11

Some combinations of ions in solution form Some combinations of ions in solution form insoluble salts. Recall the solubility rules from insoluble salts. Recall the solubility rules from Chapter 4.Chapter 4.

Even insoluble salts dissolve to a small extent.Even insoluble salts dissolve to a small extent. Substances that are considered “insoluble” are Substances that are considered “insoluble” are

more appropriately termed “slightly” or more appropriately termed “slightly” or “sparingly” soluble.“sparingly” soluble.

These substances establish an equilibrium system These substances establish an equilibrium system between the solid and its ions in solution.between the solid and its ions in solution.

Ca(OH)Ca(OH)22 equilibrium equilibrium

15.6 Solubility Equilibria15.6 Solubility Equilibria

22

15.6 Solubility Equilibria15.6 Solubility Equilibria

Solubility of insoluble salts is described by the Solubility of insoluble salts is described by the solubility product constant, Ksolubility product constant, Kspsp..

AgCl(s) AgCl(s) ⇌⇌ Ag Ag++(aq) + Cl(aq) + Cl--(aq)(aq)

KKspsp = [Ag = [Ag++][Cl][Cl--] = 1.70 x 10] = 1.70 x 10--1010

AgAg22S(s) S(s) ⇌⇌ 2Ag 2Ag++(aq) + S(aq) + S22--(aq)(aq)

KKspsp = [Ag = [Ag++]]22[S[S22--] = 1.0 x 10] = 1.0 x 10--5151

Use to determine conditions for Use to determine conditions for precipitationprecipitation

33

SolubilitySolubility

Solubility (s) = molar concentration of Solubility (s) = molar concentration of dissolved salt; ion concentrations are related dissolved salt; ion concentrations are related to this by their coefficients.to this by their coefficients.

We can solve solubility problems just like We can solve solubility problems just like ANY other equilibrium problems.ANY other equilibrium problems.

Use an I.C.E. Equilibrium Analysis!!!Use an I.C.E. Equilibrium Analysis!!!

44

SolubilitySolubility

After setting up the balanced equation, can After setting up the balanced equation, can solve for S using an ice chart.solve for S using an ice chart.

Hint: The only difference that you will see in Hint: The only difference that you will see in solubility problems is that the unknown solubility problems is that the unknown concentration is designated with an ‘s’ instead concentration is designated with an ‘s’ instead of an ‘x’.of an ‘x’.

Practice: #75Practice: #75

55

Determining Ksp from Solubility

Determining Ksp from Solubility

1.1. What is the KWhat is the Kspsp value of Calcium value of Calcium

fluoride, given that the molar solubility of fluoride, given that the molar solubility of Calcium fluoride is 2.1 x 10Calcium fluoride is 2.1 x 10-4-4 M? M?

66

Determining Ksp from Solubility

Determining Ksp from Solubility

Can solve by setting up an ICE table:Can solve by setting up an ICE table:

CaFCaF22(s) (s) ⇌⇌ Ca Ca2+2+(aq) + 2 F(aq) + 2 F--(aq)(aq)

Initial Conc.Initial Conc. 0.00 M0.00 M 0.00 M0.00 MChangeChange + s + s + 2s+ 2sEquil. Conc.Equil. Conc. s s 2s2s

KKspsp = [Ca = [Ca2+2+][F][F--]]22 = (s)(2s) = (s)(2s)22 s = 2.1 x 10s = 2.1 x 10-4-4 M so Ksp = 4s M so Ksp = 4s33= = 3.7 x 103.7 x 10-11-11

Practice #77Practice #77

77

Calculating SolubilityCalculating Solubility

1. Determine the molar solubility of iron(III) 1. Determine the molar solubility of iron(III) hydroxide. (Khydroxide. (Kspsp = 1.1 x 10 = 1.1 x 10-36-36))

88

Calculating SolubilityCalculating Solubility

Can solve by setting up an ICE table:Can solve by setting up an ICE table:

Fe(OH)Fe(OH)33(s) (s) ⇌⇌ Fe Fe3+3+(aq) + 3OH(aq) + 3OH--(aq)(aq)

Initial Conc.Initial Conc. 0.00 M0.00 M 0.00 M0.00 M

ChangeChange + s + s + 3s+ 3s

Equil. Conc.Equil. Conc. s s 3s3s

KKspsp = [Fe = [Fe3+3+][OH][OH--]]33 = (s)(3s) = (s)(3s)33 = = 1.1 x 101.1 x 10--3636

27s27s44 = 1.1 x 10 = 1.1 x 10--3636

ss44 = 4.07 x 10 = 4.07 x 10--3838

s = 4.49 x 10s = 4.49 x 10--1010 M M

99

Calculating SolubilityCalculating Solubility

If solubility, S, is known, can reverse this If solubility, S, is known, can reverse this procedure and calculate Kprocedure and calculate Kspsp::

S = 4.49 x 10S = 4.49 x 10--1010 M M[Fe[Fe33++] = 4.49 x 10] = 4.49 x 10--1010 M M[OH[OH--] = 3 x 4.49 x 10] = 3 x 4.49 x 10--1010 = 1.35 x 10 = 1.35 x 10--99 KKspsp = [Fe = [Fe33++][OH][OH --]]33 = (4.49 x 10 = (4.49 x 10--1010 )(1.35 x 10 )(1.35 x 10--99))33

so, so, KKspsp = 1.1 x 10= 1.1 x 10--3636

1010

Calculating SolubilityCalculating Solubility

Practice: #81Practice: #81

Questions about problems in Notes??Questions about problems in Notes??

1111

Comparing Solubility using KspComparing Solubility using Ksp

We can directly compare only those salts We can directly compare only those salts that have the same exponents in the that have the same exponents in the solubility product expression.solubility product expression.

If that condition is met, the lower the value If that condition is met, the lower the value of Kof Kspsp, the less soluble the salt., the less soluble the salt.

If KIf Kspsp expressions are different, K expressions are different, Kspsp can be can be

used to calculate the solubility, which can used to calculate the solubility, which can be compared for any salts.be compared for any salts.

1212

15.6 Factors that Affect Solubility

15.6 Factors that Affect Solubility

Can modify solubility to dissolve minerals Can modify solubility to dissolve minerals and ores, to precipitate ions from solution, and ores, to precipitate ions from solution, to separate and purify ions.to separate and purify ions.

Examples:Examples:remove hardness from water by adding Naremove hardness from water by adding Na22COCO33 remove Agremove Ag++ from water by adding Cl from water by adding Cl-- to to

recover Agrecover Agdissolve Cu(OH)dissolve Cu(OH)22

..CuCOCuCO33 to mine Cu to mine Cuseparate the rare earth ionsseparate the rare earth ionsseparate Useparate U44++ from Th from Th44++

separate the halide ionsseparate the halide ions

1313

Common Ion EffectCommon Ion Effect

AgCl(s) AgCl(s) ⇌⇌ Ag Ag++(aq) + Cl(aq) + Cl--(aq)(aq)Common ion effect: Adding a common ion Common ion effect: Adding a common ion

to a solution of a slightly soluble salt will to a solution of a slightly soluble salt will cause the solubility of the salt to decrease. cause the solubility of the salt to decrease.

add more Cladd more Cl-- to precipitate more Ag to precipitate more Ag++ from from solution.solution.

1414

Common Ion EffectCommon Ion Effect

Saturated solution of AgCl:Saturated solution of AgCl:KKspsp = 1.70 x 10 = 1.70 x 10--1010 = [Ag = [Ag++][Cl][Cl--] = S] = S22

S = [AgS = [Ag++] = [Cl] = [Cl--] = 1.30 x 10] = 1.30 x 10--55 M MEffect of adding 0.100 M ClEffect of adding 0.100 M Cl--

(>> 1.30 x 10(>> 1.30 x 10--55, so [Cl, so [Cl--]= 0.100 M)]= 0.100 M)1.70 x 101.70 x 10--1010 = [Ag = [Ag++](0.100)](0.100)[Ag[Ag++] = 1.70 x 10] = 1.70 x 10--99 M M

1515

pH & Insoluble Basic SaltspH & Insoluble Basic Salts

Metal hydroxide solubility depends on the pH.Metal hydroxide solubility depends on the pH.Mg(OH)Mg(OH)22(s) (s) ⇌⇌ Mg Mg22++(aq) + 2OH(aq) + 2OH--(aq)(aq)Adjust the pH to adjust the solubility.Adjust the pH to adjust the solubility.AgCN(s) AgCN(s) ⇌⇌ Ag Ag++(aq) + CN(aq) + CN--(aq)(aq)

KKspsp = 1.6 x 10 = 1.6 x 10--1414

Add HNOAdd HNO33 (why not HCl?) to dissolve: (why not HCl?) to dissolve:CNCN--(aq) + H(aq) + H33OO

++(aq) (aq) HCN(aq) + H HCN(aq) + H22O(l)O(l)K = 2.5 x 10K = 2.5 x 1099

1616

Add the two equations (multiply the Ks):Add the two equations (multiply the Ks):

AgCN(s) + HAgCN(s) + H33OO++(aq) (aq) ⇌⇌ Ag Ag++(aq) + HCN(aq) (aq) + HCN(aq)

+ H+ H22O(l)O(l)

K = 1.6 x 10K = 1.6 x 10--1414 x 2.5 x 10 x 2.5 x 1099 = 4.0 x 10 = 4.0 x 10--55

= [Ag= [Ag++][HCN]/[H][HCN]/[H33OO++]]

Adding acid shifts the equilibrium towards Adding acid shifts the equilibrium towards products.products.

pH & Insoluble Basic SaltspH & Insoluble Basic Salts

1717

General Effects of pH on Solubility

General Effects of pH on Solubility

If the anion acts like a base, salt will be If the anion acts like a base, salt will be more soluble in acidic solutions.more soluble in acidic solutions.

If the cation acts like an acid, salt will be If the cation acts like an acid, salt will be more soluble in basic solutions.more soluble in basic solutions.

If neither ion acts like an acid / base, the If neither ion acts like an acid / base, the solubility will be unaffected by pH.solubility will be unaffected by pH.

1818

15.7 Precipitation and Separation of Ions

15.7 Precipitation and Separation of Ions

To precipitate an ion, we must add sufficient To precipitate an ion, we must add sufficient common ion for Qcommon ion for Qspsp to exceed K to exceed Kspsp..

To determine if a salt will precipitate, To determine if a salt will precipitate, compare the Ksp value to the Qsp value compare the Ksp value to the Qsp value (ratio of products to reactants when not (ratio of products to reactants when not necessarily at equilibrium).necessarily at equilibrium).Q < K ; no precipitate, more will dissolveQ < K ; no precipitate, more will dissolveQ = K ; solution is saturated; no precipitate, no Q = K ; solution is saturated; no precipitate, no

more will dissolvemore will dissolveQ > K ; precipitate will formQ > K ; precipitate will form

1919

Precipitation of IonsPrecipitation of Ions

Example:Example:How much ClHow much Cl-- must be in solution to begin must be in solution to begin

precipitation of Agprecipitation of Ag++ from a 0.100 M Ag from a 0.100 M Ag++ solution. Ksolution. Kspsp = 1.70 x 10 = 1.70 x 10--1010

KKspsp = 1.70 x 10 = 1.70 x 10--1010 = (0.100)[Cl = (0.100)[Cl--]][Cl[Cl--] = 1.70 x 10] = 1.70 x 10--1010/0.100 = 1.70 x 10/0.100 = 1.70 x 10--99 M MTo dissolve AgNOTo dissolve AgNO33 in water, with no in water, with no

precipitation of AgCl, we must have very precipitation of AgCl, we must have very pure water (no Clpure water (no Cl-- in the water). in the water).

2020

Precipitation of IonsPrecipitation of Ions

Example:Example:Will a solution containing 1.0 x 10Will a solution containing 1.0 x 10--66 M Cl M Cl--

and 1.5 x 10and 1.5 x 10--44 M Ag M Ag++ form a precipitate of form a precipitate of AgCl? KAgCl? Kspsp = 1.70 x 10 = 1.70 x 10--1010

Answer: Answer:

QQspsp = (1.0 x 10 = (1.0 x 10--66)(1.5 x 10)(1.5 x 10--44) = 1.5 x 10) = 1.5 x 10--1010

Qsp < KQsp < Kspsp, so AgCl will not precipitate., so AgCl will not precipitate.

2121

Precipitation of IonsPrecipitation of Ions

Group Work:Group Work:Want to separate 0.100 M PbWant to separate 0.100 M Pb22++ from 0.100 from 0.100

M AgM Ag++, using Cl, using Cl--. How much Ag. How much Ag++ is left is left when the Pbwhen the Pb22++ begins to precipitate? begins to precipitate?

KKspsp of PbCl of PbCl22 = 1.6 x 10 = 1.6 x 10--55

KKspsp of AgCl = 1.70 x 10 of AgCl = 1.70 x 10--1010

19m09vd119m09vd119m09vd1.mov.lnk

2222

Precipitation of IonsPrecipitation of Ions

For PbClFor PbCl22, K, Kspsp = 1.6 x 10 = 1.6 x 10--55 = = (0.100)[Cl(0.100)[Cl--]]22

[Cl[Cl--] = 0.0126 M to begin precipitating PbCl] = 0.0126 M to begin precipitating PbCl22..

ForAgCl, KForAgCl, Kspsp = 1.70 x 10 = 1.70 x 10--1010 = [Ag = [Ag++](0.0126)](0.0126)

[Ag[Ag++] = 1.35 x 10] = 1.35 x 10--88 M left when PbCl M left when PbCl22 begins to begins to

precipitate.precipitate.% Ag% Ag++ left = (1.35 x 10 left = (1.35 x 10--8 8 M) / 0.100 M x 100%M) / 0.100 M x 100%

= 1.35 x 10= 1.35 x 10--55%%

Questions about problems in notes??Questions about problems in notes??