(NH3)2 (aq) Cl- (aq)

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Further tests for insoluble chlorides

• Also after adding ammonia (b), any Hg2Cl2 will
  give us a grey solid that is a mixture of black
  liquid Hg and white solid HgNH2Cl
• Hg2Cl2 (s) 2 NH3 (aq) ?
• Hg (l) HgNH2Cl (s) NH4Cl (aq)
• black white

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Further tests for insoluble chlorides

• Adding chromate (CrO42-) (c) to a Pb2 solution
  derived by heating the precipitate solutions
  (its the most soluble) will give a yellow
  precipitate - PbCrO4
• Pb2 (aq) CrO42- (aq) ?
• PbCrO4(s)

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Reactions with hydrogen sulfide

• S2- is capable of giving precipitates of many
  ions.
• H2S is a potential source of S2- in solution
  because it is a diprotic acid
• H2S (aq) H2O (l) ? H3O (aq) HS- (aq)
• Ka1 1.0 x 10-7
• HS- (aq) H2O (l) ? H3O (aq) S2- (aq)
• Ka2 1 x 10-19

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Reactions with hydrogen sulfide

• However, in acidic solution (with HCl), some of
  the precipitates dissolve, leaving behind
• PbS, HgS, Bi2S3, CuS, CdS, As2S3, SnS, Sb2S3
• In basic solution (by adding ammonia) these
  precipitates dissolve, leaving behind
• MnS, FeS, Fe(OH)3, NiS, CoS, Al(OH)3, ZnS, Cr(OH)3

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Reactions with carbonate

• Addition of carbonate ion (CO32-) in basic
  solution (usually with an ammonia-ammonium
  buffer) will precipitate the alkali earth metal
  carbonates (see Chapter 21)
• CaCO3, MgCO3, SrCO3, BaCO3

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The soluble group

• Any ions left in solution after the first four
  reaction groups are tested for are the cations of
  soluble salts
• Na, K, and NH4