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One Point Quiz

One quiz per table, list everyone’s name Agree on an answer You have two minutes

Theory of PrecipitationEdward A. Mottel

Department of Chemistry

Rose-Hulman Institute of Technology

Problem Set

Hydroxide Precipitation Reactions

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Ag+(aq) + OH–(aq)

Equilibrium Constants andSolubility Product

Ag2O(s)

What is the solubility product for silver oxide?

22 + H2O(l)

Silver ion reacts with hydroxide ionto form a brown-black oxide.

chocolatebrown

The solubility product equation correspondsto the dissolution of the solid.

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Solubility Product

2 Ag+(aq) + 2 OH–(aq) Ag2O(s) + H2O(l)

Kc =products

reactants

[Ag+]2 [OH–]2

Kc · [Ag2O(s)] [H2O(l)] = [Ag+]2 [OH–]2

[Ag2O(s)] [H2O(l)]solids areincluded in

the constant

solvent isincluded in

the constant

Ksolubility product

Ksp

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Why is the numeric value of the solubility product important?

A precipitate will form from ions whenthe ion product exceeds the solubility product.

2 Ag+(aq) + 2 OH–(aq) Ag2O(s) + H2O(l)

Ksp = [Ag+]2 [OH–]2

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If a room temperature aqueous solution has a pH of 1.0,

what is the concentration of the hydrogen ion and the hydroxide ion?

more

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Iron(III) Hydroxide

Write a chemical reactionfor the formation of this solid.

Fe(OH)3 has a solubility product (Ksp) of 6.0 x 10–38 M4.

Will a 0.10 M Fe3+ solution precipitate asiron(III) hydroxide at a pH of 1.0?

Write the solubility product equation for Fe(OH)3.

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< Ksp, Fe(OH)3

Will 0.10 M Iron(III) IonPrecipitate as a Hydroxide at pH =

1.0?

Q = [Fe3+] [OH–]3

Fe(OH)3 will not precipitate because itsion product is less than the solubility product.

= (0.10 M)(1 x 10–13 M)3

= 1 x 10–40 M4 = 6.0 x 10–38 M4

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[H+] =

[OH–]

10–pH = 1 x 10–3 M

= 1 x 10–11 M

Will 0.10 M Iron(III) IonPrecipitate as a Hydroxide at pH =

3.0?

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Iron(III) Hydroxide at pH = 3.0

Fe(OH)3 will precipitate because itsion product exceeds the solubility product.

> Ksp, Fe(OH)3

Q = [Fe3+] [OH–]3 = (0.10 M)(1 x 10–11 M)3

= 1 x 10–34 M4 = 6.0 x 10–38 M4

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Solubility Products of Hydroxidesat 25 °C

Ksp

Cd(OH)2

Co(OH)2

Cr(OH)3

Cu(OH)2

Fe(OH)3

Mn(OH)2

Ni(OH)2

Pb(OH)2

2.0 x 10–14

2.5 x 10–16

6.7 x 10–31

1.6 x 10–19

6.0 x 10–38

2.0 x 10–13

1.6 x 10–16

4.2 x 10–15

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Selective Precipitation

A solutioncontains0.10 M Fe3+

and 0.10 M Ni2+

If the solution isslowly made basic,

what will be thecomposition of thefirst solid that forms?

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Selective Precipitation

6 M NaOH, dropwise

Fe3+, Ni2+

Method: Determine the hydroxide ion concentrationat which each ion will begin to precipitate.

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At what [OH–] will 0.1 M Fe3+ begin to form iron(III) hydroxide?

[Fe3+] [OH–]3 = Ksp

[OH–] = ( Ksp

[Fe3+] ) 1/3

= (6.0 x 10–38 M4

1.0 x 10–1 M )

1/3

= 8.4 x 10–13 M

Fe(OH)3 will begin to form when itsion product exceeds its solubility product.

0.1 M Fe3+

begins to precipitatewhen OH– exceeds this

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At what [OH–] will 0.1 M Ni2+ begin to form nickel(II) hydroxide?

[Ni2+] [OH–]2 = Ksp

[OH–] = ( Ksp

[Ni2+] ) 1/2

= (1.6 x 10–16 M3

1.0 x 10–1 M )

1/2

Ni(OH)2 will begin to form when its ionproduct exceeds its solubility product.

= 4.0 x 10–8 M0.1 M Ni2+

begins to precipitatewhen OH– exceeds this

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Analysis

Fe(OH)3 begins to form at an hydroxide ion concentration of 8.4 x 10–13 M.

Ni(OH)2 begins to form at an hydroxide ion concentration of 4.0 x 10–8 M.

At what pH will each solid form?

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Selective Precipitation

14

10

6

2

pH

0.1 M Fe3+ forms Fe(OH)3 (pH = 1.9)

0.1 M Ni2+ forms Ni(OH)2 (pH = 6.6)

If the solution is slowly made basic,which solid forms first?

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Selective Precipitation

6 M NaOH, dropwise

Fe3+, Ni2+

dropwise, because we want the ions to precipitate

selectively

What is the formula of the ionwhich precipitatesfirst?

Fe(OH)3

red-brownNi2+

pale-green

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99.9% Complete Separation

Determine the hydroxide ion concentration and pHat which 99.9% of the iron(III) ion has precipitated.

In what pH range will 99.9%of the iron(III) ion precipitate

and the nickel(II) ion remain in solution?

Determine the concentration of iron(III) ionwhen 99.9% of it is precipitated.

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99.9% Complete Separation

[Fe3+] [OH–]3 = Ksp

[Fe3+] = 0.00010 M

[OH–] = (6.0 x 10–38 M4

1.0 x 10–4 M )

1/3

= 8.4 x 10–12 M

1.2 x 10–3 M

pH = 2.9

What will be the iron(III) ionconcentration if it is 99.9%

precipitated?

It started 0.10 M

[H+] =

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Selective Precipitation

0.0001 M Fe3+ forms Fe(OH)3 (pH = 2.9)

14

10

6

2

pH

0.1 M Fe3+ forms Fe(OH)3 (pH = 1.9)

0.1 M Ni2+ forms Ni(OH)2 (pH = 6.6)

If the solution is adjusted tothis pH range, more than 99.9%

of the iron(III) ion should precipitate and none of the

nickel(II) ion

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Solubility Rules

All nitrate (NO3–) salts are soluble and

highly conducting.

Ti3+(aq) + 3 NO3–(aq) Ti(NO3)3(s)

All Group I (Li+, Na+, K+, Rb+, Cs+) and ammonium salts are soluble and highly conducting.

2 Na+(aq) + SO42–(aq) Na2SO4(s)

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Solubility Rules

All chloride and bromide (Cl–, Br–) salts are soluble and highly conducting except salts of Ag+, Pb2+, Hg2

2+ and Tl+.

Cu2+(aq) + 2 Cl–(aq) CuCl2(s)

Ag+(aq) + Cl–(aq) AgCl(s)

All sulfate (SO42–) salts are soluble and highly

conducting except salts of Pb2+, Ba2+, Ca2+ and Sr2+.

Fe2+(aq) + SO42–(aq) FeSO4(s)

Ba2+(aq) + SO42–(aq) BaSO4(s)

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Solubility Rules

In almost all cases, when a salt dissolves and dissociates in solution, only one type of cation (positive ion) and one type of anion (negative ion) are formed.

N2H5+(aq) + NO3

–(aq) N2H5NO3(s)

hydrazinium ion

Sulfides

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Soluble Sulfides

Na2S 15.4 (10 °C) 1.97 M

(NH4)2S very soluble

5460 m H2S 18600 (40 °C)0.34 (100 °C) 0.10 M

g/100 g H2Omolar

solubility

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Insoluble Sulfides

Ag2SCdSCoSCuSFeSMnSNiSPbS

g/100 g H2O

6.4 x 10-16

1.4 x 10-13

2.0 x 10-10

2.8 x 10 16

1.9 x 10-8

6.2 x 10-7

5.0 x 10-10

4.9 x 10-13

molarsolubility

2.6 x 10-17

1.0 x 10-14

2.2 x 10-11

2.9 x 10-18

2.2 x 10-9

7.1 x 10-8

5.5 x 10-11

2.0 x 10-14

Ksp

6.8 x 10-50

1.0 x 10-28

5.0 x 10-22

8.7 x 10-36

4.9 x 10-18

5.1 x 10-15

3.0 x 10-21

4.2 x 10-28

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Preparation of Hydrogen Sulfide(acidic solution)

2 H+(aq) + S2–(aq) H2S(aq)

HH

CH

H

H

C N

S

+ H2O(l) H2S(aq)

thioacetamideAs an aqueous solution H2Sis called hydrosulfuric acid

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Preparation of Hydrogen Sulfide(basic solution)

H2S(aq) + 2 OH–(aq) S2–(aq) + 2 H2O(l)

0.1 M Na2S [S2–] = 5.0 x 10–2 M

0.1 M (NH4)2S [S2–] = 2.0 x 10–5 M

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Reactions of Hydrosulfuric Acid(polyprotic acid)

2 H+(aq) + S2–(aq) H2S(aq)

H+(aq) + HS–(aq) H2S(aq)

H+(aq) + S2–(aq) HS–(aq)

hydrosulfuric acid bisulfide ion

sulfide ion

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First Dissociation of H2S

H+(aq) + HS–(aq) H2S(aq)

[H+] [HS–]

[H2S]= K1 = 1.0 x 10–7 M

1 indicates it is theionization of the

first hydrogen ionAlso called Ka,1

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H+(aq) + S2–(aq) HS–(aq)

Second Dissociation of H2S

[H+] [S2–]

[HS–]= K2 = 1.3 x 10–13 M

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2 H+(aq) + S2–(aq) H2S(aq)

Double Dissociation of H2S

[H+]2 [S2–]

[H2S]= K12 = K1K2 =1.3 x 10–20 M2

This equilibrium constantcan be used for any sulfide ion

and pH calculation

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Dissociation of Hydrosulfuric Acid

H+(aq) + HS–(aq) H2S(aq) 2 H+(aq) + S2–(aq)

very acidic neutral very basic

The solubility of H2S in water decreaseswith increasing temperature.

The amount of sulfide ion in solution ispH dependent.

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What would a graph of the various sulfide containing species versus pH

look like?

pH

Mol

e F

ract

ion

acidic basic

H2S

HS–

S2–

ExcelGraph

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