class average for exam i 70. fe(oh) 3 fe 3+ (aq) + 3 oh - (aq) [fe 3+ ][oh - ] 3 = 1.1 x 10 -36...

Class average for Exam I

70

Fe(OH)3Fe3+

(aq) + 3 OH-(aq)

[Fe3+][OH-]3 = 1.1 x 10-36

[y][3y]3 = 1.1 x 10-36

If there is another source of OH- (NaOH)

that provides a higher [OH-] then that is

the value of [OH-] to be used.

pH and solubility

CaCO3(s) + H3O+(aq) Ca2+

(aq) + HCO3-(aq)

+ H2O(l)

CaCO3(s) + H3O+(aq) Ca2+

(aq) + HCO3-(aq)

+ H2O(l)

Net ionic equation.

CaCO3(s) + H3O+(aq) Ca2+

(aq) + HCO3-(aq)

+ H2O(l)

CaCO3 calcium carbonate, present in

both limestone and marble.

CaCO3(s) + H3O+(aq) Ca2+

(aq) + HCO3-(aq)

+ H2O(l)

CaCO3(s) + H2O(l) Ca2+(aq) + CO3

2-(aq)

CaCO3(s) + H3O+(aq) Ca2+

(aq) + HCO3-(aq)

+ H2O(l)

CaCO3(s) + H2O(l) Ca2+(aq) + CO3

2-(aq)

CO32-(aq) + H3O+(aq) H2O(l) +

HCO3-(aq)

Ksp = [Ca2+][CO32-]

CaCO3(s) + H3O+(aq) Ca2+

(aq) + HCO3-(aq)

+ H2O(l)

CaCO3(s) + H2O(l) Ca2+(aq) + CO3

2-(aq)

CO32-(aq) + H3O+(aq) H2O(l) +

HCO3-(aq)

Any reaction favoring the formation of HCO3-

favors the solution of a solid carbonate .

Ksp = [Ca2+][CO32-]

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

Increase solubility of Zn(OH)2 by

lowering pH.

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

Increase solubility of Zn(OH)2 by

lowering pH.

Removal of product, OH-, shifts equilibriumto right.

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[Zn2+] = ? @ pH = 7

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[Zn2+] = ? @ pH = 7

[Zn2+] = y

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[Zn2+] = ? @ pH = 7

[Zn2+] = y

(y)(2y)2 = Ksp

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[Zn2+] = ? @ pH = 7

[Zn2+] = y

(y)(2y)2 = Ksp = 4y3 = 4.5 x 10-17

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[Zn2+] = ? @ pH = 7

[Zn2+] = y

(y)(2y)2 = Ksp = 4y3 = 4.5 x 10-17

[Zn2+] = 2.2 x 10-6 M [OH-] = 4.4 x 10-6 M

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[Zn2+] = ? @ pH = 6

[Zn2+] = y

buffered

pH = 6, [H3O+] = 10-6

[OH-] = Kw

[H3O+]

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[Zn2+] = ? @ pH = 6

[Zn2+] = y

buffered

pH = 6, [H3O+] = 10-6

[OH-] = Kw

[H3O+]=

1 x 10-14

10-6= 10-8

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[OH-] = 1 x 10-8

(y)(1 x 10-8)2 = 4.5 x 10-17

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[OH-] = 1 x 10-8

(y)(1 x 10-8)2 = 4.5 x 10-17

y = 4.5 x 10-17

1 x 10-16

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[OH-] = 1 x 10-8

(y)(1 x 10-8)2 = 4.5 x 10-17

y = 4.5 x 10-17

1 x 10-16= 4.5 x 10-1

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[OH-] = 1 x 10-8

(y)(1 x 10-8)2 = 4.5 x 10-17

y = 4.5 x 10-17

1 x 10-16= 4.5 x 10-1

[Zn2+] = 0.45 M

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[Zn2+] = 0.45 MpH = 6

pH = 7 [Zn2+] = 0.0000022 M

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[Zn2+] = 0.45 MpH = 6

pH = 7 [Zn2+] = 0.0000022 M

[OH-] = 2[Zn2+]

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[Zn2+] = 0.45 MpH = 6

pH = 7 [Zn2+] = 0.0000022 M

[OH-] = 2[Zn2+] = 4.4 x 10-6 M

Zn(OH)2(s) Zn2+(aq) + 2 OH-

(aq)

[Zn2+][OH-]2 = Ksp = 4.5 x 10-17

[Zn2+] = 0.45 MpH = 6

pH = 8.6 [Zn2+] = 0.0000022 M

[OH-] = 2[Zn2+] = 4.4 x 10-6 M

Solubility of salts of

weak acids

Solubility of salts of

weak acids

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

Solubility of salts of

weak acids

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

Reduce pH, increase [H3O+]

Solubility of salts of

weak acids

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

Reduce pH, increase [H3O+]

Increase [H3O+] : equilibrium

Solubility of salts of

weak acids

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

NaA(s) Na+(aq) + A-

(aq)

conjugatebase

Solubility of salts of

weak acids

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

NaA(s) Na+(aq) + A-

(aq)

A-(aq) + H3O+

(aq) HA(aq) + H2O(l)

Solubility of salts of

weak acids

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

NaA(s) Na+(aq) + A-

(aq)

A-(aq) + H3O+

(aq) HA(aq) + H2O(l)

Reduce pH, increase [H3O+]

Solubility of salts of

weak acids

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

NaA(s) Na+(aq) + A-

(aq)

A-(aq) + H3O+

(aq) HA(aq) + H2O(l)

Reduce pH, increase [H3O+], reduce [A-]

Solubility of salts of

weak acids

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

NaA(s) Na+(aq) + A-

(aq)

A-(aq) + H3O+

(aq) HA(aq) + H2O(l)

Reduce pH, increase [H3O+], reduce [A-], increase [Na+]

Using common ions to separate

mixtures of ions.

Using common ions to separate

mixtures of ions.

Separating ions which share a

group is difficult - they have

very similar chemistries.

Using common ions to separate

mixtures of ions.

I II VII

Na+ Ca2+ Cl-

K+ Ba2+ I-

M1X M1+ + X-

M2X M2+ + X-

M1X M1+ + X-

M2X M2+ + X-

If the solution is not saturated for

either M1X or M2X, there will

be no solid present.

M1X M1+ + X-

M2X M2+ + X-

As [X-] is increased, at some point

precipitation will occur.

M1X M1+ + X-

M2X M2+ + X-

If Ksp for M1X and Ksp for M2X

differ by a large enough amount,

one will precipitate and the other

will remain in solution.

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

CaF2 Ksp = 3.9 x 10-11

BaF2 Ksp = 1.7 x 10-6

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

CaF2 Ksp = 3.9 x 10-11

BaF2 Ksp = 1.7 x 10-6

Q < Ksp no precipitate

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

CaF2 Ksp = 3.9 x 10-11

BaF2 Ksp = 1.7 x 10-6

Q = [Ba2+][F-]2 < 1.7 x 10-6

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

CaF2 Ksp = 3.9 x 10-11

BaF2 Ksp = 1.7 x 10-6

Q = [Ba2+][F-]2 < 1.7 x 10-6

Q < Ksp no precipitate

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

CaF2 Ksp = 3.9 x 10-11

BaF2 Ksp = 1.7 x 10-6

Q = [Ba2+][F-]2 < 1.7 x 10-6

Q < Ksp no precipitate

Q = [Ca2+][F-]2 > 3.9 x 10-11

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

CaF2 Ksp = 3.9 x 10-11

BaF2 Ksp = 1.7 x 10-6

Q = [Ba2+][F-]2 < 1.7 x 10-6

Q < Ksp no precipitate

Q = [Ca2+][F-]2 > 3.9 x 10-11

Q > Ksp precipitate

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

CaF2 Ksp = 3.9 x 10-11

BaF2 Ksp = 1.7 x 10-6

Q = [Ba2+][F-]2 < 1.7 x 10-6

Q < Ksp no precipitate

Q = [Ca2+][F-]2 > 3.9 x 10-11

Q > Ksp precipitate

Adjust [F-]so maximumCa2+ precipitateswith no Ba2+

precipitate.

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6

Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6

Q < Ksp no precipitate

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6

Q < Ksp no precipitate

[F-]2 < 1.7 x 10-6

[Ba2+]

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6

Q < Ksp no precipitate

[F-]2 < 1.7 x 10-6

[Ba2+]=

1.7 x 10-6

0.10

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6

Q < Ksp no precipitate

[F-]2 < 1.7 x 10-6

[Ba2+]=

1.7 x 10-6

0.10= 1.7 x 10-5

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6

Q < Ksp no precipitate

[F-]2 < 1.7 x 10-6

[Ba2+]=

1.7 x 10-6

0.10= 1.7 x 10-5

[F-] < 4.1 x 10-3

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6Ksp (BaF2) = [Ba2+][F-]2 = 1.7 x 10-6

Q < Ksp no precipitate

[F-]2 < 1.7 x 10-6

[Ba2+]=

1.7 x 10-6

0.10= 1.7 x 10-5

[F-] < 4.1 x 10-3 All Ba2+ remains in solution.

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

Ksp (CaF2) = [Ca2+][F-]2 = 3.9 x 10-11

Q > Ksp precipitate

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

[Ba2+] = [Ca2+] = 0.10 M

Ksp (CaF2) = [Ca2+][F-]2 = 3.9 x 10-11

Q > Ksp precipitate

[F-]2 > 3.9 x 10-11

0.10

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

Ksp (CaF2) = [Ca2+][F-]2 = 3.9 x 10-11

Q > Ksp precipitate

[F-]2 > 3.9 x 10-11

0.10= 3.9 x 10-10

[F-] > 2.0 x 10-5 CaF2(s) starts toprecipitate

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

[F-] > 2.0 x 10-5 CaF2(s) starts toprecipitate

[F-] < 4.1 x 10-3 All Ba2+ remains in solution.

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

[F-] > 2.0 x 10-5 CaF2(s) starts toprecipitate

[F-] < 4.1 x 10-3 All Ba2+ remains in solution.

How to reduce [Ca2+] to as low a levelas possible without BaF2 precipitation?

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

[F-] < 4.1 x 10-3 All Ba2+ remains in solution.

Adjust [F-] to 0.0041 M

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

[F-] < 4.1 x 10-3 All Ba2+ remains in solution.

Adjust [F-] to 0.0041 M

How much Ca2+ remains in solution?

CaF2(s) Ca2+(aq) + 2 F-

(aq)

BaF2(s) Ba2+(aq) + 2 F-

(aq)

CaF2(s) Ca2+(aq) + 2 F-

(aq)

Adjust [F-] to 0.0041 M

Ksp = [Ca2+][F-]2 = 3.9 x 10-11

[Ca2+] =3.9 x 10-11

[F-]2

CaF2(s) Ca2+(aq) + 2 F-

(aq)

Adjust [F-] to 0.0041 M

Ksp = [Ca2+][F-]2 = 3.9 x 10-11

[F-]2=

3.9 x 10-113.9 x 10-11

(4.1 x 10-3)2

2.3 x 10-6[Ca2+] =

[Ca2+] =

Adjust [F-] to 0.0041 M

0.10 M CaF2(s) Ca2+(aq) + 2 F-

(aq)

0.10 M BaF2(s) Ba2+(aq) + 2 F-

(aq)

[Ba2+] = 1.7 x 10-6

(4.1 x 10-3)2= 0.10 M

2.3 x 10-6[Ca2+] =

CaF2 Ksp = 3.9 x 10-11

BaF2 Ksp = 1.7 x 10-6

Metal sulfides in acidic solution

MS M2+(aq) + S2-

(aq)

Ksp = [M2+][S2-]

Metal sulfides in acidic solution

MS M2+(aq) + S2-

(aq)

Ksp = [M2+][S2-]

S2- Is a strong base

Metal sulfides in acidic solution

MS M2+(aq) + S2-

(aq)

Ksp = [M2+][S2-]

S2- Is a strong base

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Kb =[HS-][OH-]

[S2-]

Metal sulfides in acidic solution

MS M2+(aq) + S2-

(aq)

Ksp = [M2+][S2-]

S2- Is a strong base

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Kb =[HS-][OH-]

[S2-] 105

Metal sulfides in acidic solution

MS M2+(aq) + S2-

(aq)

Ksp = [M2+][S2-]

S2- Is a strong base

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Kb =[HS-][OH-]

[S2-] 105

Ksp = [M2+][HS-][OH-]

Metal sulfides in acidic solution

MS M2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [M2+][HS-][OH-]

Lower pH, lower [OH-], higher [M2+]

Ksp metal sulfides < 10-10

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

Buffer solution to pH = 2

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

Buffer solution to pH = 2

[H3O+] = 1 x 10-2

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

Buffer solution to pH = 2

[H3O+] = 1 x 10-2 [OH-] =1 x 10-14

1 x 10-2

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

Buffer solution to pH = 2

[OH-] =1 x 10-14

1 x 10-2= 1 x 10-12

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

Buffer solution to pH = 2

[OH-] = 1 x 10-12

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

Buffer solution to pH = 2

[OH-] = 1 x 10-12

H2S(aq) + H2O(l) H3O+(aq) + HS-

(aq)

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

[OH-] = 1 x 10-12

H2S(aq) + H2O(l) H3O+(aq) + HS-

(aq)

Ka =[H3O+][HS-]

[H2S]= 9.1 x 10-8

Buffer solution to pH = 2 [H2S] = 0.10 M

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

[OH-] = 1 x 10-12

H2S(aq) + H2O(l) H3O+(aq) + HS-

(aq)

Ka =(0.01)[HS-]

(0.10)= 9.1 x 10-8

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

[OH-] = 1 x 10-12

H2S(aq) + H2O(l) H3O+(aq) + HS-

(aq)

Ka =(0.01)[HS-]

(0.10)= 9.1 x 10-8

[HS-] = (0.10)(9.1 x 10-8)/(0.01) = 9.1 x 10-7

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

[OH-] = 1 x 10-12

H2S(aq) + H2O(l) H3O+(aq) + HS-

(aq)

[HS-] = 9.1 x 10-7

[Cd2+] = (7 x 10-28)/(9.1 x 10-7)(1 x 10-12)

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

[OH-] = 1 x 10-12

H2S(aq) + H2O(l) H3O+(aq) + HS-

(aq)

[HS-] = 9.1 x 10-7

[Cd2+] = (7 x 10-28)/(9.1 x 10-7)(1 x 10-12) = 8 x 10-10

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

[OH-] = 1 x 10-7

H2S(aq) + H2O(l) H3O+(aq) + HS-

(aq)

Ka =(1 x 10-7)[HS-]

(0.10)= 9.1 x 10-8

[HS-] = (0.1)(9.1 x 10-8)/(1 x 10-7) = 9.1 x 10-2

pH = 7

Metal sulfides in acidic solution

CdS(s) Cd2+(aq) + S2-

(aq)

S-2(aq) + H2O(l) HS-

(aq) + OH-(aq)

Ksp = [Cd2+][HS-][OH-] = 7 x 10-28

[OH-] = 1 x 10-7

H2S(aq) + H2O(l) H3O+(aq) + HS-

(aq)

[HS-] = 9.1 x 10-2

[Cd2+] = (7 x 10-28)/(9.1 x 10-2)(1 x 10-7) = 8 x 10-20

Complex ions

Complex ions

Metals coordinated to ligands which

can exist independantly as

molecules or ions.

Ammonia, NH3, and water are

common ligands in complex ions.

Cobalt(II) chloride

Cobalt(II) chloride

CoCl2.6H2O

CoCl2.2H2O

Cobalt(II) chloride

CoCl2.6H2O

[Co(H2O)6]2+ (Cl-)2

CoCl2.2H2O

blue

purple

magenta

2+

[Co(H2O)6]2+

Solubilities of complex ions

Equilibria involving complex ions

may involve multi-step processes

in which individual ligands are

added or removed.

Cobalt(II) chloride

CoCl2.6H2O

[Co(H2O)6]2+ (Cl-)2

CoCl2.2H2O

blue

purple

magenta

These species are quite soluble.

Ag+(aq) + NH3(aq) Ag(NH3)+

Ag+(aq) + NH3(aq) Ag(NH3)+(aq)

[Ag(NH3)+]

[Ag+][NH3]= 2.1 x 103K1 =

[Ag(NH3)+]

[Ag+][NH3]= 2.1 x 103K1 =

Ag(NH3)+(aq) + NH3(aq) Ag(NH3)2

+(aq)

Ag+(aq) + NH3(aq) Ag(NH3)+

(aq)

[Ag(NH3)+]

[Ag+][NH3]= 2.1 x 103K1 =

Ag(NH3)+(aq) + NH3(aq) Ag(NH3)2

+(aq)

Ag+(aq) + NH3(aq) Ag(NH3)+

(aq)

K2 =[Ag(NH3)2

+]

[Ag(NH3)+][NH3]= 8.2 x 103

K1, K2 > 103 - large

K1, K2 > 103 - large

Ag+ in the presence of excess NH3

is converted to 100% Ag(NH3)2+

K1, K2 > 103 - large

Ag+ in the presence of excess NH3

is converted to 100% Ag(NH3)2+.

Add 0.1 M AgNO3 to 1 L of 1 M NH3

K1, K2 > 103 - large

Ag+ in the presence of excess NH3

is converted to 100% Ag(NH3)2+.

Add 0.1 M AgNO3 to 1 L of 1 M NH3

Initial condition: [Ag(NH3)2+] = 0.10 M

[NH3] = 0.80 M

Ag(NH3)2+

( aq) Ag(NH3)+(aq) + NH3(aq)

K = 1

K2[Ag(NH3)2

+]

[Ag(NH3)+] [NH3] =

Init 0.100 M 0 0.80

Ag(NH3)2+

( aq) Ag(NH3)+(aq) + NH3(aq)

K = 1

K2[Ag(NH3)2

+]

[Ag(NH3)+] [NH3] =

Init 0.100 M 0 0.80

-y +y + y

Eq 0.100-y y 0.80 + y

Ag(NH3)2+

( aq) Ag(NH3)+(aq) + NH3(aq)

K = 1

8.2 x 103(0.100-y)

(y)(0.80+y)=

Init 0.100 M 0 0.80

-y +y + y

Eq 0.100-y y 0.80 + y

K = 1

8.2 x 103(0.100-y)

(y)(0.80+y)=

Assume y is small

8y = 1.2 x 10-4

y = 1.5 x 10-5 M/L

Ag(NH3)2+

( aq) Ag(NH3)+(aq) + NH3(aq)

Init 0.100 M 0 0.80

-y +y + y

Eq 0.100 1.5 x 10-5 0.80

Ag(NH3)+( aq) Ag+

(aq) + NH3(aq)

[Ag+][NH3]

[Ag(NH3)+]=

1

K1

= 4.8 x 10-4

Eq 0.100 1.5 x 10-5 0.80

Ag(NH3)+( aq) Ag+

(aq) + NH3(aq)

[Ag+][NH3]

[Ag(NH3)+]=

1

K1

= 4.8 x 10-4

Eq 0.100 1.5 x 10-5 0.80

[Ag+] =(1.5 x 10-5)(4.8 x 10-4)

0.80

Ag(NH3)+( aq) Ag+

(aq) + NH3(aq)

[Ag+][NH3]

[Ag(NH3)+]=

1

K1

= 4.8 x 10-4

Eq 0.100 1.5 x 10-5 0.80

[Ag+] =(1.5 x 10-5)(4.8 x 10-4)

0.80=

9 x 10-9 M/L

Ag+(aq) + NH3(aq) Ag(NH3)+

(aq)

Ag(NH3)+(aq) + NH3(aq) Ag(NH3)2

+(aq)

9 x 10-9 0.80 1.5 x 10-5

1.5 x 10-5 0.80 0.10

Formation of complex ions starting

with AgCl.

AgCl(s) Ag+(aq) + Cl-

(aq)

Ksp = 1.6 x 10-10

By adding NaCl, increase in [Cl-]

(common ion effect) causes shift

to AgCl(s)

Ksp = 1.6 x 10-10

Ag+(s) + 2 Cl-

(aq) AgCl2-

(aq)

AgCl(s) Ag+(aq) + Cl-

(aq)

[AgCl2-]

[Ag+][Cl-]2K = 1.8 x 105 =

Ksp = 1.6 x 10-10

Ag+(s) + 2 Cl-

(aq) AgCl2-

(aq)

AgCl(s) Ag+(aq) + Cl-

(aq)

[AgCl2-]

[Ag+][Cl-]2K = 1.8 x 105 =

[Ag+] = [Cl-] = 1.3 x 10-5

Ksp = 1.6 x 10-10

Ag+(s) + 2 Cl-

(aq) AgCl2-

(aq)

AgCl(s) Ag+(aq) + Cl-

(aq)

[AgCl2-] [Ag+][Cl-]2 = (1.8 x 105)

[Ag+] = [Cl-] = 1.3 x 10-5

Ksp = 1.6 x 10-10

Ag+(s) + 2 Cl-

(aq) AgCl2-

(aq)

AgCl(s) Ag+(aq) + Cl-

(aq)

[Ag+] = [Cl-] = 1.3 x 10-5

[AgCl2-] (1.3 x 10-5)3 =

4 x 10-10

= (1.8 x 105)

[AgCl2-] [Ag+][Cl-]2 = (1.8 x 105)

Ksp = 1.6 x 10-10

Ag+(s) + 2 Cl-

(aq) AgCl2-

(aq)

AgCl(s) Ag+(aq) + Cl-

(aq)

[Ag+] = 1.3 x 10-5

[AgCl2-] (1.3 x 10-5) =

2.34 M = (1.8 x 105)

[AgCl2-] [Ag+][Cl-]2 = (1.8 x 105)

Change [Cl-] 1 M

Hydrolysis by complex ions

+ H2O(l)

+ H3O+

+ H2O(l)

+ H3O+

Ka = [H3O+][Fe(H2O)5OH2+]

[Fe(H2O)63+]

=

7.7 x 10-3