Download Solubility & Complex Ions

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Hydroformylation wikipedia , lookup

Evolution of metal ions in biological systems wikipedia , lookup

Coordination complex wikipedia , lookup

Stability constants of complexes wikipedia , lookup

Metalloprotein wikipedia , lookup

Transcript
Solubility
Solubility
“Insoluble” salts are governed by
equilibrium reactions, and are really sparingly
soluble. There is a dynamic equilibrium between
the solvated ions and the crystalline solid.
Solubility
Solubility
The extent to which a substance dissolves is
its solubility. Solubility, as with all equilibria, is
highly temperature dependent.
Solubility may be expressed as grams of
compound/100 mL of H2O, or as molarity, M.
Solubility Conventions
The equilibrium constants for solubility are
based on the insoluble solid appearing on the
left side of the reaction, with the solvated ions
on the right side.
PbI2(s) ↔ Pb2+(aq) + 2I-(aq)
Ksp = [Pb2+][ I- ]2 = 1.4 x 10-8 @25oC
sp stands for solubility product.
Solubility
The molar solubility can be calculated from
the value of Ksp (and vice versa).
Since salts have differing numbers of cations
and anions, you cannot look at a table of Ksp
values and determine which compounds have
the greater solubility.
Ksp Values
Problem

Calcium phosphate has a Ksp of 1.3 x 10-32.
Calculate the molar solubility of calcium
phosphate.
The Common- Ion Effect
The solubility of a sparingly soluble salt will
be lower if one of the ions of the salt is provided
by an external source.
Silver iodide is sparingly soluble. Its
solubility in water will be greater than in a
solution of aqueous NaI. The additional iodide
ion will cause the AgI to precipitate out of
solution.
The Common- Ion Effect
pH and Solubility
Since most anions are weak bases, solubility
is highly pH dependent. The basic anion may
become protonated in acidic solutions, and thus
the “insoluble” salt will dissolve.
pH and Solubility
For example, consider the solubility of CuS.
Ksp=8.5 x 10-45. Sufide ion is the conjugate base
of HS-, a weak acid with a Ka of 1.3 x 10-13. This
means that sulfide ion is a relatively strong weak
base, and will accept protons.
pH and Solubility
Solubility Reaction:
CuS(s) ↔ Cu2+(aq) + S2-(aq) Ksp= 8.5 x 10-45
Reaction of Sulfide with Acid:
S2-(aq) + H3O+(aq) ↔ HS-(aq) + H2O(l)
Kb = Kw/Ka = (1.0 x 10-14)/1.3 x 10-13 = 7.7 x 10-2
Since sulfide is basic, CuS will dissolve in acidic
solution.
pH and Solubility
Since sulfide is basic, CuS will dissolve in
acidic solution. As acid is added, this reaction is
shifted towards the right.
S2-(aq) + H3O+(aq) ↔ HS-(aq) + H2O(l)
pH and Solubility
Since sulfide is basic, CuS will dissolve in
acidic solution. As acid is added, this reaction is
shifted towards the right.
S2-(aq) + H3O+(aq) ↔ HS-(aq) + H2O(l)
pH and Solubility
Since sulfide is basic, CuS will dissolve in
acidic solution. As acid is added, this reaction is
shifted towards the right.
S2-(aq) + H3O+(aq) ↔ HS-(aq) + H2O(l)
The decrease in [S2-] shifts this reaction
toward the right so as to make more sulfide.
CuS(s) ↔ Cu2+(aq) + S2-(aq)
pH and Solubility
CuS(s) ↔ Cu2+(aq) + S2-(aq)
The result is a large increase in solubility of
CuS in acidic solutions.
Problem: pH and Solubility

Calculate the solubility of Fe(OH)2 in buffers of
pH=3.00 and pH= 11.00. Ksp for Fe(OH)2
=1.8 x 10-15.
Predicting Precipitation Reactions
If the ions which may form a precipitate
come from two different sources (the cation
comes from one soluble salt, the anion from
another), calculate the value of Q.
Predicting Precipitation Reactions

Will a precipitate form if 10.0 mL of 0.010M
AgNO3 is added to 20.0 mL of 0.10M
Na2SO4?
1.
Determine the “insoluble” product.
Calculate the concentration of each ion of the
“insoluble” product.
Calculate Q and compare the value to Ksp.
2.
3.
Ksp Values
Selective Precipitation
Mixtures of metal ions in aqueous solution
are often separated by selective precipitation. A
solution containing an anion that will only
precipitate one of the metals while leaving the
others in solution is used.
Selective Precipitation

A solution contains 0.25M Ni(NO3)2 and 0.25M
Cu(NO3)2. Can the metal ions be separated by
slowly adding Na2CO3? (Assume that for
successful precipitation 99% of the metal ion
must be precipitated before the other metal ion
begins to precipitate, and assume no volume
change on addition of sodium carbonate.) Ksp
for NiCO3 =1.4 x 10-7, Ksp for CuCO3 =2.5 x
10-10.
Complex Ion Equilibria
A complex ion contains a central metal ion
bound to one or more ligands. A ligand is a
molecule or ion that acts as a Lewis base in
bonding to the metal ion.
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq)
Complex Ion Equilibria
The equilibrium reactions are associated with
an equilibrium constant called a formation constant
(Kf).
Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+(aq)
Kf = [Ag(NH3)2+]/[Ag+][NH3]2
Formation Constants
Most formation constants are quite large,
indicating that the reactions go very nearly to
completion.
As a result, metal metal ion salts with low
solubility can be easily dissolved if the proper
ligand is introduced into solution.
Complex Formation
Silver chloride is quite insoluble, but the
addition of ammonia will form a stable complex
ion, and cause the silver chloride to dissolve.
Ksp for AgCl = 1.77 x 10-10
Kf for Ag(NH3)2+ =
1.7 x 107
As a result, silver chloride
dissolves when aqueous
ammonia is added.