unit 8 (chp 15,17): chemical & solubility equilibrium (k eq, k c, k p, k sp ) john d. bookstaver...

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

Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

Dynamic

N2O4(g) 2 NO2(g)

(forward & reverse rxns occur simultaneously)

double arrow

Equilibrium:

Initially, forward & reverse rxns occur at different rates.(based on collisions, one slows down; other speeds up)

N2O4(g) 2 NO2(g)

Ratef

Rat

e r

At equilibrium: Rateforward = Ratereverse

Dynamic Equilibrium:

A System at EquilibriumAt equilibrium, concentrations (M, mol, P, etc.) of reactant and product remain constant.

N2O4(g) 2 NO2(g)

N2O4(g) 2 NO2(g)

HW p. 660 #2

…concentrations are ________constant

…forward and reverse rates are ________equal

At Equilibrium…

The Equilibrium Constant

The Equilibrium Constant• Consider the reaction

Keq = [C]c[D]d

[A]a[B]b

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

=

[products][reactants]


• The equilibrium constant expression (Keq) is

Kc = [C]c[D]d

[A]a[B]b

[ ] is conc. in M

K expressions do not include:pure solids(s) or pure liquids(l)

(b/c concentrations are constant)

K = [products][reactants]

• Consider the reaction

aA + bB cC + dD

In three experiments at the same temperature, equilibrium was achieved & data were collected.


Calculate Kc for each experiment at this temperature and compare the values.

Exp.Butanoic

acid(moles)

Ethanol

(moles)

Ethyl butanoate

(moles)

Water

(moles)

ProductsReactants

(Kc)

1 10.0 10.0 20.0 20.0

2 5.0 20.0 10.0 40.0

3 30.0 1.0 12.0 10.0

4

4

4same ratio(constant)

What Does the Value of K Mean?

• If K > 1, the reaction is product-favored;more productat equilibrium.

• If K < 1, the reaction is is reactant-favored;more reactantat equilibrium.

K = [products][reactants]Reactants Products

Kp =(PC)c (PD)d

(PA)a (PB)b

The Equilibrium ConstantBecause pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

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 equilibrium expression.

Kc = ?

PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq)

Kc = [Pb2+] [Cl−]2

As long as some CaCO3 or CaO remains, the amount of CO2 above the solid be constant.

CaCO3 (s) CO2 (g) + CaO(s)

Kc = [CO2]

Kp = PCO2

Kc = ?

Kp = ?

K of reverse rxn = 1/K

Kc = = 4.0 [NO2]2

[N2O4]

N2O4 2 NO2

Kc = = 1 .

(4.0)

[N2O4] [NO2]2

N2O42 NO2

↔

↔

K of multiplied reaction

= K^# (raised to power)Kc = = (4.0)2[NO2]4

[N2O4]2

4 NO22 N2O4 ↔

Manipulating K

K of combined reaction = K1 x K2 …

A + 3 D 3 B + 4 E Kovr = ?

A 3 B + 2 C K1 = 2.5

2 C + 3 D 4 E K2 = 60

Manipulating K

Kovr = (2.5)(60)

HW p. 661#14, 16, 20

Equilibrium Calculations

Equilibrium CalculationsA closed system initially containing

0.100 M H2 and 0.200 M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M. Calculate Kc at 448C for the reaction taking place, which is

H2 (g) + I2 (g) 2 HI (g)

1) find limiting reactant2) mol reactant completely to mol product, but…NOW we still have some reactant left and some product formed at equilibrium.

Reaction

Initial

Change

Equilibrium

H2 I2 2 HI+

0.100 M H2 and 0.200 M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M.Calculate Kc at 448C.

RICE Tables

Reaction

Initial

Change

Equilibrium

H2 I2 2 HI+

0.100 M H2 and 0.200 M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M.

[H2]in = 0.100 M [I2]in = 0.200 M

What Do We Know?

[HI]eq = 0.187 M

[HI]in = 0 M

0 M

0.187 M

0.100 M 0.200 M

Initial 0.100 M 0 M

Change +0.187

Equilibrium 0.187 M

Reaction

Initial 0.200 M

Change

Equilibrium

[HI] Increases by 0.187 M

H2 I2 2 HI+

0.100 M H2 and 0.200 M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M.Calculate Kc at 448C.

Reaction

Initial 0.100 M 0.200 M 0 M

Change

Equilibrium

Initial

Change –0.0935 –0.0935 +0.187

Equilibrium 0.187 M

Stoichiometry shows [H2] and [I2]decrease by half as much

H2 I2 2 HI+

0.187 M HI x 1 mol H2 = 0.0935 M H2

2 mol HI

Reaction

Initial

Change

Equilibrium

Initial 0.100 M 0.200 M 0 M

Change –0.0935 –0.0935 +0.187

Equilibrium 0.0065 M 0.1065 M 0.187 M

We can now calculate the equilibrium concentrations of all three compounds…

H2 I2 2 HI+

Calculate Kc at 448C.

Kc = [HI]2

[H2] [I2]=

(0.187)2

(0.0065)(0.1065)= 51

HW p. 662 #27, 30, 40

ICE Practicegiven 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?

Reaction

Initial

Change

Equilibrium

2 NO N2 O2+

Kc = 2.4 x 103 the initial concentration of NO is 0.200 M

What Do We Know?

Reaction

Initial 0.200 M 0 M 0 M

Change

Equilibrium

2 NO N2 O2+

Kc = 2.4 x 103 the initial concentration of NO is 0.200 M

What do we NOT know?

? ? ?

Initial 0.200 0

Change – 2x + x + x

Equilibrium

2 NO N2 O2+

Stoichiometry shows [NO] decreases by twice as much as [N2] and [O2] increases.

Reaction

Initial 0 0

Change

Equilibrium

Reaction

Initial 0

Change

Equilibrium

Initial 0.200 0 0


Equilibrium 0.200 – 2x x x

We now have the equilibrium concentrations of all three compounds…

(in terms of x)

2 NO N2 O2+

Now, what was the question again?

what are the equilibrium concentrations of NO , N2 , and O2 ?

Kc = [N2] [O2]

[NO]2

HW p. 663 # 4449 =

x(0.200 – 2x)

x = 0.099

2.4 x 103 =(x)2

(0.200 – 2x)2

9.8 – 98x = x

9.8 = 99x [N2]eq = 0.099 M [O2]eq = 0.099 M[NO]eq = 0.0020 M


√√(next slide)


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?

What Do We Know?

Reaction

Initial

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

What Do We Know?

Reaction

Initial 0.30 M 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

Reaction

Initial 0.30 M

Change

Equilibrium

Initial 0.30 M 0 M

Change – x – x + 2x

Equilibrium

Br2 Cl2 2 BrCl

Stoichiometry shows [BrCl] increases by twice as much as [Br2] and [Cl2] decrease.

+

Reaction

Initial

Change

Equilibrium

Initial 0.30 M 0.30 M 0 M

Change – x – x + 2x

Equilibrium 0.30 – x 0.30 – x 2x


(in terms of x)


what are the equilibrium concentrations of Br2, Cl2, and BrCl?

Br2 Cl2 2 BrCl+

Kc = [BrCl]2

[Br2][Cl2]

2.6 =2x

(0.30 – x)

x = 0.17

7.0 =(2x)2

(0.30 – x)2

0.78 – 2.6x = 2x

0.78 = 4.6x [Br2]eq = 0.13 M [Cl2]eq = 0.13 M[BrCl]eq = 0.34 M

Equilibrium 0.30 – x 0.30 – x 2x

√√HW p. 664

#64(next slide)


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?

Reaction

Initial

Change

Equilibrium

2 IBr I2 Br2+

Kp = 8.5 x 10–3 the initial pressure of IBr is 0.025 atm

What Do We Know?

Reaction

Initial 0.025 atm 0 atm 0 atm

Change

Equilibrium

2 IBr I2 Br2+


? ? ?

Kp = 8.5 x 10–3 the initial pressure of IBr is 0.025 atm

Reaction

Initial 0 atm

Change

Equilibrium

Initial 0.025 atm 0 atm


Equilibrium

2 IBr I2 Br2+

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




(in terms of x)

2 IBr I2 Br2+


What is the equilibrium partial pressure of IBr?

Kp = (PI2)(PBr2)

(PIBr)2

0.092 =x

(0.025 – 2x)

x = 0.0019

8.5 x 10–3 =(x)2

(0.025 – 2x)2

0.0023 – 0.184x = x

0.0023 = 1.184x (PIBr)eq = 0.021 atm


√√HW p. 664

#66(next slide)

ICE 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?

Reaction

Initial

Change

Equilibrium

Reaction

Initial … 0 atm 0 atm

Change

Equilibrium

What Do We Know?

NH4HS(s) NH3 H2S+

At equilibrium,the total pressure (PT)eq is 0.614 atm.

Kp = ?

Solid NH4HS is introduced into an evacuated flask at 24oC.

0.614 atm

What Do We Know?

Reaction

Initial … 0 atm 0 atm

Change

Equilibrium


? ? ?

NH4HS(s) NH3 H2S+

0.614 atmAt equilibrium,the total pressure (PT)eq is 0.614 atm.

Kp = ?

? ?

Initial … 0 atm

Change – x + x + x

Equilibrium

Reaction

Initial 0 atm

Change

Equilibrium

Stoichiometry shows [NH3] increases by the same amount as [H2S] increases.

NH4HS(s) NH3 H2S+

0.614 atm

Initial … 0 atm

Change – x + x + x

Equilibrium … x x

Reaction

Initial 0 atm

Change

Equilibrium


(in terms of x)

Now, what was the question again?What is Kp ?

NH4HS(s) NH3 H2S+

0.614 atm

Kp = (PNH3)(PH2S)

Kp = (x)2

Kp = (0.307)2

Kp = 0.0942

Equilibrium … x x

WS Equil Calc’s III

PT = PNH3 + PH2S

0.614 = x + x

0.614 = 2x

PT = 0.614 atm

x = 0.307

The Reaction Quotient (Q)

• A Q expression gives the same ratio as the equilibrium (K) expression, but ……for a system that is NOT at equilibrium.

Kc =

[C]c[D]d

[A]a[B]b

aA + bB cC + dD

• Calculate Q by substituting INITIAL (current) concentrations into the Q expression.

Q =[C]c[D]d

[A]a[B]bNOT

(given on exam)

(given on exam)

K

Q

ratef = rater

R P

KQ

Q =[P][R]

K

Q

If Q = K, system is at equilibrium (K).

Q = K

= K

If Q < K, too much reactant, system willshift right faster to reach equilibrium (K).

ratef > rater

R P

ratef = rater

R P

Q =[P][R]

K

Q

KQ K

Q

Q < K

Q =[P]

[R]

If Q > K, too much product, system willshift left faster to reach equilibrium (K).

ratef < rater

R P

ratef = rater

R P

Q =[P][R]

K

Q

KQ K

Q

Q > K

HW p. 662#36, 38

Q =[P][R]

Le Châtelier’s Principle

ratef > rater

R P

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 K value)

…will shift ( or ) to counteract the change.”

Because…

…the rxn goes faster in one direction (Q K)

until ratef = rater (Q = K again, R P ).

R P

(that affects collision frequency) like:

ratef < rateror

The conversion of nitrogen (N2) & hydrogen (H2) into ammonia (NH3) is tremendously significant in agriculture, where ammonia-based fertilizers are of utmost importance.

The Haber Process

N2(g) + 3 H2(g) 2 NH3(g)

N2(g) + 3 H2(g) 2 NH3(g)

K = [NH3]2

[N2][H2]3

Q = [NH3]2

[N2][H2]3

K = [NH3]2

[N2][H2]3

Q < K

Q = K

(same K)

Q = K

Addreactant:

(changes Q)

This apparatus pushes the equilibrium to the right by removing the product ammonia (NH3) from the system as a liquid.

Removeproduct:

N2(g) + 3 H2(g) 2 NH3(g)

(changes Q)

Change Volume (moles of gas)

• ↓V (↑PT) shifts to

• ↑V (↓PT) shifts to

• V has no shift if

fewer mol of gas (↓ngas)

more mol of gas (↑ngas)

N2(g) + 3 H2(g) 2 NH3(g)

equal mol of gas R & P

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

↓

• ↑PT by adding noble gas?

no shift b/c same R & P pressures

(changes Q)

Change Temperature

Changing temp. is the ONLY way to

change the value of ___.K Why?

K1 =[P][R]

K2 =[P]

[R] K2 =[P]

[R]

(larger)(smaller)(original)

T (add heat) shifts… T (add heat) shifts…

…in the _____thermic direction to ______ heat

…in the _____thermic direction to ______ heat

endoabsorb

exoproduce

Change TemperatureCoCl4

2– + 6 H2O(l) 4 Cl– + Co(H2O)62+

∆H = __heat + + heat

Remove heat

Add heat

ice

Remove product:Add product:–

increase

ratef and rater .

R

PEquilibrium occurs faster, but…at no shift (composition [P]/[R] is same).

[P][R]

K =

(same)

Catalysts

N2(g) + 3 H2(g) 2 NH3(g) ∆H = –92


Change in external factor

Shift to restore equilibrium

Reason

Increase pressure(decrease volume)

Increase temp.

Increase [N2]

Increase [NH3]

Add a catalyst

(practice)

Right

Left

Right

Left

No Shift

↑P (↓V) shifts to side of fewer moles of gas

∆H = – , adding heat shifts in the endo- dir.Adding reactant shifts right to consume it

Adding product shifts left to consume it

Catalysts change how fast, but not how far.

+ heat

∆H = –


Change in external factor

Shift to restore equilibrium

Reason

________ pressure(_______ volume)

_________ temp.

_________ PO2

_________ PH2O

Decrease H2O2

Left

Left

Right

Left

No Shift

↓P (↑V) shifts to side of more moles of gas

∆H = + , remove heat shifts in the exo- dir.Adding reactant shifts right to consume it

Removing reactant shifts left to produce it

(s) & (l) do not affect Q & K (usually)

2 H2O(g) + O2(g) 2 H2O2(l) ∆H = +108

(practice)

DecreaseIncrease

Decrease

heat +

Increase

Decrease

∆H = +

- 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)


more mol of gas (↑ngas)

R P(summary)

no shift

(H + R P) (R P + H)

WS Eq Pract. 2 #4

(changes K)

(Ptotal)

(M, PPR, PPP)

in endo dir. to use up heatin exo dir. to make more heat

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

Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

Ksp = [X+]a [Y–]b

Solubility Product Constant(Ksp)

Write the K expression for a saturated solution of PbI2 in water:

PbI2(s) Pb2+(aq) + 2 I−(aq)

XaYb(s) aX+(aq) + bY−(aq)

K =

For “insoluble” solids, the equilibrium constant,Ksp , is the solubility product constant, or the…

…product of M’s of dissolved ions at equilibrium.

[Pb2+][I−]2sp

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

Ksp is NOT the Solubility

4.2 g/L 0.027 M CaCrO4 0.027 M Ca2+

0.027 M CrO42–

CaCrO4(s) Ca2+ + CrO42–

[CaCrO4] [Ca2+] [CrO42–]

molar mass:156.08 g/mol

0.00073

HW p. 763 #46 Ksp = [Ca2+][CrO42–]

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 :

Ksp Calculations

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

If solubility (or molar solubility) is known, solve for Ksp . [PbBr2] is 0.010 M at 25oC .

HW p. 763 #48a

Ksp = (0.010)(0.020)2

Ksp = 4.0 x 10–61 PbBr2 dissociates into…

1 Pb2+ ion & 2 Br– ions

R ICE

Ksp = [Pb2+][Br–]2

(maximum that can dissolve)

(all dissolved = saturated) (any excess solid is irrelevant)

R ICE

Ksp Calculations

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 = [Pb2+][I–]2

Solubility (or molar solubility) is known, solve for Ksp .

[PbI2] = 0.0012 M

HW p. 763 #50

Ksp = (0.0012)(0.0024)2

Ksp = 6.9 x 10–9

0.54 g x 1 mol461 g

= 0.0012 mol

solubility:0.54 g/L

Molar solubility:0.0012 M

If only Ksp is known, solve for x (M).

AgCl(s) Ag+ + Cl–

x 0 M 0 M –x +x +x 0 M x x

Ksp = [Ag+][Cl–]

Ksp = x2

1.8 x 10–10 = x2

√1.8 x 10–10 = x

1.3 x 10–5 = x

Ksp for AgCl is 1.8 x 10–10 .

[AgCl] = 1.3 x 10–5 M [Ag+] = 1.3 x 10–5 M [Cl–] = 1.3 x 10–5 M

(molar solubility)

Ksp Calculations

R ICE


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

Ksp for CaSO4 is 2.4 x 10–5 .

[CaSO4] = 0.0049 M

[Ca2+] = 0.0049 M [SO4

2–] = 0.0049 M

HW p. 663 #47

(molar solubility)

Ksp Calculations

R ICE

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

[PbCl2] = 0.016 M

[Pb2+] = 0.016 M [Cl–] = 0.032 M

1.6 x 10–5 = 4x3

3√4.0 x 10–6 = x

0.016 = x


Ksp for PbCl2 is 1.6 x 10–5 .

(molar solubility)

Ksp Calculations

R ICE

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

[Cr(OH)3] = 1.6 x 10–8 M

[Cr3+] = 1.6 x 10–8 M [OH–] = 4.8 x 10–8 M

1.6 x 10–30 = 27x4

4√5.9 x 10–32 = x

1.6 x 10–8 = x


Ksp for Cr(OH)3 is 1.6 x 10–30 .

(molar solubility)

Ksp Calculations

R ICE

LaF3(s) La3+ + 3 F–

x 0 M 0 M –x +x +3x 0 M x 3x

Ksp = [La3+][F–]3

Ksp = (x)(3x)3

2 x 10–19 = 27x4

x = 9 x 10–6 M LaF3

HW p. 763 #52a Ksp Calculations(in pure H2O)

#52b

(in 0.010 M KF)

LaF3(s) La3+ + 3 F–

x 0 M –x +x +3x 0 M x 0.010 + 3x

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

0.010 M

≈ 0.010b/c K <<<1

Solubility is lower in _________________sol’n w/ common ion

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.

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?

•adding common ion shifts left (less soluble)

OR

WS Ksp #1-2

Factors Affecting Solubility

Mg(OH)2(s) Mg2+(aq) + 2 OH−(aq)

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.

Addition of HCl (acid) to this solution would cause…

Basic anions, more soluble in acidic solution.

Mg(OH)2(s) Mg2+(aq) + 2 OH−(aq)

H+

H+ NO Effect on:Cl– , Br–, I–,

NO3–, SO4

2–, ClO4–

Adding H+ would cause…

shift , more soluble.

HW p.

763 #55

Factors Affecting Solubility• Complex Ions

– Metal ions can and form complex ions with lone e– pairs in the solvent.

forming complex ions…

…increases solubility

AgCl(s) Ag+(aq) + Cl−(aq)

NH3

Ag(NH3)2+

p. 765 #59

Will a Precipitate Form?

• 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

Q = [X+]a [Y−]bWS Ksp #4 Ksp = [X+]a [Y–]b

XaYb(s) aX+(aq) + bY−(aq)

AgIO3(s) Ag+ + IO3– Ksp = [Ag+][IO3

–]

Ksp = 3.1 x 10–8

HW p. 764 #62b (OR…is Q > K ?)

Q = [Ag+][IO3–]

Q =

100 mL of 0.010 M AgNO3

10 mL of 0.015 M NaIO3

[Ag+] = ________

(0.010 M)(100 mL) = M2(110 mL)

(mixing changes M and V)M1V1 = M2V2

0.0091 M

[IO3–] = ________

(0.015 M)(10 mL) = M2(110 mL)

0.0014 M

Will a Precipitate Form?

Q = (0.0091)(0.0014)

Q = 1.3 x 10–5

Q > K , so…rxn shifts leftprec. will form

BaSO4(s) Ba2+ + SO42–

HW p. 764 #66a

(BaSO4) (SrSO4)SrSO4(s) Sr2+ + SO4

2–

Ksp = [Sr2+][SO42–]

When Will a Precipitate Form?

Ksp = [Ba2+][SO42–]

1.1 x 10–10 = (0.010)(x) 3.2 x 10–7 = (0.010)(x)

x = 1.1 x 10–8

[SO42–] = 1.1 x 10–8 M

x = 3.2 x 10–5

[SO42–] = 3.2 x 10–5 M

#66b

Ba2+ will precipitate first b/c… less SO42– is

needed to reach equilibrium (Ksp).

unit 8 (chp 15,17): chemical & solubility equilibrium (k eq, k c, k p, k sp ) john d. bookstaver...

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