solubility and complex ion equilibria. slightly soluble salts establish a dynamic equilibrium with...
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Solubility and Complex Solubility and Complex Ion EquilibriaIon Equilibria
Slightly soluble salts establish Slightly soluble salts establish a a
dynamic equilibrium with the dynamic equilibrium with the
hydrated cations and anions hydrated cations and anions in in
solution. solution.
When the solid is first added to When the solid is first added to
water, no ions are initially water, no ions are initially present. present.
As dissolution proceeds, the As dissolution proceeds, the
concentration of ions increases concentration of ions increases
until equilibrium is established. until equilibrium is established.
This occurs when the solution This occurs when the solution is is
saturated.saturated.
The equilibrium constant, the The equilibrium constant, the
KKspsp, is no more than the product , is no more than the product of of
the ions in solution. the ions in solution.
(Remember, solids do not (Remember, solids do not
appear in equilibrium appear in equilibrium expressions.)expressions.)
For a saturated solution of For a saturated solution of
AgCl, the equation would AgCl, the equation would be: be:
AgCl AgCl (s)(s) Ag Ag+ + (aq)(aq) + Cl + Cl- - (aq)(aq)
The solubility product The solubility product expression expression
would be:would be:
KKspsp = [Ag = [Ag++] [Cl] [Cl--]]
The AgClThe AgCl(s)(s) is left out since is left out since
solids are left out of solids are left out of equilibrium equilibrium
expressions (constant expressions (constant
concentrations).concentrations).
You can find loads of You can find loads of KKspsp’s on tables.’s on tables.
Find the KFind the Kspsp values & write values & write the the
KKspsp expression for the expression for the following:following:
CaFCaF2(s)2(s) Ca Ca+2+2 + 2 F + 2 F- - KKspsp = =
AgAg22SOSO4(s)4(s) 2 Ag 2 Ag++ + SO + SO44-2 -2 KKspsp = =
BiBi22SS3(s)3(s) 2 Bi 2 Bi+3+3 + 3 S + 3 S-2 -2 KKspsp = =
Determining KDetermining Kspsp From From Experimental Experimental MeasurementsMeasurements
In practice, KIn practice, Kspsp’s are determined ’s are determined
by careful laboratory by careful laboratory
measurements using various measurements using various
spectroscopic methods.spectroscopic methods.
Remember STOICHIOMETRY!!Remember STOICHIOMETRY!!
ExampleExample
Lead (II) chloride dissolves to a Lead (II) chloride dissolves to a
slight extent in water according slight extent in water according
to the equation:to the equation:
PbClPbCl22 Pb Pb+2+2 + 2Cl + 2Cl--
Calculate the KCalculate the Kspsp if the lead if the lead ion ion
concentration has been found concentration has been found to to
be 1.62 x 10be 1.62 x 10-2-2M.M.
SolutionSolution
If lead’s concentration is “x” ,If lead’s concentration is “x” ,then chloride’s concentration is then chloride’s concentration is ““2x”. 2x”.
So. . . . So. . . .
KKspsp = (1.62 x 10 = (1.62 x 10-2-2)(3.24 x 10)(3.24 x 10-2-2))22 = 1.70 x 10= 1.70 x 10-5-5
Exercise 12 Exercise 12 Calculating KCalculating Kspsp from from Solubility ISolubility ICopper(I) bromide has a Copper(I) bromide has a
measured measured
solubility of 2.0 X 10solubility of 2.0 X 10-4-4 mol/L at mol/L at
25°C. 25°C.
Calculate its KCalculate its Kspsp value. value.
SolutionSolution
KKspsp = 4.0 X 10 = 4.0 X 10-8-8
Exercise 13Exercise 13 Calculating Calculating Ksp from Solubility IIKsp from Solubility II
Calculate the KCalculate the Kspsp
value for bismuth value for bismuth
sulfide (Bisulfide (Bi22SS33), ), which which
has a solubility of has a solubility of
1.0 X 101.0 X 10-15-15 mol/L at mol/L at
25°C.25°C.
SolutionSolution
KKspsp = 1.1 X 10 = 1.1 X 10-73-73
ESTIMATING SALTESTIMATING SALT
SOLUBILITY FROM KSOLUBILITY FROM Kspsp
ExampleExample
The KThe Kspsp for CaCO for CaCO33 is 3.8 x 10 is 3.8 x 10-9-9 @ @
25°C. 25°C.
Calculate the solubility of calcium Calculate the solubility of calcium
carbonate in pure water in carbonate in pure water in
a) moles per liter a) moles per liter
b) grams per literb) grams per liter
The relative solubilities can be The relative solubilities can be deduced by comparing values of deduced by comparing values of KKspsp..
BUT, BE CAREFUL!BUT, BE CAREFUL!
These comparisons can only be These comparisons can only be made for salts having the same made for salts having the same ION:ION ratio.ION:ION ratio.
Please don’t forget solubility Please don’t forget solubility changes with temperature! changes with temperature!
Some substances become less Some substances become less soluble in cold while some soluble in cold while some become become
more soluble! more soluble!
Aragonite.Aragonite.
Exercise 14Exercise 14 Calculating Calculating Solubility from KSolubility from Kspsp
The KThe Kspsp value for copper(II) value for copper(II) iodate, iodate,
Cu(IOCu(IO33))22, is 1.4 X 10, is 1.4 X 10-7-7 at 25°C. at 25°C.
Calculate its solubility at 25°C.Calculate its solubility at 25°C.
SolutionSolution
= 3.3 X 10= 3.3 X 10-3-3 mol/L mol/L
Exercise 15Exercise 15 Solubility Solubility and Common Ionsand Common Ions
Calculate the solubility of solid Calculate the solubility of solid CaFCaF22
(K(Kspsp = 4.0 X 10 = 4.0 X 10-11-11) )
in a 0.025 in a 0.025 MM NaF solution. NaF solution.
SolutionSolution
= 6.4 X 10= 6.4 X 10-8-8 mol/L mol/L
KKspsp and the Reaction and the Reaction Quotient, QQuotient, Q
With some knowledge of the With some knowledge of the
reaction quotient, we can decidereaction quotient, we can decide
1) whether a ppt will form, AND 1) whether a ppt will form, AND
2) what concentrations of ions 2) what concentrations of ions
are required to begin the ppt. of are required to begin the ppt. of
an insoluble salt.an insoluble salt.
1. Q < K1. Q < Kspsp, the system , the system is notis not at at equil. (equil. (ununsaturated)saturated)
2. Q = K2. Q = Kspsp, the system , the system isis at at equil. (saturated)equil. (saturated)
3. Q > K3. Q > Kspsp, the system , the system is notis not at at equil. (equil. (supersupersaturatedsaturated))
Precipitates form when the Precipitates form when the
solution is supersaturated!!!solution is supersaturated!!!
Precipitation of Insoluble Precipitation of Insoluble SaltsSalts
Metal-bearing ores often contain Metal-bearing ores often contain
the metal in the form of an the metal in the form of an
insoluble salt, and, to complicate insoluble salt, and, to complicate
matters, the ores often contain matters, the ores often contain
several such metal salts.several such metal salts.
Dissolve the metal salts to obtain Dissolve the metal salts to obtain
the metal ion, concentrate in some the metal ion, concentrate in some
manner, and ppt. selectively only manner, and ppt. selectively only
one type of metal ion as an one type of metal ion as an
insoluble salt.insoluble salt.
Exercise 16Exercise 16 Determining Determining Precipitation ConditionsPrecipitation Conditions
A solution is prepared by adding A solution is prepared by adding
750.0 mL of 4.00 X 10750.0 mL of 4.00 X 10-3-3 MM Ce(NO Ce(NO33))33
to 300.0 mL of 2.00 X 10to 300.0 mL of 2.00 X 10-2-2 MM KIO KIO33. .
Will Ce(IOWill Ce(IO33))33 (K (Kspsp = 1.9 X 10 = 1.9 X 10-10-10) )
precipitate from this solution?precipitate from this solution?
SolutionSolution
yesyes
Exercise 17Exercise 17 PrecipitationPrecipitation
A solution is prepared by mixing A solution is prepared by mixing
150.0 mL of 1.00 X 10150.0 mL of 1.00 X 10-2-2 M M Mg(NO Mg(NO33))22
and 250.0 mL of 1.00 X 10and 250.0 mL of 1.00 X 10-1-1 MM NaF. NaF.
Calculate the concentrations of Calculate the concentrations of MgMg2+2+
and Fand F-- at equilibrium with solid MgF at equilibrium with solid MgF22
(K(Kspsp = 6.4 X 10 = 6.4 X 10-9-9).).
SolutionSolution
[Mg[Mg2+2+] = 2.1 X 10] = 2.1 X 10-6-6 MM
[F[F--] = 5.50 X 10] = 5.50 X 10-2-2 MM
SOLUBILITY AND THE SOLUBILITY AND THE COMMON ION EFFECTCOMMON ION EFFECT
Experiment shows that the Experiment shows that the
solubility of any salt is always solubility of any salt is always less less
in the presence of a in the presence of a “common “common
ion”.ion”.
LeChatelier’s Principle, that’s LeChatelier’s Principle, that’s why! why!
Be reasonable and use Be reasonable and use
approximations when you can!!approximations when you can!!
Just remember what Just remember what happened happened
earlier with acetic acid and earlier with acetic acid and
sodium acetate. sodium acetate.
The same idea here!The same idea here!
pH can also affect solubility. pH can also affect solubility.
Evaluate the equation to see Evaluate the equation to see who who
would want to “react” with the would want to “react” with the
addition of acid or base. addition of acid or base.
Would magnesium hydroxide be Would magnesium hydroxide be
more soluble in an acid or a more soluble in an acid or a base? base?
Why? Why?
Mg(OH)Mg(OH)2(s)2(s) Mg Mg2+2+(aq)(aq) + 2 OH + 2 OH--
(aq)(aq)
(milk of magnesia)(milk of magnesia)
Why Would I Ever Care Why Would I Ever Care About KAbout Kspsp ??? ???
Keep reading to find out ! Keep reading to find out !
Actually, very useful stuff!Actually, very useful stuff!
Solubility, Solubility, Ion Separations, and Ion Separations, and Qualitative AnalysisQualitative Analysis
……introduce you to some basic introduce you to some basic
chemistry of various ions.chemistry of various ions.
……illustrate how the principles illustrate how the principles of of
chemical equilibria can be chemical equilibria can be applied.applied.
Objective:Objective:
Separate the Separate the
following following
metal ions: metal ions:
silver, silver,
lead, lead,
cadmium and cadmium and
nickelnickel
From solubility rules, lead and From solubility rules, lead and silver silver
chloride will ppt, so add dilute chloride will ppt, so add dilute HCl. HCl.
Nickel and cadmium will stay in Nickel and cadmium will stay in
solution.solution.
Separate by filtration: Separate by filtration:
Lead chloride will dissolve in HOT Lead chloride will dissolve in HOT
water… water…
filter while HOT and those two will filter while HOT and those two will
be separate.be separate.
Cadmium and nickel are more Cadmium and nickel are more
subtle. subtle.
Use their KUse their Kspsp’s with sulfide ion. ’s with sulfide ion.
Who ppt’s first???Who ppt’s first???
Exercise 18Exercise 18 Selective Precipitation Selective Precipitation
A solution contains 1.0 X 10A solution contains 1.0 X 10-4-4 MM Cu Cu++
and 2.0 X 10and 2.0 X 10-3-3 MM Pb Pb2+2+. .
If a source of IIf a source of I-- is added gradually to is added gradually to
this solution, will PbIthis solution, will PbI22 (K (Kspsp = 1.4 X = 1.4 X
1010-8-8) or CuI (K) or CuI (Kspsp = 5.3 X 10 = 5.3 X 10-12-12) )
precipitate first? precipitate first?
Specify the concentration of ISpecify the concentration of I--
necessary to begin necessary to begin precipitation of precipitation of
each salt.each salt.
SolutionSolution
CuI will precipitate first.CuI will precipitate first.
Concentration in excess of Concentration in excess of
5.3 X 105.3 X 10-8-8 MM required. required.
**If this gets you interested, lots **If this gets you interested, lots
more information on this topic in more information on this topic in
the chapter. the chapter.
Good bedtime reading for Good bedtime reading for
descriptive chemistry! descriptive chemistry!
THE EXTENT OF LEWIS THE EXTENT OF LEWIS ACID-BASE ACID-BASE
REACTIONS: REACTIONS:
FORMATION CONSTANTSFORMATION CONSTANTS
When a metal ion (a Lewis acid) When a metal ion (a Lewis acid)
reacts with a Lewis base, a reacts with a Lewis base, a
complex ion can form. complex ion can form.
The formation of complex ions The formation of complex ions
represents a reversible equilibria represents a reversible equilibria situation. situation.
A complex ion is a charged A complex ion is a charged
species consisting of a metal species consisting of a metal ion ion
surrounded by ligands.surrounded by ligands.
A ligand is typically an anion or A ligand is typically an anion or neutral molecule that has an neutral molecule that has an unshared electron pair that can unshared electron pair that can be shared with an empty metal be shared with an empty metal ion orbital to form a metal-ligand ion orbital to form a metal-ligand bond. bond.
Some common ligands are Some common ligands are
HH22O, NHO, NH33, Cl, Cl--, and CN, and CN--. .
The number of ligands attached The number of ligands attached to to
the metal ion is the coordination the metal ion is the coordination
number. number.
The most common coordination The most common coordination
numbers are: 6, 4, 2 numbers are: 6, 4, 2
Metal ions add ligands one at a Metal ions add ligands one at a
time in steps characterized by time in steps characterized by
equilibrium constants called equilibrium constants called
formation constantsformation constants..
AgAg++ + 2NH + 2NH33 [Ag(NH [Ag(NH33))22]]+2+2
acid baseacid base
Stepwise Reactions:Stepwise Reactions:
AgAg++((aqaq)) + NH + NH3(3(aqaq)) Ag(NH Ag(NH33))++
((aqaq))
KKf1f1 = 2.1 x 10 = 2.1 x 1033
[Ag(NH[Ag(NH33))++] ] = 2.1 x 10 = 2.1 x 1033
[Ag[Ag++][NH][NH33] ]
Ag(NHAg(NH33))++ +NH +NH3(3(aqaq)) Ag(NH Ag(NH33))22++
((aqaq))
KKf2f2 = 8.2 x 10 = 8.2 x 1033
[Ag(NH[Ag(NH33))22++] ] = 8.2 x 10 = 8.2 x 1033
[Ag(NH[Ag(NH33))++][NH][NH33]]
In a solution containing Ag and In a solution containing Ag and
NHNH33, all of the species NH, all of the species NH33, Ag, Ag++, ,
Ag(NHAg(NH33))++, and Ag(NH, and Ag(NH33))22++ exist at exist at
equilibrium. equilibrium.
Actually, metal ions in aqueous Actually, metal ions in aqueous
solution are hydrated.solution are hydrated.
More accurate representations would More accurate representations would
be be
Ag(HAg(H22O)O)22++ instead of Ag instead of Ag++, and , and
Ag(HAg(H22O)(NHO)(NH33))++ instead of Ag(NH instead of Ag(NH33))++. .
The equations would be:The equations would be:
Ag(HAg(H22O)O)22++
(aq) (aq) + NH+ NH33(aq) (aq)
Ag(HAg(H22O)(NHO)(NH33))++(aq)(aq) + H+ H22OO(l)(l)
KKf1f1 = 2.1 x 10 = 2.1 x 1033
[Ag(H[Ag(H22O)(NHO)(NH33))++] ] = 2.1 x 10 = 2.1 x 1033
[Ag(H[Ag(H22O)O)22++][NH][NH33]]
Ag(HAg(H22O)(NHO)(NH33))++(aq)(aq) + NH + NH33(aq)(aq)
Ag(NHAg(NH33))22++
(aq)(aq) + 2H + 2H22OO(l)(l)
KKf2f2 = 8.2 x 10 = 8.2 x 1033
[Ag(NH[Ag(NH33))22++] ] = 8.2 x 10 = 8.2 x 1033
[Ag(H [Ag(H22O)(NHO)(NH33))++][NH][NH33]]
The sum of the equations gives The sum of the equations gives the the
overall equation, so the product overall equation, so the product of of
the individual formation constants the individual formation constants
gives the overall formation gives the overall formation constant: constant:
AgAg++ + 2NH + 2NH3 3 Ag(NH Ag(NH33))22++
or or Ag(HAg(H22O)O)22
++ + 2NH + 2NH33 Ag(NH Ag(NH33))22++ + +
2H2H22OO
KKf1 f1 x K x Kf2f2 = K = Kff
(2.1 x 10(2.1 x 1033) x (8.2 x 10) x (8.2 x 1033) = 1.7 x 10) = 1.7 x 1077
Exercise 19Exercise 19
Calculate the equilibrium Calculate the equilibrium
concentrations of Cuconcentrations of Cu+2+2, NH, NH33, and , and
[Cu(NH[Cu(NH33))44]]+2+2 when 500. mL of 3.00 M when 500. mL of 3.00 M
NHNH33 are mixed with 500. mL of 2.00 x are mixed with 500. mL of 2.00 x
1010-3-3 M Cu(NO M Cu(NO33))22. .
KKformationformation = 6.8 x 10 = 6.8 x 101212..
Solubility Solubility and and Complex Complex IonsIons
Complex ions are often insoluble Complex ions are often insoluble in in
water. water.
Their formation can be used to Their formation can be used to
dissolve otherwise insoluble salts. dissolve otherwise insoluble salts.
Often as the complex ion forms, Often as the complex ion forms,
the equilibrium shifts to the right the equilibrium shifts to the right
and causes the insoluble salt to and causes the insoluble salt to
become more soluble. become more soluble.
If sufficient aqueous ammonia If sufficient aqueous ammonia is is
added to silver chloride, the added to silver chloride, the latter latter
can be dissolved in the form of can be dissolved in the form of
[Ag(NH[Ag(NH33))22]]++..
AgClAgCl(s)(s) Ag Ag++(aq)(aq) + Cl + Cl--(aq)(aq)
KKspsp = 1.8 x 10 = 1.8 x 10-10-10
AgAg++(aq)(aq) + 2 NH + 2 NH3(aq)3(aq) [Ag(NH [Ag(NH33))22]]++
(aq)(aq)
KKformationformation = 1.6 x 10 = 1.6 x 1077
Sum:Sum:
K = KK = Kspsp x K x Kformation formation = 2.0 x 10 = 2.0 x 10-3-3 = =
{[Ag(NH{[Ag(NH33))22++}[Cl}[Cl--] ]
[NH[NH33]]22
The equilibrium constant for The equilibrium constant for
dissolving silver chloride in dissolving silver chloride in ammonia ammonia
is not large; however, if the is not large; however, if the
concentration of ammonia is concentration of ammonia is
sufficiently high, the complex ion sufficiently high, the complex ion
and chloride ion must also be high, and chloride ion must also be high,
and silver chloride will dissolve.and silver chloride will dissolve.
Exercise 20Exercise 20 Complex IonsComplex Ions
Calculate the concentrations of Calculate the concentrations of
AgAg++, Ag(S, Ag(S22OO33))--, and Ag(S, and Ag(S22OO33))223-3- in a in a
solution prepared by mixing 150.0 solution prepared by mixing 150.0
mL of 1.00 X 10mL of 1.00 X 10-3-3 MM AgNO AgNO33 with with
200.0 mL of 5.00 200.0 mL of 5.00 MM Na Na22SS22OO33..
The stepwise formation The stepwise formation equilibria are:equilibria are:
AgAg++ + S + S22OO332-2- Ag(S Ag(S22OO33))--
KK11 = 7.4 X 10 = 7.4 X 1088
Ag(SAg(S22OO33))-- + S + S22OO332- 2- Ag(S Ag(S22OO33))22
3-3-
KK22 = 3.9 X 10 = 3.9 X 1044
SolutionSolution
[Ag[Ag++] = 1.8 X 10] = 1.8 X 10-18-18 MM
[Ag(S[Ag(S22OO33))--] = 3.8 X 10] = 3.8 X 10-9-9 MM
ACID-BASE AND PPT ACID-BASE AND PPT EQUILIBRIA OF PRACTICAL EQUILIBRIA OF PRACTICAL
SIGNIFICANCESIGNIFICANCE
SOLUBILITY OF SALTS IN WATER SOLUBILITY OF SALTS IN WATER
AND ACIDSAND ACIDS
The solubility of PbS in The solubility of PbS in water:water:
PbS PbS (s)(s) Pb Pb+2+2 + S + S-2-2
KKspsp = 8.4 x 10 = 8.4 x 10-28-28
The Hydrolysis of the SThe Hydrolysis of the S-2-2 ion in Waterion in Water
SS-2-2 + H + H22O O HS HS-- + OH + OH--
KKbb = 0.077 = 0.077
Overall Process:Overall Process:
PbS + HPbS + H22O O Pb Pb+2+2 + HS + HS-- + OH + OH--
KKtotaltotal = K = Kspsp x K x Kbb = 6.5 x 10 = 6.5 x 10-29-29
May not seem like much, but it can May not seem like much, but it can
increase the environmental lead increase the environmental lead
concentration by a factor of about concentration by a factor of about
10,000 over the solubility of PbS 10,000 over the solubility of PbS
calculated from simply Kcalculated from simply Kspsp!!
Any salt containing an anion Any salt containing an anion that is that is
the conjugate base of a weak the conjugate base of a weak acid acid
will dissolve in water to a will dissolve in water to a greater greater
extent than given by the Kextent than given by the Kspsp. .
This means salts of sulfate, This means salts of sulfate,
phosphate, acetate, carbonate, phosphate, acetate, carbonate, and and
cyanide, as well as sulfide can be cyanide, as well as sulfide can be
affected. affected.
If a strong acid is added to If a strong acid is added to water-insoluble salts such as ZnS water-insoluble salts such as ZnS
or CaCOor CaCO33, then hydroxide ions from , then hydroxide ions from the anion hydrolysis is removed by the anion hydrolysis is removed by the formation of water. the formation of water.
This shifts the anion hydrolysis This shifts the anion hydrolysis further to the right; the weak acid further to the right; the weak acid is is
formed and the salt dissolves.formed and the salt dissolves.
Carbonates and many metal Carbonates and many metal sulfides along with metal sulfides along with metal hydroxides are generally soluble hydroxides are generally soluble in strong acids. in strong acids.
The only exceptions are sulfides The only exceptions are sulfides of mercury, copper, cadmium and of mercury, copper, cadmium and a few others.a few others.
Insoluble inorganic salts containing Insoluble inorganic salts containing
anions derived from weak acids anions derived from weak acids
tend to be soluble in solutions of tend to be soluble in solutions of
strong acids. strong acids.
Salts are not soluble in strong acid Salts are not soluble in strong acid
if the anion is the conjugate base of if the anion is the conjugate base of
a strong acid!!a strong acid!!