19 Heterogeneous & complex equilibria 1 19 Heterogeneous and complex equilibria CaCO 3 exists as...

19 Heterogeneous & complex equilibria 1

19 Heterogeneous and complex equilibria

CaCO3 exists as aragonite calcite

Formation of crystals such as AgCl in a solution is a heterogeneous equilibrium, because there are more than one phase,

AgCl (s) = Ag+ (aq) + Cl– (aq)

Species such as Ag(NH3)2+ & Ag(CN)2

– are complexes (or complex ions).

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


Beauty due to heterogeneous equilibria

There are many natural heterogeneous equilibria.

Please think of some!


The solubility productFor the dissolution,

CaCO3 (s) = Ca2+ (aq) + CO32- (aq)

Ksp = [Ca2+] [CO32-] = 3.8e-9 (a constant) solubility product

Substance Formula Ksp

Aluminum hydroxide Al(OH)3 4.6e-35

Barium chromate BaCrO3 1.2e-10

Calcium phosphate Ca3(PO4)2 1e-26

Iron sulfite FeS 6e-18

Lead sulfite PbS 2.5e-27

Mercury sulfite HgS 1.6e-52

Ksp = [Al3+] [OH-]3

Ksp = [Ca2+]3 [PO4]2

Which is the most and least soluble?


Solubility product

constants

Compound Ksp Compound Ksp

AgBr 5.0 x 10-13 Fe(OH)3 4.0 x 10-38

AgCl 1.8 x 10-10 FeS 6.3 x 10-18

AgI 8.3 x 10-17 HgS 1.6 x 10-52

AgIO3 3.1 x 10-8 Mg(OH)2 1.8 x 10-11

Ag3PO4 1.3 x 10-20 MgC2O4 8.6 x 10-5

Al(OH)3 2.0 x 10-32 Mn(OH)2 1.9 x 10-13

Ba(OH)2 5.0 x 10-3 MnS 2.5 x 10-13

BaSO4 1.1 x 10-10 NiS 1.0 x 10-24

Bi2S3 1.0 x 10-97 PbCl2 1.6 x 10-5

CaCO3 4.8 x 10-9 PbSO4 1.6 x 10-8

CaC2O4 4.0 x 10-9 PbS 8.0 x 10-28

CaSO4 1.2 x 10-6 SrSO4 3.2 x 10-7

CdS 8.0 x 10-27 Zn(OH)2 3.3 x 10-17

CoS 2.0 x 10-25 ZnS 1.6 x 10-23

CuS 6.3 x 10-36

Table like this is available in handbooks or data bases. Know where to find them when you need them. Write Ksp expressions.


Ksp and solubilityThe Ksp = 1.0e-6 for BaF2, what are [Ba2+] and [F¯]?

Solution:BaF2 = Ba2+ + 2 F¯

assume x 2 x

Ksp = x (2x)2 = 1.0e-6x = 3 (1.0e-6 / 4) = 6.3e-3 M

[Ba2+] = x = 6.3e-3 M molar solubility = 6.3e-3 *175 = 1.1g/L[F¯] = 2 x = 0.013 M

Checking:6.3e-3 *0.0132 = 1e-6 = Ksp

Calculate solubility of BaCrO3, Ksp = 1.2e-10

Molar mass of BaF2=175 g/mol

19-2


Concentrations of ions in solution

CaF2(s) = Ca2+(aq) + 2 F- (aq), Ksp know where to find

Ksp = 5.3e-9


Perception of a saturate solution

Ag2S = 2 Ag+ + S2–

Ksp = [Ag+]2[ S2–]


Calculate Ksp from Solubility

When 0.50 L saturated CaC2O4 solution was dried, 0.0030 g dry salt was obtained. Evaluate the Ksp.

Solution:3.0e-3 g————0.50 L

1 mol———128 g

= 4.7e-5 mol / L = [Ca2+]

CaC2O4 = Ca2+ + C2O42-

4.7e-5 4.7e-5

Ksp = 4.7e-5 * 4.7e-5 = 2.2e-9

Figure out how to calculate solubility from Ksp .

How many grams of HgS will dissolve in 1 L?


Precipitation of AgCl

Goto URL:

http:// dl.clackamas.cc.or.us/ch105-05/solubili.htm

for a demonstration if you have not seen the experiment

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


Common-ion effect on solubilityLike acid-base equilibria, presence of common ions from more than one electrolyte affects the solubility, since Ksp remains constant.

For example, the Ksp = 1.8e-10 for AgCl. The maximum [Ag+] is governed by the condition,

NaCl = Na+ + Cl-0.10 0.10

AgCl = Ag+ + Cl-[Ag+] 0.10

Thus [Ag+] * 0.10 = 1.8e-10 [Ag+] = 1.8e-9 M

Solubility of AgCl = 1.8e-10 = 1.3e-5 M in pure water is 7,454 times more.

? Should we considerAgCl = Ag+ + Cl-

x 0.10+x

19-3


Graph the common ion effect

AgCl = Ag+ + Cl-

NaCl = Na+ + Cl-


Condition for precipitationRecall: Predicting reaction directions by comparing Qc and Kc.

Same principle applies to precipitation (ppt)

Q < Ksp, unsaturated (solution)Q = Ksp, saturated (usually two phases are present)Q > Ksp, super-saturated

(unstable, often needs a seed to start the ppt)

19-5


Separation by precipitationA 10-mL solution contains 0.10 M each of Cl–, Br–, and I– ions. Micro amounts of 0.10 M AgNO3 solution is added to the system. What are the Ag+ concentrations before AgI, AgBr, and AgCl precipitate?Solution: Data sheet Ksp:AgCl 1.8e-10, AgBr 5.0e-13, and AgI 8.3e-17.

[Ag+] for AgI, AgBr, & AgCl solids to form:[I–] = 0.10 M [Ag+] = 8.3e-17/ 0.10 = 8.3e-16 AgI(s) appears

[Br–] = 0.10 M [Ag+] = 5.0e-13/ 0.10 = 5.0e-12 AgBr(s) appears [Cl–] = 0.10 M [Ag+] = 1.8e-10 / 0.10 = 1.8e-9 AgCl(s) appears

As [Ag+] increases, how do [Cl-], [Br-] and [I-] vary ? [Ag+] [Cl–] [Br–] [I–]

0 8.3e-16 0.1 0.1 0.10 AgI(s) 5.0e-12 0.1 0.10 AgBr(s) 1.7e-5 1.8e-9 0.10 AgCl(s) 2.8e-4 4.6e-8 1e-7 1.8e-3 5.0e-6 8.3e-10 0.1 ____? ____? ____?

19-6


pH H2S and solubility of metal ionsThe pH affects the equilibrium of many species, for example:

H2S = H+ + HS– Ka1 = 1e-9HS– = H+ + S2– Ka2 = 1e-14 (doubtful but adopt)H2S = 2 H+ + S2– Koverall = 1e-23

[S2–] = [H2S] *1e-23 / [H+]2 = [H2S] *1e(2*pH-23), strongly affected by pH

If [H2S] = 1.0 MpH [S2–] Ksp of MS [S2–] for [M2+] = 0.0011 1e-21 1.6e-52 HgS 1.6e-49 (ppt)2 1e-19 2.5e-27 PbS 2.5e-24 (ppt)2.52 1.1e-18 1.1e-21 ZnS 1.1e-18 (no ppt pH < 2.5) 4.39 6e-15 6e-18 FeS 6e-15 (no ppt pH < 4.4)7 1e-9 2.5e-13 MnS _____ (no ppt pH < ____)10 1e-3

11 0.1

19-7

Recalculate [s2-] at various pH if [H2S] = 0.10 M


pH, and CO2 on CaCO3 solubilityThe [CO3

2–] in a 0.0010 M H2CO3 solution is determined by,H2CO3 = H+ + HCO3

– Ka1 = 4.3e-7HCO3

– = H+ + CO32– Ka2 = 5.6e-11

H2CO3 = 2 H+ + CO32– Koverall = Ka1*Ka2 = 2.4e-17

= [H+]2 [CO32–] / [H2CO3]

[CO32–] = (0.0010*2.4e-17) [H+]–2

= 2.4e-20 * 1e(2*pH) = 2.4e(2*pH-20)

[CO32–] affects [Ca2+] due to equilibrium

CaCO3 = Ca2+ + CO32– Ksp = 8.7e–9

[Ca2+] = 8.7e–9 [CO32–]–1

= 8.7e–9 / 2.4e(2*pH–20)

= 3.6e(20 – 9–2*pH) = 3.6e(11-2*pH)

Decrease pH by 1 increases [Ca2+] by 2 order of magnitude

pH [Ca2+]8 3.6e-57 3.6e-36 0.36


Stalactites and stalagmites

stalactites

stalagmites

Rain dissolves limestone and when water drops form stalactites and stalagmites in caves. They grew about 2 cm per 1000 years.

The slightly acidic rain dissolves lime stone:CaCO3 (s) + H+ = Ca2+ (aq) + HCO3

- (aq)

When acidity is reduced, solid forms:Ca2+ (aq) + HCO3

- (aq) + OH- (aq) = CaCO3 (s) + H2O


Stalactites hanging down from the ceiling formed over hundreds of years in Mercer Cavern


Aragonite formations found at Mercer Caverns, California

The "Angel Wings" are two delicate and translucent crystalline formations, over 9 feet long and 2.5 feet wide in Mercer Caverns


Equilibrium of complexesMetal ions tend to attract Lewis bases forming coordinated complexes, or complex ions. These formation is governed by equilibrium,

metal ion ligand Step-wise formation constantAg+ (aq) + NH3 (aq) = Ag(NH3)+ (aq) K1

Ag(NH3)+ (aq) + NH3 (aq) = Ag(NH3)2+ (aq) K2

formation constantAg+ (aq) + 2 NH3 (aq) = Ag(NH3)2

+ (aq) Kf = K1 K2 = 1.7e7Ag+ (aq) + 2 S2O3

2– (aq) = Ag(S2O3)23– (aq) Kf = 2.9e13

Ag+ (aq) + 2 CN– (aq) = Ag(CN)2– (aq) Kf = 5.6e18

Dissociation constant Ag(NH3)2

+ (aq) = Ag+ (aq) + 2 NH3 (aq) Kd = 1 / Kf = 5.9e-8 Ag(S2O3)2

3– (aq) = Ag+ (aq) + 2 S2O32– (aq) Kd = 1/2.9e13 = 3.4e-14

19-8


Logical, but impractical method!!!

Evaluate [Ag+] in a solution containing 1.0 M NH3 and 0.10 M AgNO3.

Solution:Ag+ (aq) + 2 NH3 (aq) = Ag(NH3)2

+ (aq) Kf = 1.7e7 (find data)

0.1-y 1-2y y (think way)

y——————— = 1.7e7(1-2y)2(0.10-y)

y ——— = (1-2y)2 (0.1-y) By trial an error, y = 0.101.7e7

[Ag+] = 0.10 – y ~ 0 (cannot evaluate [Ag+])

0.10-y = 6e-9, too small, impractical!

5.9e-8y = 0.1 –0.8y –2y2


Concentration of ions in presence of ligands

Evaluate [Ag+] in a solution containing 1.0 M NH3 and 0.1 M AgNO3.

Solution:Ag+ (aq) + 2 NH3 (aq) = Ag(NH3)2

+ (aq) Kf = 1.7e7 (find data)

x 0.8+2x 0.1-x (think way)

0.1-x (treated as dissociation)————— = 1.7e7(0.8+2x)2 x

0.1 - x ~ 0.1; 0.8+2x ~ 0.8 (x is very small) then

0.1 0.1——— = 1.7e7 x = ————— = 9.2e-9 = [Ag+] 0.82 x 0.82 *1.7e7

3.4e7 x2 + 1.7e7 x – 0.1 = 0x =[ -1.7e7+{(1.7e7)2 + 4*0.1}] / 2 = 0 (cannot solve)

Small indeed


Complex and precipitate formation-1

What is the maximum [Cl-] before AgCl(s) forms in a solution containing 1.0 M NH3 and 0.1 M AgNO3?

Solution: Previous slide showed [Ag+] = 9.2e-9 M in a solution when [NH3] = 1.0 MKsp = 1.8e-10 for AgCl (know where to look up)

Thus, the max. [Cl-] = 1.8e-10 / 9.2e-9 = 0.02 MWhen no NH3 is present, max. [Cl-] = 1.8e-10/0.1 = 1.8e-9

What is the maximum [Br-] before AgBr(s) forms in a solution containing 1.0 M NH3 and 0.1 M AgNO3?

Ksp = 5e-13 for AgBr,Thus, the max. [Br-] = 5e-13 / 9.2e-9 = 5.4e-5 M.

0.02-------- = 368 times5.4e-5


Complex and precipitate formation-2What is the maximum [Br-] in a solution suppose to containing 0.10 M AgNO3 and 0.20 M Na2S2O3 before AgBr(s) (Ksp = 5e-13) forms?

Solution: Kf = 2.9e13 for Ag(S2O3)23–;

Ag(S2O3)23– = Ag+ + 2 S2O3

2– , K = 1/2.9e13 = 3.4e-140.1-x x 2x

4x3 ——— = 3.4e-14; x = (3.4e-14*0.1)1 / 3 = 1.5e-5 = [Ag+] (0.1–x) small

max [Br–] = Ksp/[Ag+] = 5e-13/1.5e-5 = 3.3e-8 (very low)

What is [Br–] = ? If [Na2S2O3] is increased to 1.0 M? See next page


Complex and precipitate formation-3What is the maximum [Br-] in a solution suppose to containing 0.10 M AgNO3 and 1.0 M Na2S2O3 before AgBr(s) (Ksp = 5e-13) forms?

Solution: Kf = 2.9e13 for Ag(S2O3)23– K = 1/2.9e13 = 3.4e-14

Ag(S2O3)23– = Ag+ + 2 S2O3

2–

0.1-x x 2x+0.8

(0.8+2x)2x ———— = 3.4e-14, x = 3.4e-14*0.1/0.82 = 5.3e-14 = [Ag+] (0.1-x) small

max [Br –] = 5e-13/5.3e-14 = 94 M, unrealistically large

What is the maximum [I – ] in a solution containing 0.10 M AgNO3 and 1.0 M Na2S2O3 before AgI(s) (Ksp = 8e-17) forms?

[I-] = 8e-17 / 5.3e-14 = 0.0015 M; small compare to 0.1 M


Chemical casseroleExperiment

AgNO3 solution | Cl –

AgCl (s) | 6.0 M NH3

Ag(NH3)2+ + Cl –

| Br–

AgBr (s) + NH3, Cl –

| S2O32–

Ag(S2O3)23– + Br – + NH3, Cl–

| I–

AgI (s)

Data

Solid Ksp

AgCl 1.8e-10AgBr 5.0e-13

AgI 8.0e-17

Complex Kf

Ag(NH3)2+ 1.7e7

Ag(S2O3 ) 23–

2.9e13

white

ivory

brown

June 29

19-9


Some complexes

Ligands (Lewis bases):

Negative ions, F-, Cl-, Br-, I-, OH-, CN-, SO42-, RCOO-, …

Neutral molecules: H2O, NH3, NR3, ROR, CO, C5H5, PR3, C5H5N, …Multidentate ligands: H2NCH2CH2NH2, H2NCH2CH2NHCH2CH2NH2,

Ag(NH3)2+,

Cu(CN)32- (Cu(I)), Cu(PPh3)2Br,

(NH3)2PtCl2 (cis & trans), Ni(CN)5

3-,


The heme and hemoglobin

(a) The structure of heme is a planar porphoryin ring with iron at the center. (b) Four heme units and four coiled polypeptide chains are bonded together in a molecule of hemoglobin.


Amphoteric hydroxidesAmphoteric metal hydroxides react with both acids and bases.

Examples: (M = Fe, Zn)M(H2O)4(OH)2 + H+ = M(H2O)5(OH)+

M(H2O)5(OH)+ + H+ = M(H2O)62+

M(H2O)4(OH)2 + OH– = M(H2O)3(OH)3 –

M(H2O)3(OH)3 – + OH– = M(H2O)2(OH)4

2 –

Aluminum hydroxide behave similarly:Al(H2O)3(OH)3 + H+ = Al(H2O)4(OH)2

+

Al(H2O)4(OH)2+ + H+ = Al(H2O)5(OH)2+

Al(H2O)5(OH)2+ + H+ = Al(H2O)63+

Al(H2O)3(OH)3 + OH – = Al(H2O)2(OH)4 –

Al(H2O)2(OH)4– + OH – = Al(H2O)(OH)5

2–

Al(H2O)(OH)52– + OH – = Al(OH)6

3 –


Qualitative analysis of metal ions

Mixture of metal ions add HCl

group I Ag+, Hg22+, Pb2+

filtrate of soluble chloride chloride | add 0.3 M H+ & H2S

group II Cu2+, Cd2+, Hg2+, Pb2+ filtrate of soluble metal sulfide As3+, Sb3+, Bi3+, Sn4+ | | add OH- & H2S

acid insoluble sulfide | (NH4OH) group III Mn2+, Fe2+, Co2+, Ni2+, Zn2+ filtrate of soluble metal ions base insoluble sulfide | | add CO3

2- or PO43-

Al(OH)3, Cr (OH)3 | |hydroxide |

group IV Mg2+, Ca2+, Sr2+, Ba2+ filtrate of soluble metal ions ppt as carbonates or phosphates | K+, Na+ group V


Heterogeneous and complex ion equilibria Summary

Calculate Ksp

Evaluate molar solubility (or in other units) from Ksp

Discuss concentrations of species in solution with two or more electrolytes (common ion effect, pH effect, separation by ppt)

Predicts ppt (heterogeneous equilibria)

Describe ligands, metal ions, complexes (ions), and formation constants, and dissociation constant

Apply complex formation to explain solubility

Be able to get out of traps (think in both directions)

19 Heterogeneous & complex equilibria 1 19 Heterogeneous and complex equilibria CaCO 3 exists as...

Documents

Transcript of 19 Heterogeneous & complex equilibria 1 19 Heterogeneous and complex equilibria CaCO 3 exists as...