solubility and complex ion equilibria. slightly soluble salts establish a dynamic equilibrium with...

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!!