1. Solubility Equilibrium

2. Example 16.8 Calculating Molar Solubility from K sp Calculate the molar solubility of PbCl 2 in pure water. Begin by writing the reaction by which solid PbCl 2 dissolves into its constituent aqueous ions and write the corresponding expression for K sp. Use the stoichiometry of the reaction to prepare an ICE table, showing the equilibrium concentrations of Pb 2+ and Cl – relative to S, the amount of PbCl 2 that dissolves. Substitute the equilibrium expressions for [Pb 2+ ] and [Cl – ] from the previous step into the expression for K sp. SOLUTION

3. Therefore, Example 16.8 Calculating Molar Solubility from K sp Continued Solve for S and substitute the numerical value of K sp (from Table 16.2) to calculate S. SOLUTION For Practice 16.8 Calculate the molar solubility of Fe(OH) 2 in pure water.

4. Example 16.9 Calculating K sp from Molar Solubility The molar solubility of Ag 2 SO 4 in pure water is 1.2 10 –5 M. Calculate K sp. Begin by writing the reaction by which solid Ag 2 SO 4 dissolves into its constituent aqueous ions and write the corresponding expression for K sp. Use an ICE table to define [Ag + ] and [SO 4 2– ] in terms of S, the amount of Ag 2 SO 4 that dissolves. Substitute the expressions for [Ag + ] and [SO 4 2– ] from the previous step into the expression for K sp. Substitute the given value of the molar solubility for S and compute K sp. SOLUTION For Practice 16.9 The molar solubility of AgBr in pure water is 7.3 10 –7 M. Calculate K sp.

5. Example 16.10 Calculating Molar Solubility in the Presence of a Common Ion What is the molar solubility of CaF 2 in a solution containing 0.100 M NaF? Begin by writing the reaction by which solid CaF 2 dissolves into its constituent aqueous ions and write the corresponding expression for K sp. 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 Ca 2+ and F – relative to S, the amount of CaF 2 that dissolves. SOLUTION

6. Substitute the equilibrium expressions for [Ca 2+ ] and [F – ] from the previous step into the expression for K sp. Since K sp 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 K sp (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 Ion Continued SOLUTION

7. For comparison, the molar solubility of CaF 2 in pure water is 3.32 10 –4 M, which means CaF 2 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 K sp.) For Practice 16.10 Calculate the molar solubility of CaF 2 in a solution containing 0.250 M Ca(NO 3 ) 2. Example 16.10 Calculating Molar Solubility in the Presence of a Common Ion Continued SOLUTION

8. Thermodynamics: Entropy, Gibbs Free Energy and Equilibrium

9. 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 – S solid < S liquid < S gas Reactions that have greater number of product molecules than reactant molecules Increases in Entropy:  S > 0

10. Entropy and Phase Changes H 2 O(l)  H 2 O(g) Greater entropy H 2 O(s)  H 2 O(l) Lower entropy

11. Problem: Problem: Predict whether  S system 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) + O 2 (g) 2 NH 3 (g)  N 2 (g) + 3 H 2 (g) Ag + (aq) + Cl − (aq)  AgCl(s)

12. (+)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) + O 2 (g) (+) (+)2 NH 3 (g)  N 2 (g) + 3 H 2 (g) (+) (-)Ag + (aq) + Cl − (aq)  AgCl(s) (-)

13. PROBLEM: Determine the  S o for the chemical reaction (system): 2 H 2 (g) + O 2 (g)  2 H 2 O(l)  S system =  S reaction =  (S° prod ) −  (S° react ) ∆S o = [2 S o (H 2 O(l))] - [(2 S o (H 2 )(g)) + (S o (O 2 )(g))] ∆S o = [2 mol (69.9 J/K·mol)] - [2 mol (130.7 J/K·mol) + 1 mol (205.3 J/K·mol)] ∆S o = -326.9 J/K decrease in S Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid.

14. 1 st Law: Energy (E) in the universe is conserved.  E universe =  E system +  E surroundings  E system = q + w (heat + work or P  V)  H = q (at constant P)  E =  H + P  V 2 nd Law: Total entropy (S) of the universe (total) MUST increase in a spontaneous process.  S universe =  S system +  S surroundings > 0 3 rd Law: Entropy of a substance at absolute zero (0K) for a pure substance is zero. Laws of thermodynamics

15. Spontaneity Criteria: Gibbs Free Energy,  G  S universe =  S system +  S surroundings > 0  S surroundings = -  H system / T  S universe =  S system -  H system / T > 0 T  S universe = T  S system -  H system > 0 -T  S universe = -T  S system +  H system < 0 -T  S universe =  G system  G =  H system - T  S system < 0

16.  G o Value Summary Table ∆G o = ∆H o - T∆S o Temperature INDEPENDENT ∆H o ∆S o ∆G o Reaction Spontaneity exo(–) increase(+) – Product favored Spontaneous endo(+) decrease(-) + Reactant favored Not spontaneous Temperature DEPENDENT ∆H o ∆S o ∆G o Reaction Spontaneity exo(–) decrease(-) + High temperature Not spontaneous exo(–) decrease(-) - Low temperature Spontaneous endo(+) increase(+) - High temperature Spontaneous endo(+) increase(+) + Low temperature Not spontaneous

17. Problem Problem: Calculate  G  at 25  C for CH 4 (g) + 8 O 2 (g)  CO 2 (g) + 2 H 2 O(g) + 4 O 3 (g). Given: Substance  G  kJ/mol CH 4 (g) − 50.5 O2(g)O2(g)0.0 CO 2 (g) − 394.4 H 2 O(g) − 228.6 O3(g)O3(g)163.2 Solution:  G o = (  G products o –  G reactants o )  G o = [(-394.4 kJ/mol x 1 mol) + (-228.6 kJ/mol x 2 mol)] - [(-50.5 kJ/mol x 1 mol) + 0 kJ]  G o = - 148.3 kJ

18. Problem :  H = +95.7 kJ,  S = 142.2 J/K, T = 298 K  G =  H - T  S Problem: Calculate  G and determine whether the reaction below is spontaneous. CCl 4 (g)  C (s, graphite) + 2 Cl 2 (g) Given:  H = +95.7 kJ,  S = 142.2 J/K, T = 298 K  G =  H - T  S Since  G is +, the reaction is not spontaneous at this temperature. Since  G 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

19. Problem: Problem: The ∆G o for the reaction N 2 O 4  2 NO 2 is +4.8 kJ. Calculate K for this reaction. ∆G o = -RT lnK ∆G o = -RT lnK Answer Answer: ∆G o = +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 ∆G o > 0, K < 1. When ∆G o > 0, the reaction is not spontaneous in that direction; reaction is reactant favored according to K value.