Chapter 17:Solubility & Complex

Ion Equilibria

◆ The goal of this chapter is to understand the equilibria that exist between ionic solids and their ions in solution, and factors that affect that equilibrium.

◆ write (heterogeneous) equilibrium equations & K expressionscalculate and interpret Ksp

Ksp is the solubility product constantusing Ksp, calculate solubility of salts

mol/L (molar solubility), and g/L

◆ factors that affect solubilitycommon ion effectpHformation of complex ions

◆ calculations to determine whether precipitation of a solid will occur when 2 solutions are combined

Overview of the Chapter

Solubility Equilibria◆ equilibrium between a solid and its ions in solution

ex. for calcium phosphateCa3(PO4)2 (s) ⇄ 3 Ca2+ (aq) + 2 PO43– (aq)

◆ this is a heterogeneous equilibrium

◆ equilibrium constant: Ksp solubility product constant for calcium phosphate: Ksp = [Ca2+]3[PO43–]2

◆ we can now refine our understanding of solubility rules from Chapter 4

solids that we classified as “insoluble” typically have small Ksp’s and low molar solubilities

“slightly soluble” or “sparingly soluble”

Some Solution Terminology

◆ the solubility equilibrium can also be referred to as the dissolution–precipitation equilibrium

Ca3(PO4)2 (s) ⇄ 3 Ca2+ (aq) + 2 PO43– (aq)dissolution

precipitation

◆ sol’n may be saturated, unsaturated, or supersaturatedunsaturated sol’n: more solid can dissolve

reaction continues in forward direction toward equilibrium (Q < Ksp)

supersaturated sol’n: ion [ ]’s are too high; solid will precipitate out of the solution

reaction continues in reverse direction toward equilibrium (Q > Ksp)

saturated sol’n: solution is at equilibrium (Q = Ksp)

Ion Concentrations in a Saturated Solution◆ consider 2 different preparations of a saturated sol’n

of CaF2

◆ examine the relationships between [Ca2+], [F–], Ksp

CaF2 (s) ⇄ Ca2+ (aq) + 2 F– (aq)Ksp = [Ca2+][F–]2

◆ preparation 1: put solid CaF2 in a flask; add water; wait until equilibrium is established

at equilibrium: [Ca2+] = 2.0 x 10–4 M [F–] = 4.0 x 10–4 M

Calculate Ksp.

note: with this preparation, [F–] = 2•[Ca2+] Why?

Ion Concentrations in a Saturated Solution

◆ examine the relationships between [Ca2+], [F–], KSP

CaF2 (s) ⇄ Ca2+ (aq) + 2 F– (aq)Ksp = [Ca2+][F–]2

◆ preparation 2: a sol’n with containing Ca2+ is mixed with a solution containing F–; after some time, equilibrium is established

at equilibrium: [Ca2+] = 0.038 M Ksp = 3.2 x 10–11

Determine equilibrium [F–].

note: with this preparation, [F–] ≠ 2•[Ca2+] Why not?

example:

A saturated solution of silver chromate is prepared by dissolving solid Ag2CrO4 in water, and allowing the solution to reach equilibrium.

The saturated solution has [Ag+] = 1.3 x 10–4 M.

Determine [CrO42–] and Ksp.

Ag2CrO4 (s) ⇄ 2 Ag+ (aq) + CrO42– (aq)

initial [ ] --- 0 0

∆ [ ] --- + 2x + x

equil [ ] --- 2x M x M

Using Ksp to Determine Solubility

example:

Calculate the solubility (in mol/L and g/L) of nickel (II) sulfide in water at 25°C. For NiS, Ksp = 3.0 x 10–19.

NiS (s) ⇄ Ni2+ (aq) + S2– (aq)

initial [ ] --- 0 0

∆ [ ] --- + x + x

equil [ ] --- x M x M

◆ define: x = mol of solid that dissolve per L of sol’n ∴ x is the molar solubility of the salt; units mol/L

Comparing Solubilities of Salts

◆ salts with greater solubility have higher [ions] in saturated solution

◆ solubility is related to equilibrium position

the farther to the right the equilibrium position, the greater the solubility

◆ be careful about comparing Ksp’s directly to determine relative solubilities of 2 salts . . .

must consider the salt stoichiometry and relationship between Ksp & x

Comparing Solubilities of Salts

Which has greater solubility in water at 25°C? PbCl2 or PbF2

for PbCl2 Ksp = 1.6 x 10–5

for PbF2 Ksp = 4.0 x 10–8

Which has greater solubility in water at 25°C?PbCl2 or CaSO4

for PbCl2 Ksp = 1.6 x 10–5

for CaSO4 Ksp = 6.1 x 10–5

Factors that Affect Solubility◆ the common ion effect

The molar solubility of MgF2 in water at 25°C is 2.6 x 10–4 mol/L.

Determine the molar solubility of MgF2 in 0.10 M NaF (aq). For MgF2, Ksp = 7.4 x 10–11.

MgF2 (s) ⇄ Mg2+ (aq) + 2 F– (aq)

initial [ ] --- 0 0.10

∆ [ ] --- + x + 2x

equil [ ] --- x M (0.10 + 2x) M

◆ solve x; x = molar solubility = 7.4 x 10–9 mol/L◆ the presence of a common ion (here F–) reduces the

solubility of a salt

example:

Determine the solubility of lead (II) hydroxide in a solution with pH = 10.00. For Pb(OH)2, Ksp = 1.2 x 10–15.

◆ sol’n with pH = 10.00 has [OH–] = 1.0 x 10–4 M

Factors that Affect Solubility

◆ pH of solution

In general, for an ionic compound with a basic anion, solubility will increase as the pH of the solution decreases.

H+ present reacts with the basic anion[anion] decreases

Le Chatelier’s principle predicts that equilibrium will shift to the right (in the direction of greater dissolution of solid).

example: CaCO3 (s) ⇄ Ca2+ (aq) + CO32– (aq)

calcium carbonate will be more soluble in acidic sol’n because the following reaction results in decreased [CO32–]:

CO32– (aq) + H+ (aq) → HCO3– (aq)

◆ some common examples of basic anions:CO32–, OH–, PO43–, SO42–, C2O42–, CN–, F–, S2–

◆ remember that the following anions are neutral:Cl–, Br–, I–, NO3–, ClO4–

the solubility of salts with these anions is not affected by lowering the pH of the solution

◆ complex ion formation

The solubility of an ionic compound may increase dramatically if a solution containing a Lewis base is added.

added Lewis base may react with a metal cation to form a Lewis acid-base adduct called a complex ion

formation of complex ion is an equilibrium with equilibrium constant, Kf – formation constant

Factors that Affect Solubility

example:

AgCl has very limited solubility in water and acidic solution, but will dissolve in NH3(aq):

AgCl (s) ⇄ Ag+ (aq) + Cl– (aq) Ksp = 1.6 x 10–10

Ag+ (aq) + 2 NH3 (aq) ⇄ [Ag(NH3)2]+ (aq) Kf = 1.7 x 107

AgCl (s) + 2 NH3 (aq) ⇄ [Ag(NH3)2]+ (aq) + Cl– (aq) KC = 0.0028

◆ Ag+ ions in solution react with NH3 to form the complex ion [Ag(NH3)2]+

◆ as the complex ion forms, [Ag+] in sol’n decreases

◆ as [Ag+] decreases, the solubility equilibrium position to shifts to the right ∴ more AgCl dissolves

another example of complex ion formation and effect on solubility:

◆ NH3 (aq) added to a solution of CuSO4 (aq)

◆ initially a precipitate of Cu(OH)2 (s) forms as solution becomes basic

◆ then the dissolution of Cu(OH)2 is observed after further addition of NH3 (aq) as Cu(NH3)22+ forms

Zn(OH)2 (s) ! Zn2+ (aq) + 2 OH- (aq); Ksp = 2.1 x 10-16

Zn2+ (aq) + 4 OH- (aq) ! Zn(OH)42-(aq); Kf = 2.8 x 1015

Zn(OH)2 (s) + 2 OH- (aq) ! Zn(OH)42- (aq); KC = 0.59

Calculation of Solubility After Complex Ion Formation

example (see example 17.11 in text):

Calculate the molar solubility of AgBr in 2.25 M Na2S2O3 (aq).

For AgBr, Ksp = 5.0 x 10–13; for the complex ion [Ag(S2O3)2]3–, Kf = 2.9 x 1013.

Calculation of Solubility After Complex Ion Formation

example (see example 17.11 in text):

Calculate the molar solubility of CuI in 0.88 M KCN (aq).

For CuI, Ksp = 1.1 x 10–12; for the complex ion [Cu(CN)2]–, Kf = 1.0 x 1016.

Precipitation of Ionic Solids

◆ For an ionic solid, MX the solubility equilibrium is given by:

MX (s) ⇄ Mn+ (aq) + Xn– (aq)

◆ When 2 solutions are combined – one sol’n containing Mn+, and one sol’n containing Xn– – will a precipitate of MX form?

◆ Is the system at equilibrium? If not, in what direction does the reaction proceed to reach equilibrium?

◆ Q vs K calculation

Precipitation of Ionic Solids

◆ determine concentrations of ions after solutions are combined:

[ion] = ––––––––––––––

◆ calculate the ion product, Q

◆ compare Q to Ksp:if Q = Ksp, the solution is saturated;

solution is at equilibrium

if Q > Ksp, the solution is supersaturated; precipitation of solid will occur

if Q < Ksp, the solution is unsaturated; precipitation will not occur - more solid will dissolve

mol iontotal sol’n volume

example:

250.0 mL of 0.0062 M AgNO3 (aq) and 250.0 mL of 0.00014 M Na2CO3 are combined. Will a precipitate of Ag2CO3 form? For Ag2CO3, Ksp = 8.1 x 10–12.

Ag2CO3 (s) ⇄ 2 Ag+ (aq) + CO32– (aq)

Q = [Ag+]2[CO32–]

example:

Determine the minimum concentration of carbonate ion required to cause the precipitation of silver carbonate from a 5.8 x 10–4 M solution of AgNO3. For Ag2CO3, Ksp = 8.1 x 10–12.

Ag2CO3 (s) ⇄ 2 Ag+ (aq) + CO32– (aq)

Ksp = [Ag+]2[CO32–]

◆ solve for [CO32–] present in a saturated solution of Ag2CO3;

[CO32–] at equilibrium OR when Q = Ksp

◆ any greater [CO32–] will result in Q > Ksp and precipitation of solid