Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Unit 8 (Chp 15,17): Chemical & Solubility Equilibrium (Keq , Kc , Kp , Ksp ) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall

(forward & reverse rxns occur simultaneously) Dynamic Equilibrium: (forward & reverse rxns occur simultaneously) N2O4(g) 2 NO2(g) double arrow

Dynamic Equilibrium: N2O4(g) 2 NO2(g) Initially, forward & reverse rxns occur at different rates. (based on collisions, one slows down; other speeds up) At equilibrium: Rateforward = Ratereverse N2O4(g) 2 NO2(g) Ratef Rater

At Equilibrium N2O4(g) 2 NO2(g) At equilibrium, concentrations (M, mol, P, etc.) of reactant and product remain constant. N2O4(g) 2 NO2(g)

At Equilibrium… N2O4(g) 2 NO2(g) …forward and reverse rates are ________ …concentrations are ________ equal constant N2O4(g) 2 NO2(g) HW p. 660 #2

The Equilibrium Constant (Keq)

The Equilibrium Constant Consider the reaction aA + bB cC + dD At equilibrium… Ratef = Rater kf [A]a[B]b = kr [C]c[D]d kf [C]c[D]d kr [A]a[B]b = Keq = [C]c[D]d [A]a[B]b [products] [reactants]

The Equilibrium Constant Consider the reaction aA + bB cC + dD The equilibrium constant expression (Keq) is Kc = [C]c[D]d [A]a[B]b K = [products] [reactants] [ ] is conc. in M K expressions do not include: pure solids(s) or pure liquids(l) (b/c concentrations are constant)

The Equilibrium Constant In three experiments at the same temperature, equilibrium was achieved & data were collected. Exp. Butanoic acid (moles) Ethanol Ethyl butanoate Water Products Reactants (Kc) 1 10.0 20.0 2 5.0 40.0 3 30.0 1.0 12.0 4 4 Dingle apnotes13 Task 13a 4 same ratio (constant) Calculate Kc for each experiment at this temperature and compare the values.

What Does the Value of K Mean? [products] [reactants] Reactants  Products If K > 1, the reaction is product-favored; more product at equilibrium. If K < 1, the reaction is reactant-favored; more reactant at equilibrium.

The Equilibrium Constant Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written Kp = (PC)c (PD)d (PA)a (PB)b Example: 3 Fe(s) + 4 H2O(g) ↔ Fe3O4(s) + 4 H2(g)

Heterogeneous Equilibria The concentrations of solids and liquids do not appear in the K expression. PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq) Kc = [Pb2+] [Cl−]2 Kc = ?

CaCO3 (s) CO2 (g) + CaO(s) Kc = ? Kc = [CO2] Kp = ? Kp = PCO2 As long as some CaCO3 or CaO remains, the amount of CO2 above the solid be constant. Kc = ? Kc = [CO2] Kp = ? Kp = PCO2

↔ ↔ ↔ Manipulating K N2O4 2 NO2 N2O4 2 NO2 K of reverse rxn = 1/K Kc = = 4.0 [NO2]2 [N2O4] N2O4 2 NO2 Manipulating K ↔ Kc = = 1 . (4.0) [N2O4] [NO2]2 N2O4 2 NO2 K of reverse rxn = 1/K ↔ Kc = = (4.0)2 [NO2]4 [N2O4]2 4 NO2 2 N2O4 K of multiplied reaction = K^# (raised to power)

K of combined reaction = K1 x K2 … Manipulating K A  3 B + 2 C K1 = 2.5 2 C + 3 D  4 E K2 = 60 A + 3 D  3 B + 4 E Kovr = ? Kovr = (2.5)(60) K of combined reaction = K1 x K2 … HW p. 661 #14, 16, 20

Equilibrium Calculations

Equilibrium Calculations A closed system initially containing 0.100 M H2 and 0.200 M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M. Calculate Kc at 448C for the reaction taking place, which is H2 (g) + I2 (g) 2 HI (g) find limiting reactant mol reactant completely to mol product, but… NOW we still have some reactant left and some product formed at equilibrium.

RICE Tables Reaction Initial Change Equilibrium H2 I2 2 HI + 0.100 M H2 and 0.200 M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M. Calculate Kc at 448C.

What Do We Know? H2 I2 2 HI + [H2]in = 0.100 M [I2]in = 0.200 M [HI]in = 0 M Reaction Initial Change Equilibrium H2 I2 2 HI + 0.100 M 0.200 M 0 M 0.187 M [HI]eq = 0.187 M 0.100 M H2 and 0.200 M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M.

[HI] Increases by 0.187 M H2 I2 2 HI + Initial 0.100 M 0 M Change +0.187 Equilibrium 0.187 M Reaction Initial 0.200 M Change Equilibrium H2 I2 2 HI + 0.100 M H2 and 0.200 M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M. Calculate Kc at 448C.

Stoichiometry shows [H2] and [I2] decrease by half as much Initial Change –0.0935 +0.187 Equilibrium 0.187 M Reaction Initial 0.100 M 0.200 M 0 M Change Equilibrium H2 I2 2 HI + +0.187 0.187 M 0.187 M HI x 1 mol H2 = 0.0935 M H2 2 mol HI

Kc = [HI]2 [H2] [I2] = (0.187)2 (0.0065)(0.1065) = 51 We can now calculate the equilibrium concentrations of all three compounds… Initial 0.100 M 0.200 M 0 M Change –0.0935 +0.187 Equilibrium 0.0065 M 0.1065 M 0.187 M Reaction Initial Change Equilibrium H2 I2 2 HI + Calculate Kc at 448C. HW p. 662 #27,30,40,34 Kc = [HI]2 [H2] [I2] = (0.187)2 (0.0065)(0.1065) = 51

RICE Practice given K, solve for x HW p. 663 #43 At 2000oC the equilibrium constant for the reaction 2 NO(g) ↔ N2(g) + O2(g) is Kc = 2.4 x 103 . If the initial concentration of NO is 0.200 M , what are the equilibrium concentrations of NO , N2 , and O2 ?

What Do We Know? 2 NO N2 O2 + Kc = 2.4 x 103 Reaction Initial Change Equilibrium 2 NO N2 O2 + 0.200 M 0 M Kc = 2.4 x 103 the initial concentration of NO is 0.200 M

? ? ? What Do We Know? What do we NOT know? 2 NO N2 O2 + Reaction Initial 0.200 M 0 M Change Equilibrium 2 NO N2 O2 + ? ? ? Kc = 2.4 x 103 the initial concentration of NO is 0.200 M

Stoichiometry shows [NO] decreases by twice as much as [N2] and [O2] increases. Reaction Initial Change Equilibrium Initial 0.200 M 0 M Change Equilibrium 2 NO N2 O2 + – 2x + x

Now, what was the question again? We now have the equilibrium concentrations of all three compounds… (in terms of x) Initial 0.200 Change – 2x + x Equilibrium Reaction Initial Change Equilibrium 2 NO N2 O2 + 0.200 – 2x x Now, what was the question again? what are the equilibrium concentrations of NO , N2 , and O2 ?

√ √ Kc = [N2] [O2] [NO]2 Kc = 2.4 x 103 2.4 x 103 = (x)2 (0.200 – 2x)2 Equilibrium 0.200 – 2x x Kc = [N2] [O2] [NO]2 Kc = 2.4 x 103 √ 2.4 x 103 = (x)2 (0.200 – 2x)2 √ HW p. 663 # 44 x (0.200 – 2x) 49 = (next slide) 9.8 – 98x = x [N2]eq = 0.099 M [O2]eq = 0.099 M [NO]eq = 0.0020 M 9.8 = 99x x = 0.099

RICE Practice given K, solve for x HW p. 663 #44 For the equilibrium Br2(g) + Cl2(g) ↔ 2 BrCl(g) at 400 K, Kc = 7.0. If 0.30 mol of Br2 and 0.30 mol of Cl2 are introduced into a 1.0 L container at 400 K, what will be the equilibrium concentrations of Br2, Cl2, and BrCl?

the initial concentrations of Br2 and Cl2 are 0.30 M What Do We Know? Reaction Initial Change Equilibrium Br2 Cl2 2 BrCl + 0.30 M 0 M Kc = 7.0 [Br2]in = 0.30 M [Cl2]in = 0.30 M the initial concentrations of Br2 and Cl2 are 0.30 M

the initial concentrations of Br2 and Cl2 are 0.30 M What Do We Know? What do we NOT know? Reaction Initial 0.30 M 0 M Change Equilibrium Br2 Cl2 2 BrCl + ? ? ? Kc = 7.0 [Br2]in = 0.30 M [Cl2]in = 0.30 M the initial concentrations of Br2 and Cl2 are 0.30 M

Stoichiometry shows [BrCl] increases by twice as much as [Br2] and [Cl2] decrease. Reaction Initial 0.30 M Change Equilibrium Initial 0.30 M 0 M Change Equilibrium Br2 Cl2 2 BrCl + – x + 2x

Now, what was the question again? We now have the equilibrium concentrations of all three compounds… (in terms of x) Initial 0.30 M 0 M Change – x + 2x Equilibrium Reaction Initial Change Equilibrium Br2 Cl2 2 BrCl + 0.30 – x 2x Now, what was the question again? what are the equilibrium concentrations of Br2, Cl2, and BrCl?

√ √ Kc = [BrCl]2 [Br2][Cl2] Kc = 7.0 7.0 = (2x)2 (0.30 – x)2 2x Equilibrium 0.30 – x 2x Kc = [BrCl]2 [Br2][Cl2] Kc = 7.0 √ 7.0 = (2x)2 (0.30 – x)2 √ HW p. 664 #64 2x (0.30 – x) 2.6 = (next slide) 0.78 – 2.6x = 2x [Br2]eq = 0.13 M [Cl2]eq = 0.13 M [BrCl]eq = 0.34 M 0.78 = 4.6x x = 0.17

RICE Practice given K, solve for x HW p. 664 #64 For the equilibrium 2 IBr(g) ↔ I2(g) + Br2(g) Kp = 8.5 x 10–3 at 150oC. If 0.025 atm of IBr is placed in a 2.0 L container, what is the partial pressure of this substance after equilibrium is reached?

What Do We Know? 2 IBr I2 Br2 + Kp = 8.5 x 10–3 Reaction Initial Change Equilibrium 2 IBr I2 Br2 + 0.025 atm 0 atm Kp = 8.5 x 10–3 the initial pressure of IBr is 0.025 atm

? ? ? What Do We Know? What do we NOT know? 2 IBr I2 Br2 + Reaction Initial 0.025 atm 0 atm Change Equilibrium 2 IBr I2 Br2 + ? ? ? Kp = 8.5 x 10–3 the initial pressure of IBr is 0.025 atm

Stoichiometry shows [NO] decreases by twice as much as [N2] and [O2] increases. Reaction Initial 0 atm Change Equilibrium Initial 0.025 atm 0 atm Change Equilibrium 2 IBr I2 Br2 + – 2x + x

Now, what was the question again? We now have the equilibrium concentrations of all three compounds… (in terms of x) Initial 0.025 atm 0 atm Change Equilibrium Reaction Initial 0 atm Change Equilibrium 2 IBr I2 Br2 + – 2x + x 0.025 – 2x x Now, what was the question again? What is the equilibrium partial pressure of IBr?

√ √ Kp = (PI2)(PBr2) (PIBr)2 8.5 x 10–3 = (x)2 (0.025 – 2x)2 x Equilibrium 0.025 – 2x x Kp = (PI2)(PBr2) (PIBr)2 Kp = 8.5 x 10–3 √ 8.5 x 10–3 = (x)2 (0.025 – 2x)2 √ x (0.025 – 2x) HW p. 664 #66 0.092 = (next slide) 0.0023 – 0.184x = x 0.0023 = 1.184x (PIBr)eq = 0.021 atm x = 0.0019

RICE Practice HW p. 664 #66 Solid NH4HS is introduced into an evacuated flask at 24oC. The following reaction takes place: NH4HS(s) ↔ NH3(g) + H2S(g) At equilibrium the total pressure in the container is 0.614 atm. What is Kp for this equilibrium at 24oC?

What Do We Know? NH4HS(s) NH3 H2S + Solid NH4HS is introduced into an evacuated flask at 24oC. Reaction Initial Change Equilibrium Reaction Initial … 0 atm Change Equilibrium NH4HS(s) NH3 H2S + At equilibrium, the total pressure (PT)eq is 0.614 atm. Kp = ? 0.614 atm

? ? ? ? ? What Do We Know? What do we NOT know? NH4HS(s) NH3 H2S + Reaction Initial … 0 atm Change Equilibrium NH4HS(s) NH3 H2S + ? ? ? ? ? At equilibrium, the total pressure (PT)eq is 0.614 atm. Kp = ? 0.614 atm

Stoichiometry shows [NH3] increases by the same amount as [H2S] increases. Initial … 0 atm Change – x + x Equilibrium Reaction Initial 0 atm Change Equilibrium NH4HS(s) NH3 H2S + 0.614 atm

Now, what was the question again? We now have the equilibrium concentrations of all three compounds… (in terms of x) Reaction Initial 0 atm Change Equilibrium Initial … 0 atm Change – x + x Equilibrium x NH4HS(s) NH3 H2S + 0.614 atm Now, what was the question again? What is Kp ?

Kp = (PNH3)(PH2S) PT = 0.614 atm PT = PNH3 + PH2S Kp = (x)2 Equilibrium … x Kp = (PNH3)(PH2S) PT = 0.614 atm PT = PNH3 + PH2S Kp = (x)2 0.614 = x + x 0.614 = 2x Kp = (0.307)2 x = 0.307 Kp = 0.0942 WS Equil Calc’s III

The Reaction Quotient (Q) aA + bB cC + dD Kc = [C]c[D]d [A]a[B]b (given on exam) A Q expression gives the same ratio as the equilibrium (K) expression, but … …may NOT be at equilibrium. Q = [C]c[D]d [A]a[B]b NOT (given on exam) Calculate Q by substituting INITIAL (current) concentrations into the Q expression.

If Q = K, system is at equilibrium (K). [P] [R] = K R P ratef = rater Q K Q K K Q Q = K If Q = K, system is at equilibrium (K).

Q R P [R] K Q K K [P] Q = [R] R P Q ratef > rater R P ratef = rater Q = [P] [R] Q K Q K K Q Q < K If Q < K, too much reactant, system will shift right faster to reach equilibrium (K).

Q Q = [P] K Q K K [P] Q = [R] R P HW p. 662 #36, 38 ratef < rater R P Q = [P] [R] ratef = rater Q K Q K K Q Q > K If Q > K, too much product, system will shift left faster to reach equilibrium (K).

Le Châtelier’s Principle

Le Châtelier’s Principle: “Systems at equilibrium disturbed by a change amount (M , PP) of reactants or products volume (V) of container temperature (T) (changes Keq value) …will shift ( or ) to counteract the change.” (that affects collision frequency) like: Because… R P R P ratef > rater ratef < rater or …the rxn goes faster in one direction (Q  K) until ratef = rater (Q = K again, R P ).

The Haber Process N2(g) + 3 H2(g)  2 NH3(g) The conversion of nitrogen (N2) & hydrogen (H2) into ammonia (NH3) is tremendously significant in agriculture, where ammonia-based fertilizers are of utmost importance. N2(g) + 3 H2(g)  2 NH3(g)

[NH3]2 [NH3]2 K = [N2][H2]3 [NH3]2 [N2][H2]3 K = Add reactant: (changes Q) N2(g) + 3 H2(g)  2 NH3(g) Q = K K = [NH3]2 [N2][H2]3 Q < K Q = [NH3]2 [N2][H2]3 Q = K K = [NH3]2 [N2][H2]3 (same K)

N2(g) + 3 H2(g)  2 NH3(g) Remove product: (changes Q) This apparatus pushes the equilibrium to the right by removing the product ammonia (NH3) from the system as a liquid.

Change Volume (moles of gas) N2(g) + 3 H2(g)  2 NH3(g) (changes Q) ↓V (↑PT) shifts to ↑V (↓PT) shifts to V has no shift if fewer mol of gas (↓ngas) more mol of gas (↑ngas) equal mol of gas R & P ↑PT by adding noble gas? no shift b/c same R & P pressures 1. 2 NO(g)  N2(g) + O2(g) , ↑V will shift ____. 2. N2O4(g)  2 NO2(g) , ↓V will shift ____. 3. CaCO3(s)  CO2(g) + CaO(s) , shift  by _V. NOT  ↓

[P] [R] K2 = Change Temperature [P] K1 = [R] Changing temp. is the ONLY way to change the value of ___. K Why? T (add heat) shifts… T (remove heat) shifts… …in the _____thermic direction to ______ heat endo …in the _____thermic direction to ______ heat exo produce absorb (original) K2 = [P] [R] (smaller) K2 = [P] [R] (larger) K1 = [P] [R] N2(g) + 3 H2(g)  2 NH3(g) DH = –

CoCl42– + 6 H2O(l)  4 Cl– + Co(H2O)62+ Change Temperature CoCl42– + 6 H2O(l)  4 Cl– + Co(H2O)62+ heat + + heat – ∆H = __ Add product: Remove product: ice Add heat Remove heat

[P] [R] K = increase ratef and rater . Catalysts (same) R Equilibrium occurs faster, but… at no shift (composition [P]/[R] is same). P

Change in external factor Shift to restore equilibrium Le Châtelier’s Principle (practice) N2(g) + 3 H2(g)  2 NH3(g) ∆H = –92 + heat Change in external factor Shift to restore equilibrium Reason Increase pressure (decrease volume) Increase temp. Increase [N2] Increase [NH3] Add a catalyst ↑P (↓V) shifts to side of fewer moles of gas Right  ∆H = – ∆H = – , adding heat shifts in the endo- dir.  Left Adding reactant shifts right to consume it Right  Adding product shifts left to consume it  Left Catalysts change how fast, but not how far. No Shift

Change in external factor Shift to restore equilibrium Le Châtelier’s Principle (practice) 2 H2O(g) + O2(g)  2 H2O2(l) ∆H = +108 heat + Change in external factor Shift to restore equilibrium Reason ________ pressure (_______ volume) _________ temp. _________ PO2 _________ PH2O Decrease H2O2 Decrease ↓P (↑V) shifts to side of more moles of gas  Left Increase ∆H = + ∆H = + , remove heat shifts in the exo- dir.  Left Decrease Adding reactant shifts right to consume it Right  Increase Removing reactant shifts left to produce it Decrease  Left (s) & (l) do not affect Q & K (usually) No Shift

Le Châtelier’s Principle (summary) R  P (M, PPR, PPP) add R or P: remove R or P: volume: ↓V shifts to ↑V shifts to temp. ↑T shifts ↓T shifts catalyst: shift away faster (consume) shift toward faster (replace ) fewer mol of gas (↓ngas) (Ptotal) more mol of gas (↑ngas) (changes K) (H + R  P) (R  P + H) in endo dir. to use up heat in exo dir. to make more heat no shift WS Eq Pract. 2 #4

Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Unit 8 (Chp 15, 17): Chemical & Solubility Equilibria (Keq , Kc , Kp , Ksp ) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall

Solubility Product Constant (Ksp) Write the K expression for a saturated solution of PbI2 in water: [Pb2+][I−]2 PbI2(s)  Pb2+(aq) + 2 I−(aq) K = sp XaYb(s)  aX+(aq) + bY−(aq) Ksp = [X+]a [Y–]b For “insoluble” solids, the equilibrium constant, Ksp , is the solubility product constant, or the… …product of M’s of dissolved ions at equilibrium.

Ksp is NOT the Solubility grams of solid (s) dissolved in 1 L (g/L) mol of solid (s) dissolved in 1 L (M) product of conc.’s (M) of ions(aq) at equilibrium solubility: molar solubility: Ksp : XaYb  aX+ + bY– [XaYb] [X+] [Y–] molar solubility molar conc.’s of ions Ksp = [X+]a[Y–]b

Ksp is NOT the Solubility molar solubility: Ksp : grams of solid (s) dissolved in 1 L (g/L) mol of solid (s) dissolved in 1 L (M) product of conc.’s (M) of ions (aq) at equilibrium 4.2 g/L 0.027 M CaCrO4 0.027 M Ca2+ 0.027 M CrO42– 0.00073 molar mass: 156.08 g/mol CaCrO4(s)  Ca2+ + CrO42– [CaCrO4] [Ca2+] [CrO42–] Ksp = [Ca2+][CrO42–] HW p. 763 #46

1 PbBr2 dissociates into… Ksp Calculations HW p. 763 #48a If solubility (or molar solubility) is known, solve for Ksp . [PbBr2] is 0.010 M at 25oC . (maximum that can dissolve) R I C E PbBr2(s)  Pb2+ + 2 Br– 0.010 M 0 M 0 M –0.010 +0.010 +0.020 0 M 0.010 M 0.020 M Ksp = [Pb2+][Br–]2 (all dissolved = saturated) (any excess solid is irrelevant) Ksp = (0.010)(0.020)2 1 PbBr2 dissociates into… 1 Pb2+ ion & 2 Br– ions Ksp = 4.0 x 10–6

Ksp Calculations HW p. 763 #50 Solubility (or molar solubility) is known, solve for Ksp . Ksp = (0.0012)(0.0024)2 [PbI2] = 0.0012 M R I C E Ksp = [Pb2+][I–]2 PbI2(s)  Pb2+ + 2 I– 0.0012 M 0 M 0 M –0.0012 +0.0012 +0.0024 0 M 0.0012 M 0.0024 M Ksp = 6.9 x 10–9 solubility: 0.54 g/L Molar solubility: 0.0012 M 1 mol 461 g 0.54 g x = 0.0012 mol

Ksp Calculations If only Ksp is known, solve for x (M). Ksp for AgCl is 1.8 x 10–10 . Ksp = [Ag+][Cl–] Ksp = x2 R I C E AgCl(s)  Ag+ + Cl– x 0 M 0 M –x +x +x 0 M x x 1.8 x 10–10 = x2 √1.8 x 10–10 = x 1.3 x 10–5 = x (molar solubility) [AgCl] = 1.3 x 10–5 M [Ag+] = 1.3 x 10–5 M [Cl–] = 1.3 x 10–5 M

Ksp Calculations If only Ksp is known, solve for x (M). HW p. 663 #47 If only Ksp is known, solve for x (M). Ksp for CaSO4 is 2.4 x 10–5 . R I C E CaSO4(s)  Ca2+ + SO42– x 0 M 0 M –x +x +x 0 M x x Ksp = [Ca2+][SO42–] Ksp = x2 2.4 x 10–5 = x2 √2.4 x 10–5 = x 0.0049 = x (molar solubility) [CaSO4] = 0.0049 M [Ca2+] = 0.0049 M [SO42–] = 0.0049 M

Ksp Calculations If only Ksp is known, solve for x (M). Ksp for PbCl2 is 1.6 x 10–5 . R I C E PbCl2(s)  Pb2+ + 2 Cl– x 0 M 0 M –x +x +2x 0 M x 2x Ksp = [Pb2+][Cl–]2 Ksp = (x)(2x)2 Ksp = 4x3 (molar solubility) 1.6 x 10–5 = 4x3 3√4.0 x 10–6 = x 0.016 = x [PbCl2] = 0.016 M [Pb2+] = 0.016 M [Cl–] = 0.032 M

Ksp Calculations If only Ksp is known, solve for x (M). Ksp for Cr(OH)3 is 1.6 x 10–30 . R I C E Cr(OH)3(s)  Cr3+ + 3 OH– x 0 M 0 M –x +x +3x 0 M x 3x Ksp = [Cr3+][OH–]3 Ksp = (x)(3x)3 Ksp = 27x4 1.6 x 10–30 = 27x4 4√5.9 x 10–32 = x 1.6 x 10–8 = x (molar solubility) [Cr(OH)3] = 1.6 x 10–8 M [Cr3+] = 1.6 x 10–8 M [OH–] = 4.8 x 10–8 M

Ksp Calculations (in pure H2O) (in 0.010 M KF) LaF3(s)  La3+ + 3 F– HW p. 763 #52a #52b (in pure H2O) (in 0.010 M KF) LaF3(s)  La3+ + 3 F– x 0 M 0 M –x +x +3x 0 M x 3x LaF3(s)  La3+ + 3 F– x 0 M –x +x +3x 0 M x 0.010 + 3x 0.010 M ≈ 0.010 b/c K <<<1 Solubility is lower in _________________ sol’n w/ common ion Ksp = [La3+][F–]3 Ksp = (x)(3x)3 2 x 10–19 = 27x4 x = 9 x 10–6 M LaF3 Ksp = [La3+][F–]3 Ksp = (x)(0.010 + 3x)3 2 x 10–19 = (x)(0.010)3 x = 2 x 10–13 M LaF3

Factors Affecting Solubility Common-Ion Effect (more Le Châtelier) If a common ion is added to an equilibrium solution, the equilibrium will shift left and the solubility of the salt will decrease. OR adding common ion shifts left (less soluble) BaSO4(s)  Ba2+(aq) + SO42−(aq) BaSO4 would be least soluble in which of these 1.0 M aqueous solutions? Na2SO4 BaCl2 Al2(SO4)3 NaNO3 most soluble? WS Ksp #1-2

Factors Affecting Solubility Mg(OH)2(s)  Mg2+(aq) + 2 OH−(aq) Addition of HCl (acid) to this solution would cause… Greater solubility because… Added acid (H+) reacts with OH– thereby removing product causing a shift to the right to dissolve more solid Mg(OH)2. 75

Basic anions, more soluble in acidic solution. HW p. 763 #55 Basic anions, more soluble in acidic solution. H+ NO Effect on: Cl– , Br–, I–, NO3–, SO42–, ClO4– Adding H+ would cause… shift  , more soluble. H+ Mg(OH)2(s)  Mg2+(aq) + 2 OH−(aq)

Factors Affecting Solubility Complex Ions Metal ions can and form complex ions with lone e– pairs in the solvent.

AgCl(s)  Ag+(aq) + Cl−(aq) forming complex ions… …increases solubility p. 765 #59 Ag(NH3)2+ NH3 AgCl(s)  Ag+(aq) + Cl−(aq)

Will a Precipitate Form? XaYb(s)  aX+(aq) + bY−(aq) Ksp = [X+]a [Y–]b Q = [X+]a [Y−]b WS Ksp #4 In a solution, If Q = Ksp, at equilibrium (saturated). If Q < Ksp, more solid will dissolve (unsaturated) until Q = Ksp . (products too small, proceed right→) If Q > Ksp, solid will precipitate out (saturated) until Q = Ksp . (products too big, proceed left←) HW p. 764 #62b, 66

Will a Precipitate Form? (OR…is Q > K ?) HW p. 764 #62b AgIO3(s)  Ag+ + IO3– Ksp = [Ag+][IO3–] Ksp = 3.1 x 10–8 100 mL of 0.010 M AgNO3 10 mL of 0.015 M NaIO3 Q = [Ag+][IO3–] (mixing changes M and V) M1V1 = M2V2 Q = (0.0091)(0.0014) Q = [Ag+] = ________ (0.010 M)(100 mL) = M2(110 mL) 0.0091 M Q = 1.3 x 10–5 Q > K , so… rxn shifts left prec. will form [IO3–] = ________ (0.015 M)(10 mL) = M2(110 mL) 0.0014 M

Which Will Precipitate First? HW p. 764 #65 (AgI) (PbI2) AgI(s)  Ag+ + I– PbI2(s)  Pb2+ + 2 I– Ksp = [Ag+][I–] = 8.3 x 10–17 Ksp = [Pb2+][I–]2 = 7.9 x 10–9 AgI will precipitate first b/c… less I– is needed to reach equilibrium at a smaller Ksp. Ksp = [Ag+][I–] Ksp = [Pb2+][I–]2 8.3 x 10–17 = (2.0 x 10–4)(x) 7.9 x 10–9 = (1.5 x 10–3)(x)2 x = 4.2 x 1013 x = 2.3 x 10–3 [I–] = 4.2 x 10–13 M [I–] = 2.3 x 10–3 M