16.10

Complex Ion Equilibria and Solubility

A complex ion can increase the solubility of a salt.

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

Kf =[Ag(NH3)2

+][Ag+][NH3]2

AgCl (s) Ag+ (aq) + Cl- (aq) Ksp = 1.6 x 10-10

Consider

What is we add NH3 (Lewis base) ?

Lewis acid

Lewis base Complex ion

= Formation constant = 1.7 x 107

Huge !!!Ag+ is effectively removed from solution,

allowing more AgCl to dissolve.

AgCl (s) + 2 NH3 (aq) Ag(NH3)2+ (aq) + Cl- (aq)

Le Chatelier’s Principle

What is the molar solubility of AgCl in 0.10 M NH3 ?

Assume s = 5.0 x 10-3 Ms << 0.010

Initial (M)Change (M)

Equilibrium (M)

0.00+s

0.00+s

s s

AgCl (s) + 2 NH3 (aq) Ag(NH3)2+ (aq) + Cl- (aq)

-s

( - s)

0.10- 2s

0.10 -2s

AgCl (s) Ag+ (aq) + Cl- (aq) Ksp = 1.6 x 10-10

Ag+ (aq) + 2 NH3 (aq) Ag(NH3)2+ (aq) Kf = 1.7 x 107

K = Ksp Kf = (1.6 x 10-10)(1.7 x 107) = 2.8 x 10-3

Kf = [Ag(NH3)2

+] [Cl-][NH3]2

= 2.8 x 10-3 = (s)(s)(0.10 -2s)2

Check: (0.0050/0.10) x 100% = 5.0 %

Molar solubility = s

What happens when we mix two solutions?Dilution

Molarity = moles / L

# moles = (conc.)(volume) = (Molarity)(L)

mole1 = C1V1 = mole2 = C2V2

So, C2 = C1V1 / V2

Reaction?

same

Dilute to new volume

Qsp = Ksp Saturated solutionQsp < Ksp Unsaturated solution No precipitate

Qsp > Ksp Supersaturated solution Precipitate will form

A solution contains 0.10 M AgNO3 (aq). When a 10.0 ml sample is mixed with 10.0 ml tap water (containing 1.0 x 10-5 M Cl-), will a white precipitate form?

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

Ksp = 1.6 x 10-10 = [Ag+] 0 [Cl-] 0

Since Qsp > Ksp, a precipitate will form

Qsp = 0.10 M 10.0 ml 1.0 x 10-5M 10.0 ml 20.0 ml 20.0

ml

Qsp = 2.5 x 10-7 > 1.6 x 10-10 = Ksp

[Ag+] 0 [Cl-] 0

What pH is required to keep [Cu2+] below 0.0010 M in a saturated solution of Cu(OH)2 ? Ksp,Cu(OH)2 = 2.2 x 10-20

Cu(OH)2 (s) Cu2+ (aq) + 2 OH- (aq)

Ksp = 2.2 x 10-20 = [Cu2+] [OH-] 2

Ksp = 2.2 x 10-20 = (0.0010 M) (c2)

How can we control the concentration of an ion in solution?

At Equilibrium s0.0010 M c

c = 4.7 x 10-9 = [OH-]

pH = 14 + log[OH-] = 14 + (-8.33) = 5.67

=> if pH is higher than 5.67, [OH-] > 4.7 x 10-9 M and [Cu2+] will be less than 0.0010 M; at lower pH more Cu2+ will be in solution

What concentration of NH4+ is required to prevent the

precipitation of Co(OH)2 in a solution of 0.10 M Co(NO3 )2

and 0.30 M NH3 ?(Assume NH4+ / NH3 only acts as a

buffer.) Ksp,Co(OH)2 = 2.5 x 10-10 , Kb,NH3 = 1.8 x 10-5

Co(OH)2 (s) Co2+ (aq) + 2 OH- (aq)Ksp = 2.5 x 10-10 = [Co2+] [OH-] 2 = ( 0.10 M) [OH-] 2

Kb,NH3 = 1.8 x 10-5 = [NH4+][OH-] = [NH4

+](5.0 x 10-5 M)

=> Buffer will fix pH => fix [OH-]

=> Find [NH4+] that sets [OH-] too low to cause Co(OH)2 ppt.

[OH-] = 5.0 x 10-5 or less to prevent ppt.

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

[NH3] 0.30 M

=> [NH4+] = 0.11 M

Calculate the concentration of the free metal ion (Hg2+) in the presence of a ligand.Starting concentrations: [Hg2+] 0 = 0.010 M [I-] 0 = 0.78 M Hg2+ + 4 I- HgI4

2- Kf = 1.0 x 10+30

=> Assume complete reaction with I-

Kf = 1.0 x 10+30 = [HgI42-]

[Hg2+][I-]4Large

Hg2+ + 4 I- HgI42-

start

end

0.010 M 0.78 M 0.00 M

0.00 M 0.78 - 4(0.010)= 0.74 M 0.010 M

But Kf 8 Cannot be zero !!

=> Assume x << 0.010 < 0.74

Kf = 1.0 x 10+30 = [HgI42-] = 0.10 - x

[Hg2+][I-]4

Hg2+ + 4 I- HgI42-

At Equilibrium x 0.74 + 4x 0.10 -x

(x)(0.74 + 4x) 4

(1.0 x 10+30)(0.74)4 = 3.0 x 10+31 = 1 0.010 x

=> x = 3.3 x 10-32 M = [Hg2+]

16.11

Qualitative Analysis of

Cations

16.11