5&6 analisis kation

Analisis Kation

19-2

Figure 19.16

The general procedure for separating ions in qualitative analysis

Add precipitating

ionC

entr

ifuge

Add precipitating

ion

Cen

trifu

ge

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

19-3

A qualitative analysis scheme for separating cations into five ion groups

Add 6M HCl

Cen

trifu

geAcidify to pH 0.5;

add H2S

Cen

trifu

ge

Add NH3/NH4

+ buffer(pH 8)

Cen

trifu

ge

Add (NH4)2HPO4

Cen

trifu

ge


19-4

Figure 19.17 A qualitative analysis scheme for separating cations into five ion groups

Add 6M HCl

Cen

trifu

geAcidify to pH 0.5;

add H2S

Cen

trifu

ge

Add NH3/NH4

+ buffer(pH 8)

Cen

trifu

ge

Add (NH4)2HPO4

Cen

trifu

ge


19-5

Using pH and complexation to Separate IonsFor Qualitative Analysis

19-6

Hot water

Pb2+AgCl(s); Hg2Cl2(s)

Pb C rO PbC rOaqueous aqueous so lid2

42

4

, ,

ppt

CrO aqueous42

,

19-7

NH3

HgNH2Cl(s)

Ag(NH3)6+

Hot water

ppt

CrO aqueous42

,

Pb2+AgCl(s); Hg2Cl2(s)

Ag NH AgNH

AgNH NH Ag NH

aqueous aqueous aqueous

aqueous aqueous aqueous

3 3

3 3 3 2

19-8

Figure 19.18Step 1

Add NH3(aq)

Cen

trifu

ge

Cen

trifu

ge

Step 2 Add HCl

Step 3 Add NaOH

Cen

trifu

geStep 4

Add HCl, Na2HPO4

Step 5 Dissolve in

HCl and add KSCN

A qualitative analysis scheme for Ag+,Al3+,Cu2+, and Fe3+


Valdosta State University

Background


General Unknown

Group IIAcid Insoluble Metal Sulfides

Group IIIAlkaline Insoluble Metal Sulfides and Hydroxides

Group IVSoluble Metal

Ions

Group IInsoluble Metal Chlorides

and Ammonia

HCl

HCl / H2S

NH3 / H2S


Background


• For this experiment, the group III ions are Fe3+, Ni2+, Mn2+, Al3+ and Zn2+.

• These ions initially precipitate as either metal sulfides (in an alkaline environment) or metal hydroxides.

• This requires the chemist to generate a small quantity of sulfide ion to precipitate the metals.

• A convenient source of S2- is thioacetamide, which decomposes when heated to give hydrogen sulfide (H2S) which yields S2- in chemical reactions.

• A reagent that is made and consumed in the same flask is said to be produced in situ.


Background – Hydrogen Sulfide


CC

N

SH

HH

H

H

CC

O

OH

HH

H

+ 2 H2O + H+(aq) + NH4+(aq) + H2S(g)

H2S(aq) + 2 H2O(l) 2 H3O+(aq) + S2-(aq)

• The addition of base to the second reaction consumes the hydronium ion and drives the reaction to the right, increasing the concentration of S2-(aq).

Thioacetamide


Background – Group III Separation Scheme


Group III Unknown

NiS, FeS, MnS, ZnS, Fe(OH)3, Al(OH)3

Group IV ions

Ni2+, Fe3+, Mn2+, Zn2+, Al3+

Waste

Al(OH)4-, Zn(OH)4

2-Fe(OH)3, Ni(OH)2, Mn(OH)2

MnO4-

purple

Fe(OH)3

Divide sample

Ni(NH3)62+

Ni(DMG)2

strawberry red ppt.

Fe(SCN)63-

blood red

Al(OH)3 Zn(NH3)42+

Al(OH)3aluminoncherry red ppt.

K2Zn3[Fe(CN)6]2white ppt.

NH3, H2S, heat

HCl, HNO3, heat

6 M NaOH

NaBiO3

Conc.NH3

HCl / NH4SCN H2DMG

HNO3

NH3

aluminon, NH3

K4Fe(CN)6

HNO3




Group III unknown

NiS, FeS, MnS, Fe(OH)3, Al(OH)3, ZnS

Group IV ions

NH3, H2S, Heat

A – Preparation of Group III cations

• The group III ions are initially separated from the bulk solution by precipitation as either insoluble metal sulfides or hydroxides.

Ni2+(aq) + S2(aq) NiS(s) (black) Fe2+(aq) + S2(aq) FeS(s) (black) Zn2+(aq) + S2(aq) ZnS(s) (white) Mn2+(aq) + S2(aq) MnS(s) (pink) Al3+(aq) + 3 OH(aq) Al(OH)3(s) (white, gel)




Group III unknown


Group IV ions

NH3, H2S, Heat


• Since iron has two common oxidation states, its chemistry in this step is more complex.

• If iron(III) is present it is reduced to iron(II) and elemental sulfur in produced.

2 Fe3+(aq) + H2S (aq) 2 Fe2+(aq) + S(s) + 2 H+(aq)




Group III unknown


Group IV ions

NH3, H2S, Heat


• Alternately, the iron(III) can combine with the hydroxide ion and precipitate as iron(III) hydroxide.

Fe3+(aq) + 3 OH-(aq) Fe(OH)3 (rust color)




NiS, FeS, MnS, Fe(OH)3, Al(OH)3

Ni2+, Fe3+, Mn2+, Zn2+, Al3+Waste

HCl, HNO3, Heat

B1 – Separation of Group III cations

• Following the precipitation, the metal ions are combined with acid to form the free (and soluble) metal ions.

3NiS(s) +8H+(aq) + 2NO3(aq) 3Ni2+(aq) + 2NO(g) + 3S(s) + 4H2O(l)

FeS(s) + 2 H+ (aq) Fe2+(aq) + H2S(aq)

3Fe2+(aq) + 4H+(aq) + NO3(aq) 3Fe3+(aq) + NO(g) + 2H2O(l)

MnS(s) + 2 H+(aq) Mn2+(aq) + H2S(aq)

ZnS(s) + 2 H+(aq) Zn2+(aq) + H2S(aq)

Al(OH)3(s) + 3 H+(aq) Al3+(aq) + H2O(l)




Ni2+, Fe3+, Mn2+, Zn2+, Al3+

Al(OH)4-, Zn(OH)4


6 M NaOH


• Aluminum and zinc ions are amphoteric.• This means that at high acid or base

concentrations, these metals form soluble complexes, but precipitate at moderate pH.

• Iron, manganese and nickel form insoluble hydroxides at high pH.

Fe3+(aq) + 3 OH(aq) Fe(OH)3(s) (rust-color)

Ni2+(aq) + 2 OH(aq) Ni(OH)2(s) (green)

Mn2+(aq) + 2 OH(aq) Mn(OH)2(s) (light brown)




Ni2+, Fe3+, Mn2+, Zn2+, Al3+

Al(OH)4-, Zn(OH)4


6 M NaOH


• Aluminum and zinc ions are amphoteric.• This means that at high acid or base

concentrations, these metals form soluble complexes, but precipitate at moderate pH.

• Iron, manganese and nickel form insoluble hydroxides at high pH.

Al3+(aq) + 3 OH(aq) Al(OH)3(s) (white, gelatinous)

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




Ni2+, Fe3+, Mn2+, Zn2+, Al3+

Al(OH)4-, Zn(OH)4


6 M NaOH


Excess Acid

Al(OH)3(s) + 3H+(aq) Al3+ + 3 H2O(aq)

Zn(OH)2(s) + 2H+(aq) Zn2+ + 2 H2O(aq)




Ni2+, Fe3+, Mn2+, Zn2+, Al3+

Al(OH)4-, Zn(OH)4


6 M NaOH


Excess Base

Al(OH)3(s) + OH(aq) Al(OH)4-(aq)

Zn(OH)2(s) + 2OH(aq) Zn(OH)42-(aq)




Fe(OH)3, Ni(OH)2, Mn(OH)2

Fe3+, Ni2+, Mn2+

C1 – Test for Mn2+, Fe3+, Ni2+

HNO3, KNO2• The precipitate is redissolved by adding

acid to the precipitate.• The addition of nitric acid neutralizes the

sodium hydroxide and regenerates the free cations.

• There is no easy method which will allow Mn2+, Fe3+ and Ni2+ to be separated; therefore, the sample is divided.

DivideSample




DivideSample

MnO4-

purple

NaBiO3

C2 – Test for Mn2+

• If sodium bismuthate is added to a solution containing manganese(II), a redox reaction occurs resulting in the formation of the purple permanganate ion.

14H+(aq) + 2Mn2+(aq) + 5BiO3-(s) 2 MnO4

-(aq) + 5Bi3+(aq) + 7H2O(l)

Fe3+, Ni2+, Mn2+




DivideSample

D1 – Separation of Fe3+ and Ni2+

Fe(OH)3 Ni(NH3)62+

Conc. NH3

• The nickel and iron ions can be separated by the addition of ammonia.

• The increased pH causes the formation of the insoluble iron(III) hydroxide.

• The nickel ion combines with ammonia to form a soluble complex ion, hexaamminenickel(II).

Fe3+(aq) + 3NH3(aq) + 3H2O(l) 3NH4+(aq) + Fe(OH)3(s) (brown)

Ni2+(aq) + 6NH3(aq) Ni(NH3)6

2+(aq) (blue)

Fe3+, Ni2+, Mn2+




DivideSample

D2 – Test for Fe3+

Fe(OH)3 Ni(NH3)62+

Conc. NH3

Fe(SCN)63-

blood red

HCl / NH4SCN

• The presence of the iron(III) ion is confirmed by the addition of ammonium thiocyanate.

• If iron(III) is present, a blood red solution forms.

Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq) blood red

Fe3+, Ni2+, Mn2+




DivideSample

E – Test for Ni2+

Fe(OH)3 Ni(NH3)62+

Conc. NH3

Fe(SCN)63-

blood red

HCl / NH4SCN

Ni(DMG)2

strawberry red ppt.

H2DMG

• The presence of the nickel ion is confirmed by the addition of dimethylglyoxime.

• Dimethylglyoxime combines with the nickel ion to form a complex which forms a strawberry red precipiate.

Ni(NH3)62+(aq) + 2 HC4H7N2O2(aq) 4NH3(aq) + 2NH4

+(aq) + Ni(C4H7N2O2)2(s) (red)

Fe3+, Ni2+, Mn2+




Al(OH)4-, Zn(OH)4

2-F1 – Separation of Al3+ and Zn2+

Al(OH)3 Zn(NH3)42+

NH3

HNO3

• Careful control of pH allows for the separation of aluminum and zinc ions.

• The solution is made very slightly basic.• At these conditions, the aluminum ion

precipitates as aluminum hydroxide.• The zinc ion remains in solution.

Al3+(aq) + 3 NH3(aq) + 3 H2O(l) 3 NH4+(aq) + Al(OH)3(s)

Zn2+(aq) + 4 NH3(aq) Zn(NH3)42+(aq)




Al(OH)4-, Zn(OH)4

2-

F2 – Test for Al3+

Al(OH)3 Zn(NH3)42+

NH3

HNO3

Al(OH)3 aluminoncherry red ppt.

NH3, aluminon

• A successful test for aluminum requires that the previous reactions and their pH control were properly performed.

• If not, false positive tests result.• The test for aluminum requires the free

aluminum ion to react with ammonia in the presence of a reagent called aluminon and form a red precipitate.

• Be careful, if there is iron or zinc left in the sample, a red precipitate will form resulting in a false positive.

Al3+(aq) + 3 NH3(aq) + 3 H2O + aluminon(aq) 3 NH4+(aq) + Al(OH)3aluminon(s) (red)




Al(OH)4-, Zn(OH)4

2-

F2 – Test for Al3+

Al(OH)3 Zn(NH3)42+

NH3

HNO3


NH3, aluminon

• To confirm that the red precipitate is the aluminum complex, ammonium carbonate is added.

• If the red color does not fade, aluminum is present.




Al(OH)4-, Zn(OH)4

2-

G – Test for Zn2+

Al(OH)3 Zn(NH3)42+

NH3

HNO3


NH3, aluminon

K2Zn3[Fe(CN)6]2

white ppt.

K4Fe(CN)6

• To test for the zinc ion, a solution of potassium hexacyanoferrate(II) is added to the test solution.

• If zinc is present a white precipitate forms.

• The exact color of the precipitate can vary depending on the presence of other ions.

• If iron is present the color can change to yellow, green or blue.

3 Zn2+(aq) + 2 K+(aq) + 2 Fe(CN)64 (aq) K2Zn3[Fe(CN)6]2 (s)




Zn2+ Al3+ Ni2+ Fe3+ Mn2+


Background


Group IV Unknown

BaCrO4

yellow ppt.

Flame Test Na+

orange-yellow flame

Ca2+, Mg2+

HC2H3O2, K2CrO4

Ba2+ Flame Testapple - green

BaSO4

white ppt.

6M HCl

6M H2SO4

CaC2O4

white ppt.Mg2+

MgNH4PO46H2Owhite ppt.

(NH4)2C2O4

NH3(aq), NaH2PO4, heat

Flame Testred-orange

6M HCl

Flame Test K+

lavender flame


Background – Group IV Separation Scheme


A – Flame test for Na+ and K+

• Insoluble salts of sodium and potassium are not known.

• One method of determining the presence of these ions is the flame test.

Group IV Unknown

Flame Test Na+

orange-yellow flameFlame Test K+

lavender flame

Na K




B – Test for Ba2+

• The formation of a yellow precipitate on the addition of potassium chromate indicates the presence of the barium ion.

HC2H3O2,K2CrO4

BaCrO4

yellow ppt.

Ba2+

6M HCl

Flame Testapple - green

BaSO4

white ppt.

6M H2SO4

Ba2+(aq) + K2CrO4(aq) BaCrO4(s) + 2K+(aq)




B – Test for Ba2+

• A flame test (apple green) is used to confirm the presence of the ion.

HC2H3O2,K2CrO4

BaCrO4

yellow ppt.

Ba2+

6M HCl


BaSO4

white ppt.

6M H2SO4




B – Test for Ba2+

• The final test for barium is the formation of a white precipitate on the addition of a sulfate to the solution.

HC2H3O2,K2CrO4

BaCrO4

yellow ppt.

Ba2+

6M HCl


BaSO4

white ppt.

6M H2SO4

Ba2+(aq) + H2SO4(aq) BaSO4(s) + 2H+(aq)




C – Test for Ca2+

• The calcium ion is separated from magnesium by precipitating calcium oxalate.

Ca2+, Mg2+

CaC2O4

(NH4)2C2O4

Mg2+

Flame testred-orange

6M HCl

Ca2+(aq) + (NH4)2C2O4(aq) CaC2O4(s) + 2NH4

+(aq)




C – Test for Ca2+

• A flame test (red-orange flame) is used to confirm the presence of the calcium ion.

Ca2+, Mg2+

CaC2O4

(NH4)2C2O4

Mg2+


6M HCl




C – Test for Mg2+

• The magnesium ion is precipitated as a white solid (ammonium phosphate) in an alkaline solution.

Ca2+, Mg2+

CaC2O4

K2C2O4

Mg2+


6M HCl

Mg2+(aq) + NH3(aq) + HPO42-(aq) MgNH4PO4(s)

MgNH4PO4–6H2Owhite ppt.

NH3(aq)Na2HPO4(aq)

Sample Problem 19.12 Separating Ions by Selective Precipitation

SOLUTION:

PROBLEM: A solution consists of 0.20M MgCl2 and 0.10M CuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20.

PLAN: Both precipitates are of the same ion ratio, 1:2, so we can compare their Ksp

values to determine which has the greater solubility.

It is obvious that Cu(OH)2 will precipitate first so we calculate the [OH-] needed for a saturated solution of Mg(OH)2. This should ensure that we do not precipitate Mg(OH)2. Then we can check how much Cu2+ remains in solution.

Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = 6.3x10-10

Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = 2.2x10-20

[OH-] needed for a saturated Mg(OH)2 solution =

Ksp

[Mg2]

6.3x10 10

0.20= 5.6x10-5M

5&6 analisis kation

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