CHEM FINAL

Solubility equilibrium

A dynamic equilibrium where dissolution and precipitation occur at equal rates.

Insoluble compound

Compound with very low solubility, does not mean zero solubility

Saturated solution

Solution containing maximum amount of dissolved solute possible

Solubility rules

Patterns used to predict whether ionic compounds dissolve

Generally soluble salts

Group 1 salts and nitrates are generally soluble

Common soluble halides

Many chlorides, bromides, and iodides are soluble, except for important ions such as Ag+, Pb2+, and Hg2^2+.

Poorly soluble anions

Many carbonates, phosphates, sulfides, and hydroxides are poorly soluble unless paired with certain cations.

Ksp

The equilibrium constant for the dissolution of a sparingly soluble ionic solid

Ksp rule for pure solids

Pure solids are omitted from equilibrium expressions because their effective concentration is constant.

Molar solubility x for CaF2

If CaF2 has molar solubility x, then [Ca2+] = x and [F-] = 2x.

Ksp from CaF2 molar solubility x

Ksp = (x)(2x)^2 = 4x^3.

Calculate Ksp from solubility

1) Write the balanced dissolution equation. 2) Relate ion concentrations to molar solubility using coefficients. 3) Substitute into Ksp. 4) Solve.

Calculate Ksp from g/L solubility

Convert g/L to mol/L using molar mass, use stoichiometry to find ion concentrations, then substitute those concentrations into Ksp.

CaF2 calculation example

If [CaF2] dissolved = 2.15×10^-4 M, then [F-] = 2(2.15×10^-4) = 4.30×10^-4 M. Ksp = (2.15×10^-4)(4.30×10^-4)^2 ≈ 3.98×10^-11.

Common-ion effect

Decrease in solubility that occurs when an ion already present in the dissolution equilibrium is added.

Common-ion effect direction

Adding a common ion shifts the dissolution equilibrium left and decreases solubility.

AgCl + NaCl

NaCl adds Cl-, a common ion for AgCl, so AgCl solubility decreases.

AgCl + AgNO3

AgNO3 adds Ag+, a common ion for AgCl, so AgCl solubility decreases.

AgCl + NaNO3

NaNO3 supplies neither Ag+ nor Cl-, so there is no common-ion effect in the simplified treatment.

Compare AgCl in 0.10 M vs 0.05 M NaCl

AgCl is more soluble in 0.05 M NaCl because the lower common-ion concentration causes a weaker common-ion effect.

Qsp

ion product expression using current ion concentrations instead of equilibrium concentrations

Q < K

The solution is unsaturated, no precipitate forms until equilibrium is restored

Q = K

The solution is saturated and at equilibrium.

Q > K

too many dissolved ions, precipitate forms until equilibrium is restored

Precipitation calculation

1) Write the possible ionic dissolution equation. 2) Write Qsp. 3) Use the current ion concentrations. 4) Compare Qsp with Ksp. 5) Qsp > Ksp means precipitate forms.

Qsp example

For CaCO3, Qsp = [Ca2+][CO3^2-]. If the calculated Qsp is greater than the tabulated Ksp, CaCO3 precipitates.

Selective precipitation

Separating ions by adding a reagent that causes one ion to precipitate before another because their Ksp values differ.

ICE table for Ksp

An ICE table can be used to determine equilibrium ion concentrations when initial concentrations and Ksp are given.

Coupled equilibria

Two or more equilibrium reactions that share a reactant or product, so a shift in one can cause a shift in another.

Le Chatelier coupled-equilibrium rule

If one equilibrium removes a dissolved product of another equilibrium, the first equilibrium shifts to replace what was removed.

Atmospheric CO2 and ocean acidity

More atmospheric CO2 → more dissolved CO2 → more H2CO3 → more H3O+ → lower pH and greater ocean acidity.

Daytime coral-reef pH

Photosynthesis removes dissolved CO2 during the day, decreasing H3O+ and increasing pH.

Nighttime coral-reef pH

Respiration releases CO2 at night, increasing H3O+ and decreasing pH.

Acid added to a salt with a basic anion

H3O+ consumes the basic anion directly or indirectly, lowering its concentration and shifting dissolution right, so solubility often increases.

Sodium acetate in acid

Added H3O+ removes OH-, acetate hydrolysis to consume more CH3COO-, which pulls more CH3COONa into solution.

Key Ksp memory rule

More common ion → less soluble. Remove a dissolved ion → more solid dissolves. Qsp > Ksp → precipitate.

Spontaneous process

A process that occurs naturally under given conditions without requiring continuous outside energy.

Nonspontaneous process

A process that requires continuous energy input under the given conditions.

Spontaneous does not mean fast

Spontaneity describes thermodynamic favorability, not reaction rate, not reaction rate, can be extremely slow

Thermodynamics vs kinetics

Thermodynamics asks whether a process is favored, kinetics asks how quickly the process occurs

Activation energy

The energy barrier that may need to be overcome to start a spontaneous process.

Entropy S

A measure of the dispersal/randomness of matter and energy, more accurately related to the number of possible arrangements or microstates

Microstate

One possible arrangement of particles in a system.

Microstates and entropy

More possible microstates means greater entropy.

Phase entropy order

Solid < liquid < gas.

Why gases have high entropy

Gas particles have much more freedom of movement and occupy a larger available space.

Positive ∆S

Entropy increases

Negative ∆S

Entropy decreases, matter or energy becomes more organized or concentrated

Processes with positive ∆S

Melting, boiling, sublimation, dissolving, expansion, heating, and producing more gas particles.

Processes with negative ∆S

Freezing, condensation, deposition, compression, cooling, and producing fewer gas particles.

Gas-mole shortcut for ∆S

For reactions involving gases, an increase in the number of gas particles generally means ∆S > 0, a decrease usually means ∆S<0

Entropy example: 2SO2 + O2 → 2SO3

There are 3 mol gas on the left and 2 mol gas on the right, so ∆S is expected to be negative.

Entropy of NaCl(s) → Na+(aq) + Cl-(aq)

ions become dispersed through water, so entropy increases

Entropy of larger/complex molecules

For substances in the same phase, larger, heavier, and more complex molecules generally have greater entropy.

Entropy equation for a reversible process

∆S = qrev/T, with T in kelvin and qrev in joules.

Standard reaction entropy equation

∆S°rxn = ΣnS°products − ΣnS°reactants.

∆S calculation procedure

1) Balance the reaction. 2) Multiply every tabulated S° by its coefficient. 3) Add products. 4) Add reactants. 5) Subtract reactants from products.

∆S calculation example

For A + 2B → C, ∆S°rxn = S°C − [S°A + 2S°B].

Second Law of Thermodynamics

spontaneous process causes the total entropy of the universe to increase

Universe entropy equation

∆Suniv = ∆Ssystem + ∆Ssurroundings.

Spontaneous universe criterion

∆Suniv > 0.

Heat-flow direction

Heat spontaneously flows from hot to cold because this increases the overall dispersal of energy.

Third Law of Thermodynamics

The entropy of a perfect crystalline substance at 0 K is zero.

Entropy at 0 K

For a perfect crystal at absolute zero, S = 0 and there is one possible microstate.

Gibbs free energy

A thermodynamic quantity used to determine spontaneity under specified conditions.

Gibbs equation

∆G = ∆H − T∆S.

Temperature units in Gibbs equation

Temperature must be in kelvin when using ∆G = ∆H − T∆S.

∆G < 0

The process is spontaneous under the stated conditions.

∆G > 0

the process is nonspontaneous under the stated conditions

∆G = 0

The system is at equilibrium.

∆H < 0 and ∆S > 0

always spontaneous because ∆G is negative at every temperature

∆H > 0 and ∆S < 0

Never spontaneous because ∆G is positive at every temperature.

∆H < 0 and ∆S < 0

Spontaneous at low temperature, temperature determines whether ∆H outweighs the positive -T∆S contribution

∆H > 0 and ∆S > 0

Spontaneous at high temperature because T∆S can become large enough to overcome positive ∆H.

Temperature where spontaneity changes

At the transition point ∆G = 0, so T = ∆H/∆S, with ∆H and ∆S in matching units.

Temperature threshold example

If ∆H = 25.0 kJ/mol and ∆S = 75.0 J/(mol·K), convert ∆H to 25,000 J/mol, then T = 25,000/75.0 = 333 K.

∆G°

Gibbs free energy change under standard-state conditions.

∆G

Gibbs free energy change under the current, possibly nonstandard conditions.

∆G°f

Standard free energy of formation: the free-energy change when 1 mol of a substance forms from its elements in their standard states.

∆G°f for an element

0 for an element in standard state

Standard reaction free energy

∆G°rxn = Σn∆G°f(products) − Σn∆G°f(reactants).

Free energy and equilibrium constant

∆G° = −RT ln K.

R for ∆G° = −RT ln K

R = 8.314 J/(mol·K).

K > 1 and ∆G°

K > 1 means ∆G° < 0 and products are favored under standard conditions.

K < 1 and ∆G°

K < 1 means ∆G° > 0 and reactants are favored under standard conditions.

K = 1 and ∆G°

K = 1 means ∆G° = 0 and neither side is favored under standard conditions.

Calculate K from ∆G°

Rearrange ∆G° = −RT ln K: ln K = −∆G°/(RT), then K = e^[−∆G°/(RT)]. Use matching energy units.

Oxidation

Loss of electrons by a species.

Reduction

Gain of electrons by a species.

OIL RIG

Oxidation Is Loss, reduction is gain

Redox reaction

A reaction in which oxidation and reduction occur together because electrons lost by one species must be gained by another.

Oxidized species

species that loses electrons, increases oxidation state

Reduced species

The species that gains electrons and decreases its oxidation state.

Oxidation half-reaction example

Cd(s) → Cd2+(aq) + 2e−.

Reduction half-reaction example

Ni2+(aq) + 2e− → Ni(s).

Galvanic cell

An electrochemical cell that uses a spontaneous redox reaction to produce electrical energy.

Anode in a galvanic cell

The electrode where oxidation occurs, electrons leave

Cathode in a galvanic cell

electrode where reduction occurs, electrons flow toward electrode

AN OX, RED CAT

ANode = OXidation; REDuction = CAThode

Electron flow

Electrons flow through the external circuit from anode to cathode.