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Question 1

Solubility Equilibria: Saturated Solutions

Answer 1

When a slightly soluble solid salt and its aquated ions are at equilibrium

Question 2

solubility product (Ksp).

Answer 2

equilibrium constant

in saturated: Q = K

Question 3

Solubility and precipitation

Answer 3

Solubility the concentration of a substance in a saturated solution at a specified temperature

Solubility product (Ksp) the equilibrium constant in a saturated solution of a slightly soluble salt

Calculations of solubilities from solubility products gives accurate estimates only if the cation and/or the anion are only involved in an equilibrium
reaction between the solid salt and its aquated ions in saturated solution.

Question 4

Solubility and Precipitation in Barium Sulfate

Answer 4

To reduce toxicity, potassium sulfate is added to reduce solubility through common ion effect

Question 5

pH-Dependence of Solubility of Salts Whose Anions Are Bases

Answer 5

become more soluble as the pH of solution lowers. The weaker the anion is as a base, the more pH-dependent the solubility of such salts.

Question 5

Calcium carbonate and acidified water

Answer 5

the cause of deterioration of marble statues in
many parts of the world. The marble, a form of calcium carbonate, has reacted over many
years with H3O+(aq) ions in “acid rain”—rain with dissolved acidic pollutants derived from oxidation in the atmosphere of SO2 and NO2.

Question 6

Industrial methods for extracting metals from ores

Answer 6

• Dissolve metal salts to obtain the metal ions in solution.
• Separate the solution from insoluble materials.
• Add a reagent to precipitate selectively a salt of only the desired metal.
• Reduce the cation to the metal either chemically or electrochemically

Question 7

Mineral precipitation reactions

Answer 7

Fluorite (calcium fluoride)

Black hematite (iron (III) oxide)

brown geothite (a mixture of iron (III) oxide and iron (III) hydroxide).

• often insoluble salts and are formed through geological precipitation reactions.

Question 8

Deciding If a Salt Will Precipitate: Q versus Ksp

Answer 8

– If Q = Ksp , the solution is saturated and at equilibrium.
– If Q < Ksp , the solution is unsaturated and more can dissolve.
– If Q > Ksp , the solution is supersaturated, and precipitation would bring the system to equilibrium.

Question 9

Precipitation When Reagent Solutions Are Mixed

Answer 9

one of which contains the cation of a slightly soluble salt, and the other containing the anion, we can decide whether salt will precipitate, or what concentration would be needed for it to occur.

Question 10

How do we know what concentration to increase one of the ions to induce saturation?

Answer 10

Adjust the concentration of the added ion until Q = Ksp.

Question 11

Separation of Metal Cations by Selective Precipitation

Answer 11

When two metals have salts of the same anion with sufficiently different solubilities, they may be separated by selective precipitation.

• In a solution containing aquated cations of both metals, if the concentration of the aquated anion is raised incrementally, as the solution becomes saturated with respect to the most soluble salt, the less soluble salt selectively precipitates from solution.

Question 12

Complexation Reactions

Answer 12

involve bonding between a metal ion and a species (ligand) which donates a lone pair of electrons.

Products are complex ions.

Question 13

Lewis acid-base model

Answer 13

The analysis of complexation reactions leads to a more general acid-base model

Question 14

Lewis base / ligand

Answer 14

The lone pair donor

Question 15

Lewis acid

Answer 15

recipient of electron pair

Question 16

acid-base adduct

Answer 16

The product of an acid-base reaction in the Lewis model

Question 17

Aquated Metal Ions

Answer 17

the result of competition between water and other Lewis bases to bond with the metal.

Question 18

Competitive Equilibria

Answer 18

• Salts that are only slightly soluble in water can dissolve as a result of reaction between the metal cation and a Lewis base to form a complex ion.
• This can be regarded as the result of two competing reactions for the cation: formation of a solid lattice vs. formation of the complex ion.
• Equilibrium constants for formation of complex ions, called formation constants (Kf ), are measures of stability of the complex ions.

Question 19

Competition: Precipitation vs. Complexation.

Answer 19

For example:
Beginning with a precipitate of AgCl(s), adding an aqueous ammonia solution dissolves the precipitate as the soluble complex ion [Ag(NH3)2]+(aq) is formed.

Question 20

Coordination complexes

Answer 20

contain a metal atom or ion, bonded to which are molecules or ions called ligands or Lewis bases (which supply the electron pair).

• The net charge on a coordination complex is the sum of the charges on the metal and its attached groups.

Question 21

coordination number.

Answer 21

The number of atoms to which the metal atom is bonded

Question 22

Complex Stability

Answer 22

The equilibrium constant for a reaction in which a metal ion complex is formed from the aqua complex of the metal ion is called the formation constant ß

The larger the formation constant, the more
stable is the complex

Complex ions with polydentate ligands are more stable than complexes that have the same number of donor atoms in monodentate ligands.

Question 23

chelate effect

Answer 23

due to differences in entropy of formation of the complexes

Question 24

Labile substances

Answer 24

Complexes that undergo rapid substitution of ligand species

Question 25

Inert substances

Answer 25

Complexes undergo slow substances

Question 26

Crystal-field theory

Answer 26

focuses on the electrons in the d orbitals of the transition metal ion, and the effect of ligands on their energies.

• It assumes the metal-ligand bond is due to electrostatic attractions between the positively charged metal ion and the donor atom.
• This is a very simplified view of the bonding, but provides us with an entry point to consider some otherwise complex topics (which will be addressed in more detail in level 3 and 4 courses)

Question 27

Crystal-field theory: d-orbital energy splitting

Answer 27

Before bonding, the five 3d orbitals on the metal ion have equal energy (degenerate).

The effect of bonding on these orbitals can be imagined by considering the ligands as point charges on x, y, and z-axes around the metal.

Question 28

ligand-field splitting (Δo).

Answer 28

difference in energy of the orbitals upon bonding

Example: Splitting in an octahedral complex.

Question 29

Spectrochemical series

Answer 29

lists ligands in order of their ability to interact with metal d orbitals and bring about ligand-field splitting.

• The magnitude of the ligand-field splitting affects the colour of the metal complex.

Question 30

Geometry of the complex

Answer 30

defines the order and energies of d orbitals as the position of the ligands around the metal ion will influence which d-orbitals are most affected by donation from the ligand

EXAMPLE: Tetrahedral and square planar complexes.

Question 31

Magnetism of transition metal complexes

Answer 31

related to the magnitude of ligand-field splitting.

– Strong-field ligands yield low-spin electron configurations. Electrons pair up low-energy orbitals.

– Weak-field ligands yield high-spin configurations. Electrons are able to populate high-energy orbitals instead of pairing.