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16.10. AgCl ( s ) Ag + ( aq ) + Cl - ( aq ). Complex Ion Equilibria and Solubility. A complex ion can increase the solubility of a salt. Consider. K sp = 1.6 x 10 -10. What is we add NH 3 (Lewis base) ?. Ag + ( aq ) + 2 NH 3 ( aq ) Ag(NH 3 ) 2 + ( aq ).

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slide2

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

Complex Ion Equilibria and Solubility

A complex ion can increase the solubility of a salt.

Consider

Ksp= 1.6 x 10-10

What is we add NH3 (Lewis base) ?

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

Lewis acid

Lewis base

Complex ion

[Ag(NH3)2+]

= Formation constant = 1.7 x 107

Kf =

[Ag+][NH3]2

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

slide3

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

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

Ksp= 1.6 x 10-10

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

Kf = 1.7 x 107

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

0.10

0.00

0.00

Initial (M)

-s

- 2s

+s

+s

Change (M)

( - s)

0.10 -2s

s

s

Equilibrium (M)

Molar solubility = s

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

[Ag(NH3)2+] [Cl-]

(s)(s)

= 2.8 x 10-3 =

Kf =

[NH3]2

(0.10 -2s)2

Assume

s << 0.010

s = 5.0 x 10-3 M

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

slide4

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

No precipitate

Unsaturated solution

Qsp < Ksp

Qsp = Ksp

Saturated solution

Precipitate will form

Qsp > Ksp

Supersaturated solution

slide5

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

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?

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

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

20.0 ml 20.0 ml

[Ag+] 0

[Cl-] 0

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

Since Qsp > Ksp, a precipitate will form

slide6

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

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)

s

0.0010 M

At Equilibrium

c

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

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

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

slide7

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

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

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

Co(OH)2 (s) Co2+(aq) + 2 OH-(aq)

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

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

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

Kb,NH3= 1.8 x 10-5 = [NH4+][OH-] = [NH4+](5.0 x 10-5 M)

[NH3] 0.30 M

=> [NH4+] = 0.11 M

slide8

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- HgI42-

Kf= 1.0 x 10+30

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

Large

[Hg2+][I-]4

=> Assume complete reaction with I-

start

0.010 M

0.78 M

0.00 M

Hg2+ + 4 I- HgI42-

0.78 - 4(0.010)

= 0.74 M

end

0.00 M

0.010 M

Cannot be zero !!

But Kf

8

slide9

Hg2+ + 4 I- HgI42-

At Equilibrium

x

0.10 -x

0.74 + 4x

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

[Hg2+][I-]4

(x)(0.74 + 4x) 4

=> Assume x << 0.010 < 0.74

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

x

0.010

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