Solubility Equilibrium

Example 16.8 Calculating Molar Solubility from Ksp

Calculate the molar solubility of PbCl2 in pure water.

Begin by writing the reaction by which solid PbCl2 dissolves into its constituent aqueous ions and write the corresponding expression for Ksp.

Use the stoichiometry of the reaction to prepare an ICE table, showing the equilibrium concentrations of Pb2+ and Cl– relative to S, the amount of PbCl2 that dissolves.

Substitute the equilibrium expressions for [Pb2+] and [Cl–] from the previous step into the expression for Ksp.

SOLUTION

Therefore,

Example 16.8 Calculating Molar Solubility from Ksp

Continued

Solve for S and substitute the numerical value of Ksp (from Table 16.2) to calculate S.

SOLUTION

For Practice 16.8Calculate the molar solubility of Fe(OH)2 in pure water.

Example 16.9 Calculating Ksp from Molar SolubilityThe molar solubility of Ag2SO4 in pure water is 1.2 10–5 M. Calculate Ksp.

Begin by writing the reaction by which solid Ag2SO4 dissolves into its constituent aqueous ions and write the corresponding expression for Ksp.

Use an ICE table to define [Ag+] and [SO42–] in terms of

S, the amount of Ag2SO4 that dissolves.

Substitute the expressions for [Ag+] and [SO42–] from

the previous step into the expression for Ksp. Substitute the given value of the molar solubility for S and compute Ksp.

SOLUTION

For Practice 16.9The molar solubility of AgBr in pure water is 7.3 10–7 M. Calculate Ksp.

Example 16.10 Calculating Molar Solubility in the Presence of a Common IonWhat is the molar solubility of CaF2 in a solution containing 0.100 M NaF?

Begin by writing the reaction by which solid CaF2 dissolves into its constituent aqueous ions and write the corresponding expression for Ksp.

Use the stoichiometry of the reaction to prepare an ICE table showing the initial concentration of the common ion. Fill in the equilibrium concentrations of Ca2+ and F– relative to S, the amount of CaF2 that dissolves.

SOLUTION

Substitute the equilibrium expressions for [Ca2+] and [F–] from the previous step into the expression for Ksp. Since Ksp is small, we can make the approximation that 2S is much less than 0.100 and will therefore be insignificant when added to 0.100 (this is similar to the x is small approximation that we have made for many equilibrium problems).

Solve for S and substitute the numerical value of Ksp (from Table 16.2) to calculate S.

Note that the calculated value of S is indeed small compared to 0.100, so our approximation is valid.

Example 16.10 Calculating Molar Solubility in the Presence of a Common IonContinued

SOLUTION

For comparison, the molar solubility of CaF2 in pure water is 3.32 10–4 M, which means CaF2 is over 20,000 times more soluble in water than in the NaF solution. (Confirm this for yourself by calculating the solubility in pure water from the value of Ksp.)

For Practice 16.10Calculate the molar solubility of CaF2 in a solution containing 0.250 M Ca(NO3) 2.

Example 16.10 Calculating Molar Solubility in the Presence of a Common IonContinued

SOLUTION

Thermodynamics:Entropy, Gibbs Free Energy and

Equilibrium

• Molecular complexity– The larger the molecule, the more spatial movements.

• Temperature elevation/ energy flow– Hot flows to cold

• Reactions whose products are in more random state– Ssolid < Sliquid < Sgas

• Reactions that have greater number of product molecules than reactant molecules

Increases in Entropy: S > 0

Entropy and Phase ChangesH2O(l) H2O(g)Greater entropy

H2O(s) H2O(l)Lower entropy

Problem:Problem:Predict whether Ssystem is (+) or (−) for each of the following.

• Heating air in a balloon

• Water vapor condensing

• Separation of oil and vinegar salad dressing

• Dissolving sugar in tea

• 2 HgO(s) 2 Hg(l) + O2(g)

• 2 NH3(g) N2(g) + 3 H2(g)

• Ag+(aq) + Cl−(aq) AgCl(s)

• Heating air in a balloon (+)(+)

• Water vapor condensing (-)(-)

• Separation of oil and vinegar salad dressing (-)(-)

• Dissolving sugar in tea (+)(+)

• 2 HgO(s) 2 Hg(l) + O2(g) (+)(+)

• 2 NH3(g) N2(g) + 3 H2(g) (+)(+)

• Ag+(aq) + Cl−(aq) AgCl(s) (-)(-)

PROBLEM: PROBLEM: Determine the So for the chemical reaction (system):

2 H2(g) + O2(g) 2 H2O(l)

SSsystemsystem = = SSreactionreaction = = ((SS°°prodprod) − ) − ((SS°°reactreact))

∆So = [2 So (H2O(l))] - [(2 So (H2)(g)) + (So (O2)(g))]

∆So = [2 mol (69.9 J/K·mol)] - [2 mol (130.7 J/K·mol) + 1 mol (205.3 J/K·mol)]

∆So = -326.9 J/K

• Note that there is a decrease in decrease in SS because 3 mol of gas give 2 mol of liquid.

1st Law: Energy (E) in the universe is conserved.• Euniverse = Esystem +Esurroundings • Esystem = q + w (heat + work or PV)• H = q (at constant P)• E = H + PV

2nd Law: Total entropy (S) of the universe (total) MUST increase in a spontaneous process.

• Suniverse = Ssystem + Ssurroundings > 0

3rd Law: Entropy of a substance at absolute zero (0K) for a pure substance is zero.

Laws of thermodynamics

Spontaneity Criteria: Gibbs Free Energy, Spontaneity Criteria: Gibbs Free Energy, GG

Suniverse = Ssystem + Ssurroundings > 0

Ssurroundings = -Hsystem/ T

Suniverse = Ssystem - Hsystem/ T > 0

TSuniverse = TSsystem - Hsystem > 0

-TSuniverse = -TSsystem + Hsystem < 0

-TSuniverse = Gsystem

G = Hsystem - TSsystem < 0

GGoo Value Summary Table Value Summary Table

∆∆GGoo = ∆ = ∆HHoo - - TT∆∆SSoo

Temperature INDEPENDENT

∆∆HHoo ∆∆SSoo ∆∆GGoo ReactionReaction SpontaneitySpontaneity

exo(–) increase(+) – Product favored Spontaneous

endo(+) decrease(-) + Reactant favored Not spontaneous

Temperature DEPENDENT

∆∆HHoo ∆∆SSoo ∆∆GGoo ReactionReaction SpontaneitySpontaneity

exo(–) decrease(-) + High temperature Not spontaneous

exo(–) decrease(-) - Low temperature Spontaneous

endo(+) increase(+) - High temperature Spontaneous

endo(+) increase(+) + Low temperature Not spontaneous

ProblemProblem: Calculate G at 25 C for

CH4(g) + 8 O2(g) CO2(g) + 2 H2O(g) + 4 O3(g).

Given:Substance G kJ/mol

CH4(g) −50.5O2(g) 0.0

CO2(g) −394.4H2O(g) −228.6O3(g) 163.2

Solution:

Go = (Gproductso – Greactants

o)

Go = [(-394.4 kJ/mol x 1 mol) + (-228.6 kJ/mol x 2 mol)] - [(-50.5 kJ/mol x 1 mol) + 0 kJ]

Go = - 148.3 kJ

ProblemProblem: Calculate G and determine whether the reaction below is

spontaneous.

CCl4(g) C (s, graphite) + 2 Cl2(g)

Given: : HH = +95.7 kJ, = +95.7 kJ, SS = 142.2 J/K, = 142.2 J/K, TT = 298 K = 298 K

GG = = HH - - TTSS

• Since Since GG is +, the reaction is not spontaneous at this temperature. is +, the reaction is not spontaneous at this temperature. • To make it spontaneous, we need to increase the temperature.To make it spontaneous, we need to increase the temperature.

Solution: G = +95.7 kJ – {(298 K)(0.1422 J/K)}G = 53.5 kJ

Problem: Problem: The ∆Go for the reaction N2O4 2 NO2 is +4.8 kJ.

Calculate K for this reaction.

∆∆GGoo = - = -RT RT lnlnKK

AnswerAnswer: ∆Go = +4800 J = - (8.31 J/K)(298 K) ln K

ln K = - (4800 J/{(8.31 J/K(298 K)}

ln K = - 1.94

Take the antilog of both sides :

K = e-1.94

K = 0.14

• When ∆Go > 0, K < 1.

• When ∆Go > 0, the reaction is not spontaneous in that direction; reaction is reactant favored according to K value.