Transcript Acid-Base Equilibria and Solubility Equilibria

Solubility Equilibria
Chapter 16
Solubility Equilibrium:
Chemical equilibrium between
a saturated solution of ions
and the remaining solid.
•Solid salts dissociate into
ions
•Ions precipitate and reform
the solid salt.
AgCl(s) ⇌
+
−
Ag (aq)+Cl (aq)
Equilibrium Expression?
+
−
• Keq= [Ag ]·[Cl ] = Ksp
[AgCl]
Solubility Product Ksp
• product
of concentrations
of each ion raised to the
powers of their
coefficients.
Ksp= [Ag+]·[Cl−]
for previous example
Solubility Product:
c+
bBbCc ⇌ bB + cC
Ksp=
c+
b
bc
[B ] • [C ]
Like all equilibrium constants, this is a
constant for that particular temperature.
Problem #1:
•
What is the solubility
product expression for
CaSO4?
Ksp=
2+
2−
[Ca ]·[SO4 ]
Problem #2:
•
What is the solubility
product expression for
Cu3(PO4)2 ?
Ksp=
2+
3
3−
2
[Cu ] ·[PO4 ]
Problem #3:
•
What are the concentrations
of silver and chloride ions in a
saturated solution of AgCl?
−10
Ksp= 1.8 x 10
+
−
[Ag ]·[Cl ]
2
Ksp=
=x
−10
x = SQRT (1.8x10 )
−5
+
=1.34 x10 M Ag and
−
Cl
Problem #4:
Silver sulfide (Ag2S) has a solubility
−17
of 3.4x10 M at 25°C. Calculate
Ksp for Ag2S.
Ag2S ⇌ 2Ag+ + S2−
+
−
Ksp= [Ag ]·[Cl ] =
(2∙(3.4x10−17))2∙(3.4x10−17)
−49
=1.57x10
AP Exam
Know the effect the addition of a common
ion will have on solubility.
 Highly UNLIKELY that you will need to do
calculations like Problem #7.
 Common Ion Effect is also seen with pH:
Ca(OH)2(s) ⇌ Ca2+(aq)+ 2OH−(aq)

◦ Adding a base (OH −) will do what?
◦ Adding an acid (H+) will do what?
←
→
Ion Product (Q):
•
•
•
•
used to determine if a reaction is at
equilibrium.
Calculate as for Ksp using the current
concentrations.
If Q > Ksp the reaction proceeds to the left
(precipitates - forming more solid reactant)
If Q ≤ Ksp then no precipitation occurs.
Problem:
•
Will a ppt. form if 20 mL
0.01M BaCl2 is added to 20
mL 0.005M Na2SO4?
(Ksp table p.A24)
Possible products: BaSO4, NaCl
Ksp= 1.5x10−9
[Ba+ ] = .0002mol/.04L=.005M
[SO42−]=.0001mol/.04L= .0025M
(.005)(.0025)=1.25x10−5 …is > Ksp
Solubility Rules – ions
soluble in water
•
alkali metal ions & ammonium
(NH4+)
• nitrate (NO3-), chlorate (ClO4-, ClO3-)
and acetate (C2H3O2-) ions
• halide ions (Cl-, Br-, I-) EXCEPT if
combined with Ag+, Pb2+, or Hg22+
• sulfates (SO42-) EXCEPT if
combined with Pb2+, Sr2+, or Ba2+
Insoluble in water
(OH -) & sulfides (S2-)
EXCEPT if combined with group I
2+
2+
2+
ions, Ca , Sr , and Ba , or
+
ammonium (NH4 ).
• phosphate (PO43-), carbonate
22(CO3 ), & sulfite (SO3 ) ions
EXCEPT if combined with group I
+
ions or NH4 .
• hydroxides
Problem:
What is the solubility of solid PbI2 in
a 0.0333 M NaI solution?
What are the concentrations of the
ions in solution? Ksp = 1.4x10−8
PbI2(s) ⇌ Pb2+(aq) + 2 I−(aq)
Ksp = [Pb2+] [I−]2 = x(0.0333 +2x)2
x=1.4x10−8/0.00111= 1.26x10−5 M
x = 1.26x10−5 M Pb2+(aq)
0.0333 + 2x = 0.0333 M I−(aq)
Selective
Precipitation
of Cations
Complex Ion
A charged species of a metal ion surrounded by
ligands (Lewis base).
Ligands
Neutral molecule or ion having a lone electron pair that can
be used to form a covalent bond to a metal ion.
Coordination Number

Number of ligands attached to the metal ion.
◦ Usually 6 or 4 ligands, but can also be 2 or 8.
Complex Ion Equilibria
Metal ions add ligands one at a time:
Be2+(aq) + F–(aq) ⇌ BeF+(aq)
K1 = 7.9 x 104
BeF+(aq) + F–(aq) ⇌ BeF2(aq)
K2 = 5.8 x 103
BeF2(aq) + F–(aq) ⇌ BeF3– (aq)
K3 = 6.1 x 102
BeF3– (aq) + F–(aq) ⇌ BeF42– (aq)
K4 = 2.7 x 101
Be2+(aq) + 4F–(aq) ⇌ BeF42– (aq)
Formation constants
Using Hess’s Law we add the equations.
This K = K1 x K2 x K3 x K4 = 7.55x1012
Problem:
Calculate the solubility of silver chloride in 10.0 M
ammonia given the following information:
AgCl(s) ⇌ Ag+(aq) + Cl−(aq)
Ksp = 1.6 x 10−10
Ag+ + NH3 ⇌ AgNH3+
K1 = 2.1 x 103
AgNH3+ + NH3 ⇌ Ag(NH3)2+ K2 = 8.2 x 103
AgCl(s)+ 2NH3(aq)⇌ Ag(NH3)2+(aq)+ Cl−(aq)
+
K=
K=
Ag NH 3 2 Cl −
𝑁𝐻3 2
𝑥2
10.0 − 2𝑥
2
= Ksp ∙K1 ∙K2 = 0.00276
Square root of both sides. Solve for x.
x = 0.475 M which is the solubility of
AgCl in this 10M NH3 solution.
Comparison: This is much more soluble than in water, where
the solubility = SQRT Ksp = 1.3x10−5 M
Complex Ions and Solubility

To increase solubility of an insoluble ionic solid:
 Acidify the solution if the anion is a good base.
 Add a ligand (Lewis base that forms stable
complex ions with the cation) if the anion is not
sufficiently basic.
Transition Metals
Ch.21
3d Transition Metals
 Scandium – chemistry strongly resembles lanthanides
 Titanium – excellent structural material (light weight)
 Vanadium – mostly in alloys with other metals
 Chromium – important industrial material
 Manganese – production of hard steel
 Iron – most abundant heavy metal
 Cobalt – alloys with other metals
 Nickel – plating more active metals; alloys
 Copper – plumbing and electrical applications
 Zinc – galvanizing steel
Equilibria
Equilibrium reaction: aA + bB ⇌ cC + dD
Keq = [C]c · [D]d
[A]a · [B]b
Kw?
Ka?
Kb ?
Ksp?
Review Questions
What is a buffer?
 What equations do we use with buffers?
 How do you make the most efficient
buffer?
 Finding pH of strong acid?
 What are the strong acids?
 Finding pH of a strong base?
 What are the strong bases?
 Finding pH of weak acid?
 How do you get Ka from Kb?

Review Questions





Write the equilibrium reaction of a weak acid
(HF) in water:
Write the equilibrium reaction of a weak base
(NH3) in water:
What is the common ion effect?
How do you find the pH with a common ion?
Titrations: describe shape of curve, initial pH, pH
as add titrant, pH at equivalence point, post-eq.pt.
pH for:
◦
◦
◦
◦
◦
◦
Strong Acid to a Strong Base
Strong Base to a Strong Acid
Strong Base to a Weak Acid
Strong Acid to a Weak Base
Weak Base to a Strong Acid
Weak Acid to a Strong Base
Review Questions

pH of:
◦
◦
◦
◦
◦
excess strong acid? Strong base?
pH of weak acid? Weak base?
pH in a buffer?
pH of SA-SB at equivalence point?
pH of SA-WB at equivalence point?
Ksp
 Solubility of a salt
 Determining which salt is most soluble
