Chapter 17:
Solubility & Complex
Ion Equilibria
◆
Solubility Equilibria
equilibrium between a solid and its ions in solution
ex. for calcium phosphate
Ca3(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”
Overview of the Chapter
◆
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 expressions
calculate and interpret Ksp
Ksp is the solubility product constant
using Ksp, calculate solubility of salts
mol/L (molar solubility), and g/L
◆
factors that affect solubility
common ion effect
pH
formation of complex ions
◆
calculations to determine whether precipitation of a solid
will occur when 2 solutions are combined
Some Solution Terminology
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the solubility equilibrium can also be referred to as
the dissolution–precipitation equilibrium
dissolution
Ca3(PO4)2 (s) ⇄ 3 Ca2+ (aq) + 2 PO43– (aq)
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◆
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
precipitation
◆
sol’n may be saturated, unsaturated, or supersaturated
unsaturated 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
◆
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?
◆
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?
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
xM
Comparing Solubilities of Salts
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.
◆
◆
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
define: x = mol of solid that dissolve per L of sol’n
∴ x is the molar solubility of the salt; units mol/L
NiS (s)
⇄
◆
Ni2+
(aq)
+
S2–
(aq)
initial [ ]
---
0
0
∆[]
---
+x
+x
equil [ ]
---
xM
xM
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
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)
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
⇄
Mg2+ (aq)
+
2 F– (aq)
initial [ ]
---
0
0.10
∆[]
---
+x
+ 2x
equil [ ]
---
xM
(0.10 + 2x) M
◆
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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
Factors that Affect Solubility
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
◆
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)
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Factors that Affect Solubility
◆
complex ion formation
The solubility of an ionic compound may increase
dramatically if a solution containing a Lewis base is
added.
some common examples of basic anions:
CO32–, OH–, PO43–, SO42–, C2O42–, CN–, F–, S2–
added Lewis base may react with a metal
cation to form a Lewis acid-base adduct
called a complex ion
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
formation of complex ion is an equilibrium
with equilibrium constant, Kf – formation
constant
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
◆
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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:
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NH3 (aq) added to a solution
of CuSO4 (aq)
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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
Calculation of Solubility After Complex Ion Formation
example (see example 17.11 in text):
example (see example 17.11 in text):
Calculate the molar solubility of AgBr in 2.25 M
Na2S2O3 (aq).
Calculate the molar solubility of CuI in 0.88 M KCN
(aq).
For AgBr, Ksp = 5.0 x 10–13; for the complex ion
[Ag(S2O3)2]3–, Kf = 2.9 x 1013.
For CuI, Ksp = 1.1 x 10–12; for the complex ion
[Cu(CN)2]–, Kf = 1.0 x 1016.
Precipitation of Ionic Solids
Precipitation of Ionic Solids
◆
For an ionic solid, MX the solubility equilibrium is
given by:
MX (s) ⇄
◆
◆
◆
Mn+
(aq) +
Xn–
Is the system at equilibrium? If not, in what
direction does the reaction proceed to reach
equilibrium?
determine concentrations of ions after solutions are
combined:
mol ion
[ion] = ––––––––––––––
total sol’n volume
(aq)
When 2 solutions are combined – one sol’n
containing Mn+, and one sol’n containing Xn– – will a
precipitate of MX form?
Q vs K calculation
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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
example:
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.
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)
Ag2CO3 (s) ⇄ 2 Ag+ (aq) + CO32– (aq)
Q = [Ag+]2[CO32–]
Ksp = [Ag+]2[CO32–]
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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