salt analysis

The Chemistry of the Qualitative Analysis Scheme

The Analysis of Group I Precipitation of Group I

Hydrochloric acid is the reagent which serves as the precipitant for the metal ions of Group I. The chlorides of mercury (I), silver, and lead are the only ones of the cations under consideration in this qualitative scheme which are insoluble in acid solutions. The equations for the precipitation of the metal ions of Group I are

Hg2++ + 2 Cl- ⇔ Hg2Cl2

Ag+ + Cl- ⇔ AgCl Pb++ + 2 Cl- ⇔ PbCl2

The solubility products and colors of the chlorides of this group are given in Table QA.1. These values show that the solubility product of lead chloride is about 94 000 times greater than that of silver chloride and about 8 x 1010 times greater than that of mercury (I) chloride. Thus, it can be seen that the mercury (I) ion and the silver ion are in effect completely removed from solution when hydrochloric acid is added to a solution containing these ions. On the other hand, lead is not completely precipitated by chloride, and some of it is carried through to Group II.

A slight excess of hydrochloric acid is used in the precipitation of the group to prevent the precipitation of bismuth oxychloride by hydrolysis of its chloride salt.

Bi3+ + Cl- + H2O ⇔ BiOCl + 2 H+ The hydrogen ions of the hydrochloric acid keep the above equilibria shifted to the left. In addition, a slight excess of chloride ions insures more complete precipitation of Hg2Cl2, AgCl, PbC12, which is due to the common ion effect (Zumdahl, 708). For example, in a solution which is 0.3 M in chloride ions, the common ion effect will reduce the concentration of Ag+ from 1.1 x 10-5 M in a saturated solution of AgCl to 4 x 10-10 M. On the other hand, a large excess of chloride ion must be avoided because silver chloride and lead chloride tend to react with chloride ions, forming soluble salts of complex ions. The equations are

AgCl + Cl- ⇔ AgCl2-

PbC12 + 2 Cl- ⇔ PbC14-

Separation and Identification of the Lead Ion

Lead chloride may fail to precipitate at all if the lead ion concentration is too low or if the temperature is too high. For this reason a test for lead should be made on the centrifugate from the original Group I separation.

The fact that lead chloride is about three times as soluble in hot water as in cold water enables one to separate it from silver and mercury (I) chlorides by extraction of the mixed chlorides with hot water. The presence of the lead ion may be confirmed by precipitation of the slightly soluble, yellow chromate. The equation is

Pb++ + CrO42 ⇔ PbCrO4

Separation and Identification of the Mercury (I) Ion

When aqueous ammonia is added to a mixture of silver chloride and mercury (I) chloride, the silver chloride

dissolves by forming soluble diamminesilver chloride. The equation is

-1- 16 Cation Analysis

AgCl + 2 NH3 ⇔ Ag(NH3)2

+ + Cl- In the presence of ammonia, mercury (I) chloride undergoes auto-oxidation reduction with the formation of finely divided metallic mercury, which is black, and mercury (II) amido chloride, HgNH2CI, which is white. The equation is

Hg2C12 + 2NH3 ⇔ Hg + HgNH2Cl +NH4

+ + Cl- The formation of this black (or gray) precipitate is sufficient proof of the presence of mercury (I) ions.

Identification of the Silver Ion

The presence of the silver ion may be confirmed by treating the aqueous ammonia extract of Hg2Cl2 and AgCl with nitric acid. This causes the reprecipitation of white silver chloride. The hydrogen ion supplied by the nitric acid unites with the free ammonia in equilibrium with the diamminesilver ion and causes the equilibrium, Ag(NH3)2

+ ⇔ Ag+ + 2 NH3, to shift to the right. The free silver ion is then precipitated by the chloride ion which is present in the solution. The net reaction is the sum of three equilibria, as shown below.

Ag(NH3)2+ ⇔ Ag+ + 2 NH3

2 NH3 + 2 H+ ⇔ 2 NH4+

Ag+ + Cl- ⇔ AgCl

Ag(NH3)2+ + Cl-+ 2 H+ ⇔ AgCl + 2 NH4

+

The Analysis of Group II

The solution to be analyzed may be a Group II known or unknown, or it may be the solution from the Group I separation. In either case proceed according to (1) below. See page 478 for the Group II flow sheet.

Precipitation of Group II Sulfides

1. Conditions for the separation of Group II from Group III. Ions of lead, bismuth, copper, and tin form

sulfides which are insoluble in solutions which are 0.3 M in hydrogen ion. Lead sulfide is the most soluble of the sulfides of Group II and zinc sulfide is the least soluble of the sulfides of Group III. Therefore, the separation of the sulfides of Group II from those of Group III is assured if the sulfide ion concentration of the solution is controlled in such a way that the lead is completely precipitated as the sulfide without exceeding the solubility product of zinc sulfide. In a solution which is saturated with hydrogen sulfide and which is 0.3 M in hydrogen ion, the sulfide ion concentration is just right to effect the separation of lead from zinc and, hence, Group II from Group III.

The hydrogen sulfide required for the precipitation of the Group II cations is supplied by thioacetamide, CH3CSNH2, which hydrolyzes in hot acidic solutions with the formation of ammonium acetate and hydrogen sulfide.

CH3CSNH2 + 2 H2O ⇔ + CH3COO- + NH4+ + H2S

The hydrogen sulfide formed ionizes in aqueous solution according to the equation

H2S ⇔ 2 H+ + S2-


Hydrochloric acid furnishes hydrogen ions which repress the ionization of hydrogen sulfide, and this causes a decrease in the sulfide ion concentration of the solution. (The ionization of hydrogen sulfide and its control is discussed in detail beginning on page 745 of the text.) The precipitation of the Group II sulfides may be represented by the following equations:

Pb++ + H2S ⇔ PbS + 2 H+ 2 Bi3+ + 3 H2S ⇔ Bi2S3 + 6 H+

Cu2+ + H2S ⇔ CuS + 2 H+ Sn4+ +H2S ⇔ SnS2 + 4 H+

It should be noted that the acidity of the solution is increased by such reactions since the hydrogen ion is one of the products of the sulfide precipitation reactions.

2. Mechanism of the precipitation of the sulfides. The concentration of the sulfide ion in a solution saturated

with hydrogen sulfide and 0.3 M in hydrogen ions is 1.2 x 10-22 mole per liter (Chapter 15). Because there are 6.023 x 1023 ions in one mole, only about 70 sulfide ions are present in one liter of solution. Yet, practically all the Pb2+ is precipitated immediately when hydrogen sulfide is passed into an acidic solution of a lead salt. The precipitation is so rapid that it is inconceivable that the reaction is one of simple union of lead and sulfide ions (Pb2+ + S- ⇔ PbS) when the sulfide ion is present in such a small concentration. However, the concentration of the hydrosulfide ion, HS-, in a saturated solution of hydrogen sulfide containing 0.3 M hydrogen ion is comparatively large relative to the low concentration of the sulfide ion, as seen from the following calculations:

[H+][HS-] = (0.3)[HS-] = 9 x 10-8 [H2S] 0.1

[HS-] = (0.1 x 9 x10-8) = 3 x 10-8

0.3 Thus, we find that the concentration of HS- is 3 x 10-8 mole per liter, or about 1014 times larger than that of the sulfide ion (1.2 x 10-22) . Therefore, it is reasonable to assume that an unstable intermediate hydrosulfide, Pb(HS)2, could be formed which immediately breaks down to form PbS and H2S, as shown by the following equations:

Pb2+ + 2 HS- ⇔ Pb(HS)2 ⇔ PbS + H2S Such processes are known for the formation of oxides through the precipitation of unstable hydroxides. A notable example is the formation of silver oxide according to the reaction

2 Ag+ + 2 OH- ⇔ 2 AgOH ⇔ Ag2O + H2O Equilibrium is concerned only with the final result and not with the mechanism by which the result is obtained. This means that calculations involving the precipitation of metal sulfides are valid when concentrations of the sulfide ion are employed even though this ion may not be involved directly in the mechanism for the precipitation of the metal sulfide. 3. Colors of the sulfides of Group II. The sulfides of lead (II), and copper (II), are black. The sulfides of tin (IV) is yellow, that of bismuth is dark brown or black.

Tin (II) sulfide, SnS, forms a gelatinous precipitate which does not dissolve in sodium sulfide solutions, whereas SnS2 is readily soluble in this reagent. For this reason tin (II) is oxidized to tin (IV) by adding nitric acid and heating the original solution before the sulfides of the Group II ions are precipitated. Separation of the Group II Ions into the A and B Subdivisions


The separation of the Group II ions into subdivisions is based on the fact that SnS2 forms a soluble complex

sulfides in alkaline sulfide solutions, whereas PbS, Bi2S3, and CuS, do not form similar complex ions.

The reagent used in this separation is sodium sulfide, which is made by the action of hydrogen sulfide (formed by the hydrolysis of thioacetamide) upon sodium hydroxide. The equation is

(2 Na+) + 2 OH- + H2S ⇔ (2 Na+) + S2- + 2 H2O The presence of the hydroxide ion in high concentration, supplied by an excess of sodium hydroxide, reduces the extent of hydrolysis of the sulfide ion.

S2- + H2O ⇔ HS- + OH- This provides a relatively high concentration of sulfide ion required for the dissolution of the Division B sulfides. The equation is

SnS2+ S2- ⇔ SnS32- (thiostannate)

The cations whose sulfides fail to dissolve in the presence of a high concentration of sulfide ions constitute Division A of Group II. They are Pb2+, Bi3+ and Cu2+. The Separation and Identification of Lead

The undissolved sulfides of Division A (PbS, Bi2S3, and CuS) are washed with a solution of ammonium nitrate. This is used, rather than pure water, to decrease the tendency of the sulfides to become colloidal. In the colloidal condition these sulfides would pass into the centrifugate and be lost in part. The sulfides are then brought into solution by treatment with hot dilute nitric acid. The acid oxidizes the sulfide ion to elementary sulfur, and causes the metal sulfides to dissolve by holding the sulfide ion concentration at such a low level that the product of the concentration of the ions is less than the solubility product constant. The two equilibria involved in the system, in the case of lead sulfide, are given by the equations

PbS ⇔ Pb2+ + S2-

3S2-+8H++2NO3- ⇔ 3S + 2NO + 4H2O

and the net reaction is

3 PbS + 8 H+ + 2 NO3- ⇔ 3Pb++ + 3 S + 2 NO +4H2O

Lead is separated from bismuth, and copper by precipitating the sulfate, PbSO4, which is white. The sulfates of

the other two metals are soluble. The nitric acid in the solution must first be removed because the sulfate ion is converted largely to the hydrogen sulfate ion by hydrogen ions in high concentration, SO4

2-+ H+ ⇔ HSO4-.

Under these conditions the concentration of sulfate ion is too low to allow the solubility product constant of lead sulfate to be exceeded. The removal of the nitric acid is accomplished by adding sulfuric acid and evaporating the solution to the point where sulfur trioxide fumes appear.

H2SO4 ⇔ H2O + SO3 The more volatile nitric acid will have been expelled before the SO3 fumes appear. Upon dilution of the sulfuric acid solution with water, sulfate ions are formed and the lead precipitates as the sulfate. The equations are:

H2SO4 ⇔ H+ + HSO4-

HSO4- ⇔ H+ + SO4

2- Pb++ + SO4

2 ⇔ PbSO4


The lead sulfate is then dissolved in a solution of ammonium acetate with the formation of slightly ionized lead acetate. The equation is

PbSO4 + 2 OAc- ⇔ Pb(OAc)2 + SO42-

When chromate ions are added to this acetate solution, yellow lead chromate, PbCrO4, precipitates.

Pb(OAc)2 + CrO42- ⇔ PbCrO4 + 2 OAc-

Separation and Identification of Bismuth

Upon the addition of aqueous ammonia to the solution from the lead sulfate separation, the hydroxides of bismuth (white), and copper (pale blue) are first precipitated. The equations are

Bi3+ + 3 NH3 + 3 H2O → Bi(OH)3 + 3 NH4+

Cu2+ + 2 NH3 +2 H2O → Cu(OH)2 +2 NH4+

When an excess of aqueous ammonia is added, the hydroxide of copper dissolves by forming a tetrammine complex, whereas the bismuth (III) hydroxide does not dissolve.

Bi(OH)3 + NH3 → no reaction Cu(OH)2+ 4 NH3 → Cu(NH3)4

2+ + 2 OH- After it is separated from the deep blue colored complex of copper, the bismuth hydroxide is treated with sodium stannite, a strong reducing agent. The bismuth is reduced to the elementary state, in which condition it appears black. The equation is

2 Bi(OH)3 + 3 Sn(OH)3- + 3 OH- → 2 Bi + 3 Sn(OH)6

2- Stannite Stannate

Because sodium stannite is unstable and darkens upon standing, it must be prepared just prior to use, and the test for bismuth as a black residue must be observed immediately. Sodium stannite is prepared by treating a solution of tin (II) chloride with an excess of sodium hydroxide. The equations are

Sn2+ + 2 OH- ⇔ Sn(OH)2 (white) Sn(OH)2 + excess OH- ⇔ Sn(OH)3

- (colorless) The darkening of the stannite solution upon standing is due to the formation of black tin (II) oxide. The equation is

Sn(OH)3- → SnO + H2O + OH-

Identification of Copper

The presence of copper is apparent from the deep blue color exhibited by the complex ion, Cu(NH3)42+.

However, if the blue color is too faint to discern, as in the case of a trace of copper, the formation of pink copper (II) hexacyanoferrate (II) serves as a delicate test for the ion. The equation for the reaction is

2 Cu(NH3)42+ + Fe(CN)6

4- → Cu2Fe(CN)6 + 8 NH3 Reprecipitation of the Sulfides of the Division B Ions


To reprecipitate the sulfide of tin from the solution containing its complex thiosalt, it is necessary to increase

the concentration of the cations to a value such that, for example,

[Hg2+][S2-] > Ksp This is accomplished by increasing the extent of dissociation of the complex ions through the addition of hydrochloric acid.

SnS32- ⇔ Sn4+ + 2 S2-

2 S2-+ 4 H+ ⇔ 2 H2S net reaction SnS3

2-+ 4 H+ ⇔ Sn4+ + 2 H2S↑ As seen from the above equations, the added acid displaces the equilibria in the direction of an increased concentration of Sn4+ because of the formation and escape of the weak, gaseous electrolyte H2S. This results in an increase in the concentration of the cation to such an extent that, for example, the product of the concentration of the tin (IV) ion and the sulfide ion becomes larger than the solubility product constant, and reprecipitation of the sulfide occurs. The net reaction for the reprecipitation is given by the following equation:

SnS3-2 + 2 H+ ⇔ SnS2 +H2S

The tin(IV) present is then reduced to tin(II) by reaction with iron metal in an acidified solution. Separation of Tin

To dissolve SnS2, the concentrations of the cations and sulfide ion must be made such that

[Sb3+]2 [S2-]3 < Ksp, and [Hg++][S2-] > Ksp The addition of 6 M HCl reduces the sulfide ion concentration (S2- + 2 H+ ⇔ 2 H2S), and the concentrations of antimony and tin ions (Sb3+ + 4 Cl- ⇔ SbC14

-, and Sn4+ + 6 Cl- ⇔ SnCl62-) to values such that these sulfides

dissolve. The net reactions are given by the equations

Sb2S3 + 6 H+ + 8 Cl- → 2 SbCl4- + 3 H2S

SnS2 + 4 H+ + 6 Cl- → SnCl62- + 2 H2S

The sulfides of mercury and arsenic remain undissolved in the presence of 6M HCl due to the very small values of their solubility product constants.

To be exact, aluminum actually reduces tin to the metallic state. When all of the aluminum is gone, the metallic tin dissolves in the hydrochloric acid, forming SnC13

- ions and liberating hydrogen. Use is made of the fact that divalent tin is a strong reducing agent in the confirmatory test for the ion. SnCh- will reduce mercury (II) to mercury (I), or metallic mercury, depending upon the amount of the tin (II) present in solution.

SnCl3- + 2 HgC14

2- → SnCl6 + Hg2Cl2 + 3 Cl- SnC13

- + Hg2C12 + Cl- → SnC162- + 2 Hg

Antimony metal reacts with nitric acid to form the insoluble oxide, Sb4O6. This oxide is soluble in oxalic acid, H2C2O4, forming antimony (III) oxyhydrogen oxalate, (SbO)HC2O4


4 Sb + 4 H+ + 4 NO3- → Sb4O6 + 4 NO + 2 H2O

Sb4O6 + 4 H2C2O4 → 4(SbO)HC2O4 + 2 H2O The presence of antimony is then confirmed by precipitating it as the orange-red sulfide, Sb2S3.

2 (SbO)HC2O4 + 3 H2S - Sb2S3 + 2 H2C2O4 + 2 H2O

The Analysis of Group III

The solution to be analyzed may be a Group III known or unknown, or it may be the solution from the Group II separation. See page 491 for the Group III flow sheet. Precipitation of the Group III Ions

Analytical Group III contains the metallic ions, Ni2+, Co2+, Mn2+, Fe3+, Al3+, Cr3+, and Zn2+. This experiment will not analyze Cr3+ or Co2+. These ions are not precipitated by hydrochloric acid (the Group I precipitant), nor are they precipitated by sulfide ions in solutions which are 0.3 M in hydrogen ion (the Group II precipitant). However, an ammonium sulfide solution precipitates Ni2+, Mn2+, Fe2+, and Zn2+ as sulfides and Al3+ as the hydroxide.

The concentration of sulfide ions in the 0.3 M hydrogen ion solution of the Group II precipitation is too small to allow the solubility products of the Group III sulfides to be exceeded. Hydrogen sulfide in an aqueous solution of ammonia has a much higher sulfide ion concentration, due to the formation of ammonium sulfide, according to the equation

2 NH3 + H2S ⇔ 2 NH4+ + S2-

The resultant sulfide ion concentration is sufficiently large so that the solubility products of the sulfides of

nickel, manganese, iron, and zinc are exceeded and precipitation occurs. Likewise, the hydroxyl ion concentration of the ammonium sulfide solution is great enough to precipitate the hydroxide of aluminum. The solubility product of aluminum hydroxide is lower than those of the corresponding sulfides, thus favoring the precipitation of the former.

Hydrogen sulfide is produced for the precipitation of this group by the acidic hydrolysis of thioacetamide. Upon the addition of aqueous ammonia the following precipitations occur.

Ni++ + S2- ⇔ NiS (brown or black) Mn2+ + S2- ⇔ MnS (flesh colored)

Fe2+ + S2- ⇔ FeS (black) Zn2+ + S2- ⇔ ZnS (white)

Al3+ + 3 OH- ⇔ Al (OH)3 (white)

In the event that iron is present in the original solution in the Fe3+ state, it will be reduced by hydrogen sulfide in the acidic solution, according to the equation

2 Fe3+ + H2S → 2 Fe2+ + S + 2 H+

Of the cations remaining in solution, only the magnesium ion, of Group V, forms an insoluble hydroxide. The hydroxyl ion concentration is kept just low enough in the aqueous ammonia to prevent the precipitation of magnesium hydroxide by having ammonium chloride present. The ammonium ions from ammonium chloride exert the common ion effect with the ammonium ions from the ionization of ammonia, thus repressing the ionization of ammonia and reducing the hydroxyl ion concentration of the solution.


Some evidence regarding the cations present in a solution containing the Group III ions can be obtained by

noting the colors of the ions: Ni2+, pale green; Zn2+, colorless; Mn2+, colorless in low concentrations; Fe2+, pale green; Fe3+, reddish-brown; Al3+, colorless. Due to the phenomenon of complementary colors, a solution containing certain combinations of colored ions may appear colorless. Separation of Cobalt and Nickel

Although hydrogen sulfide will precipitate the sulfides of cobalt and nickel completely only from basic solution, these sulfides are but slightly soluble in dilute HCl It appears that these sulfides precipitate in a form which is soluble in acid, but change rapidly into other crystalline modifications which have much lower solubility products. The Ksp of α CoS is 7 x 10-23, of β CoS is 1.9 x 10-27, of α NiS is 3 x 10-21, and of β NiS is 1 x 10-26. Use is made of these facts in separating CoS and NiS from MnS, FeS, Al(OH)3, and ZnS, the latter four precipitates being dissolved by 1 M HCl

MnS + 2 H+ ⇔ Mn2+ + H2S FeS + 2 H+ ⇔ Fe2+ + H2S ZnS + 2 H+ ⇔ Zn2+ + H2S

Al(OH)3+ 3 H+ ⇔ Al3+ + 3 H2O

Identification of Cobalt and Nickel

The sulfides of cobalt and nickel readily dissolve in a mixture of nitric and hydrochloric acids due to the higher hydrogen ion concentration and the oxidation of the sulfide ion to sulfur. The equations are

3 CoS + 8 H+ + 2 NO3- ⇔ 3 Co2+ + 2 NO + 3 S+ 4 H2O

3 NiS + 8 H+ + 2 NO3- ⇔ 3 Ni2+ + 2 NO + 3 S + 4 H2O

After boiling the solution containing the cobalt and nickel ions to remove the oxides of nitrogen which would destroy the reagents used in the confirmation of these ions, an excess of aqueous ammonia is added. Ammonia in excess reacts with cobalt and nickel ions to form hexamminemetal complexes.

Co2+ + 6 NH3 ⇔ Co(NH3)6++ (pink)

Ni2+ + 6 NH3 ⇔ Ni(NH3)6++ (blue)

In the ammoniacal solution, Ni++ will react with dimethylglyoxime to form a very insoluble, bright red complex

compound. Co++ forms a brown colored soluble complex with dimethylglyoxime which does not interfere with the test.

A concentrated solution of ammonium thiocyanate is used in the identification of the cobalt ion, with which the thiocyanate ion forms a complex ion, Co(SCN)4

2-. that has a characteristic blue color.

Co(NH3)6++ + 4 SCN- -. Co(SCN)4

2- + 6NH3


When iron (III) ions are present, they interfere with this test by forming a bright red complex ion of the formula Fe(SCN)++. This interference may be avoided by converting the iron to the colorless and very stable hexafluoferrate (III) ion, FeF6

3-. The equation is

Fe3+ + 6 F- ⇔ FeF63-

A second confirmatory test for cobalt involves its oxidation by the nitrite ion from Co2+ to Co3+ and the

precipitation of the complex yellow salt, K3Co(NO2)6, tripotassium. hexanitrocobaltate (III).

Co2+ + 6 NO2 → Co(NO2)64-

Co(NO2)64- + NO2

- + 2 H+ → Co(NO2)63- +NO +H2O

3 K+ + Co(NO2)63- → K3Co(NO2)6

Separation of Iron and Manganese from Aluminum, and Zinc

The solution containing the ions of the above metals is treated with nitric acid to remove the sulfide ion by oxidizing it to sulfur, and to oxidize Fe2+ to Fe3+

The equation for the oxidation of the iron is

3Fe2+ + 4H+ +NO3

- → 3Fe3+ + NO +2H2O It is desirable to have the iron in the +3 condition because Fe(OH)3 is less soluble and less gelatinous in character than is Fe(OH)2.

In order to separate iron and manganese from aluminum, and zinc, the solution is first treated with sodium

hydroxide and then with hydrogen peroxide. The precipitations which first occur, as sodium hydroxide is added to the solution, are described by the following equations:

Mn2+ + 2 OH- ⇔ Mn(OH)2 (white) Fe 3+ + 3 OH- ⇔ Fe(OH)3 (reddish brown)

Al3+ 3 OH- ⇔ Al(OH)3 (white) Zn2+ + 2 OH- ⇔ Zn(OH)2 (white)

The precipitate is not separated, and the addition of an excess of NaOH dissolves the amphoteric hydroxides of aluminum, and zinc, but not the hydroxides of iron and manganese.

Al(OH)3 + OH- ⇔ AI(OH)4- (aluminate) (colorless)

Zn(OH)2 + 2 OH- ⇔ Zn(OH)42- (zincate) (colorless)

Hydrogen peroxide is added to the resulting mixture to oxidize manganese (II) hydroxide to a mixture of manganese dioxide and manganese (III) hydroxide, both of which are much less soluble than Mn(OH)2

Mn(OH)2+ H2O2 → +MnO2+2H2O 2 Mn(OH)2+ H2O2 → 2 Mn(OH)3

The mixture containing MnO2 and Mn(OH)2 is brown. Separation and Identification of Manganese and Iron

The residue containing MnO2, Mn(OH)3, and Fe(OH)3 is dissolved by a mixture of HNO3 and NaNO2. The nitrite ion in acid solution reduces the manganese to the divalent condition. The iron (III) hydroxide is dissolved by nitric acid. The equations for these reactions are


2 Mn(OH)3 + 4 H+ + NO2 -→ 2 Mn2+ + NO3

- + 5 H2O MnO2 + 2 H+ + NO2

- -→ Mn2+ + NO3- + H2O

Fe(OH)3 + 3 H+ -→ Fe3+ + 3 H2O

The presence of Fe3+ is confirmed by its reaction with the thiocyanate ion, SCN-, to produce a blood red complex ion, Fe(SCN)++, or one of the possible complexes containing from one to six SCN- groups.

Fe3+ + SCN- ⇔ Fe(SCN)2+ This test for iron may be conducted on a solution containing Mn2+ without interference by this ion. It should be noted that there are traces of Fe3+ existing in many reagents. Therefore a light pink color in this test may be due to iron as an impurity. Blank tests can be run on the reagents used in the test to establish this point.

Manganese may be identified in the presence of iron by oxidizing it to the permanganate ion by means of sodium bismuthate in nitric acid. The MnO4

- ion is purple but in dilute solutions it may appear pink.

2 Mn2+ + 5 BiO3- + 14 H+ → . 2 MnO4

-+ 5 Bi3+ + 7 H2O Separation and Identification of Aluminum

Acetic acid is added to the solution containing the Al(OH)4-, CrO4

2-, and Zn(OH)42-. The aluminum and zinc

portray their basic character in the acidic solution and are converted to their cation forms.

Al(OH)4- + 4 H+ ⇔ Al3+ + H2O

Zn(OH)42- + 4 H+ ⇔ Zn++ + 4 H2O

Acids react with yellow chromate, CrO4

2-, and convert it in part to the dichromate, Cr2O72-, the extent of the

conversion depending upon the concentration of thehydrogen ion.

2 CrO42-+ 2 H+,--± Cr2O7

2- + H2O

The addition of excess aqueous ammonia precipitates aluminum as the hydroxide, converts Cr207_ to Cr04-, and complexes the zinc as the soluble Zn(NH3)4

++ ion.

Al3+ + 3 NH3 + 3 H2O ⇔ Al(OH)3 + 3 NH4+

Cr2O72- + 2 OH- ⇔ 2 CrO4

2-+ H2O Zn++ + 4 NH3 ⇔ Zn(NH3)4

++

The presence of aluminum is confirmed by dissolving the hydroxide in acetic acid and adding the aluminum reagent and (NH4)2CO3. Aurin tricarboxylic acid (aluminon) imparts a red color to the Al(OH)3 which is formed. Identification of Zinc

The zinc ions are precipitated by sulfide ions from a thioacetamide solution, white zinc sulfide being formed. The chromate ion may oxidize part of the sulfide ions used in the test to elementary sulfur, but ZnS is soluble in hydrochloric acid whereas sulfur is not. A second confirmatory test for zinc may be made by boiling all of the hydrogen sulfide out of the solution and precipitating the zinc as the hexacyanoferrate (II), which is white.

2 K+ + Zn++ + Fe(CN)64- → K2ZnFe(CN)6


The Analysis of Group IV The solution to be analyzed may be a Group IV known or unknown, or it may be the solution from the Group

III separation. The Group IV flow sheet is given on page 502. Precipitation of the Group IV Ions

Analytical Group IV contains the metallic ions, Ba++, Sr++, and Ca++. These metals form chlorides, sulfides, and hydroxides which are soluble under the conditions which prevail in the precipitation of Groups I, II, and III. The carbonates of barium, strontium, and calcium precipitate in aqueous ammonia solutions containing ammonium carbonate.

The concentration of the carbonate ion in a solution of ammonium carbonate is too low to effect a complete precipitation of barium, strontium, and calcium due to the partial hydrolysis of the carbonate ion.

CO32- + H2O ⇔ HCO3

- + OH- This hydrolysis is repressed by increasing the hydroxide ion concentration of the solution, the additional hydroxide ions being supplied by the buffer pair - aqueous ammonia and ammonium chloride. These conditions permit a carbonate concentration high enough to precipitate BaCO3, SrCO3, and CaCO3, but not MgCO3. At the same time the hydroxide ion concentration in this buffered solution is not high enough to precipitate Mg(OH)2

In order that the solution may have the required ammonium ion concentration, the centrifugate from the Group III separation, which contains ammonium salts, is first evaporated to dryness and then ignited to expel the ammonium salts present. The equations are

NH4Cl → NH3↑ + HCl↑ NH4NO3 → NH3↑ + HNO3↑

NH3 + HNO3 → N2O↑ + 2 H2O↑ The necessary ammonium ion concentration is then obtained by adding the Group IV reagent which consists of ammonium carbonate.. ammonium chloride, and aqueous ammonia in the required concentrations. The equations for the precipitations of the Group IV carbonates are

Ba++ + CO32- ⇔ BaCO3 (white)

Sr++ + CO32- ⇔ SrCO3 (white)

Ca++ + CO32- ⇔ CaCO3 (white)

The Separation and Identification of Barium

1. The dissolution of the carbonates. The carbonates of barium, strontium, and calcium are dissolved readily by strong acids such as hydrochloric acid or by weak acids such as acetic acid. The latter acid is used in these procedures. The equilibria involved, as applied to barium carbonate, are given by the following equations:

BaCO3 ⇔ Ba++ +CO3

2- HOAc ⇔ H+ + OAc- CO3

2-+ H+ ⇔ HCO3-

HCO3- + H+ ⇔ H2CO3

H2CO3 ⇔ H2O + CO2↑ If the concentration of the hydrogen ion supplied by acetic acid is sufficiently high, the quantity of H2CO3 formed will exceed that in a saturated solution of CO2 at atmospheric pressure and CO2 will escape from solution. The


loss of carbon dioxide from the system in this way causes all the above equilibria to be shifted to the right until the precipitate dissolves completely. The net reaction is given by the equation

BaCO3 + 2 HOAc ⇔ Ba++ + 2 OAc- + H2O + CO2↑

Similar equations may be written for the dissolution of the carbonates of strontium and calcium in acetic acid.

2. The separation of barium from strontium and calcium by precipitation of barium chromate. By taking advantage of the fact that barium chromate is less soluble than strontium chromate and calcium chromate, barium can be separated from strontium and calcium. The solubility product constants are 2.4 x 10-10 for BaCrO4 and 3.6 x 10-1 for SrCrO4- Calcium chromate is relatively very soluble. The problem in. this separation, then, is to control the concentration of the chromate ion so that the concentration of the barium ion will be reduced to at least 0.0001 M, and strontium chromate will not be precipitated from a solution 0.1 M in the strontium ion. Let us first calculate the concentration of chromate ion required to reduce the barium ion concentration to 0.0001 M. [Ba++][CrO4

2-] = 2 x 10-10

[CrO4

2-] = 2 x 10-10 = 2 x 10-1 = 2 x 10-6 [Ba++] 1 x 10-4

Now let us determine the maximum concentration of chromate ion that we can have without precipitating any strontium chromate when the concentration of the strontium ion is 0.1 M.

[Sr++][CrO42-] = 3.6 x 10-5

[CrO4

2-] = 3.6 x 10-5 = 3.6 x 10-5 = 3.6 x 10-4 [Sr++] 10-1

Thus we find that the concentration of the chromate ion must be held between 2 x 10-6 M and 3.6 x 10-4 M to cause a nearly complete precipitation of BaCrO4, but not the precipitation of any SrCrO4.

In a solution containing chromate ions, we have the following equilibrium:

Cr2O72- + H2O ⇔ 2 CrO4

2- + 2 H+ The concentration of chromate ions in a solution made by dissolving a dichromate in water can be controlled by adjusting the hydrogen ion concentration. Applying the law of chemical equilibrium to this system, we have

[CrO42-]2[H+]2 = 2.4 x 10-16

[Cr2O72-]

It is readily seen from this expression that the chromate ion concentration is a function of the hydrogen ion concentration. The more acidic the solution, the lower will be the chromate ion concentration; and the more basic the solution, the higher will be the chromate ion concentration. By substituting in the above expression, it can be shown that for a solution which is about 0.01 M in dichromate ions, the hydrogen ion concentration must be adjusted to about 1 x 10-5 M (pH = 5) to maintain the chromate ion concentration at about 5 x 10-4 M. As shown above, this concentration of chromate ions is approximately the value needed for the separation of Ba++ from Sr++. A buffered solution composed of acetic acid and ammonium acetate is used to obtain the required hydrogen ion concentration.

As chromate ions are removed from the system through precipitation of barium chromate, the concentration of these ions does not change appreciably. This is true because the excess of dichromate serves as a reservoir for chromate ions; and as CrO4

2- is used up in precipitating BaCrO4, the equilibrium Cr2O72- + H2O ⇔ 2 CrO4

2- + 2


H+ shifts to the right with the generation of more CrO42- from the relatively large supply of Cr2O7

2-. The hydrogen ions that are formed unite with the acetate ions of the buffered solution, and a constant pH is maintained.

3. The dissolution of barium chromate. Even though barium chromate is precipitated from a weakly acidic solution, it is readily dissolved by strong acids. The hydrogen ion concentration of solutions of strong acids is sufficiently high to cause a reduction in the chromate ion concentration below the value required to maintain a saturated solution of barium chromate. Thus

[Ba++][CrO42-] < Ksp

and the barium chromate dissolves. A solution of 12 M HCl is used in this scheme to bring about the dissolution of BaCrO4

4. The identification of barium. The presence of the barium ion in the solution is confirmed by precipitating it as barium sulfate, with sulfuric acid being used as the precipitant.

Additional evidence for the presence of the barium ion is obtained using the flame test. When heated in the Bunsen flame, barium salts impart to it a yellow-green color. The Separation and Identification of Strontium

1. The separation of strontium from calcium. Although the chromate ion concentration in an acetic acid-ammonium acetate buffered solution is too low to exceed the solubility product of strontium chromate, this compound will precipitate when the solution is made basic with aqueous ammonia. The equilibria are

Cr2O72- + H2O ⇔ 2 CrO4

2- + 2 H+ 2 OH- + 2 H+ ⇔ 2 H2O

and the net reaction is Cr2O7

2- + 2 OH- ⇔ 2 CrO4 + H2O The resulting concentration of chromate ions is large enough so that the solubility product constant of strontium chromate is exceeded and precipitation occurs. The complete precipitation of strontium chromate is insured by further reducing its solubility through the addition of enough ethyl a1cohol to the solution to make it 50 per cent alcohol by volume. Calcium chromate is soluble in this solution.

2. The identification of strontium. The formation of a fine yellow crystalline precipitate of the chromate serves to confirm the presence of strontium. A crimson color imparted to the Bunsen flame is characteristic of the strontium ion. The Identification of the Calcium

The solution from the strontium chromate separation contains the calcium ion. The addition of ammonium

oxalate will precipitate calcium oxalate, CaC2O4, a white crystalline salt.

Ca++ + C2O42- ⇔ CaC2O4

Calcium ions impart a brick red color to the Bunsen flame.

The Analysis of Group V

The solution for analysis may be a Group V known or unknown or it may be the solution from the Group IV separation.


Removal of Traces of Calcium and Barium "

The cations comprising Group V (Na+, K+, Mg++, and NH4+) do not form precipitates with the reagents used to

separate Group I, II, III, and IV, and the ions of this group have no precipitant which is common to all four ions.

If ammonium salts were present in excessive quantities during the precipitation of the carbonates of the Group IV ions, then trace amounts of barium and calcium ions may have come through into the Group V solution. Because these ions will interfere with the identification of the Group V ions, they must be removed. Ammonium oxalate is added to precipitate the calcium ions and ammonium sulfate to precipitate the barium ions. Identification of Sodium

After the removal of the traces of barium and calcium, the sodium ion maybe detected by the formation of the triple salt, sodium zinc uranyl acetate hexahydrate, which is a yellow crystalline compound.

Na+ + Zn2+ + 3 UO2++ + 9 OAc- + 6 H2O ⇔ NaZn(UO2)3(OAc)9 • 6 H2O The Sodium Reagent is a saturated solution of zinc acetate and uranyl acetate in acetic acid. Ammonium, magnesium, and potassium ions do not interfere with the test. The sodium ion imparts a yellow color to the Bunsen flame. Separation and Identification of Magnesium

To a second portion of the solution for analysis is added an excess of aqueous ammonia and disodium hydrogen phosphate. The formation of a white crystalline precipitate, MgNH4PO4, is evidence of the presence of the magnesium ion. The equation is

Mg++ + NH4+ + HPO4

2- ⇔ MgNH4PO4 + H+ Note that the hydrogen ion is one product of the reaction. Completeness of precipitation of the magnesium ammonium phosphate is assured by making the solution basic with aqueous ammonia, which combines with the hydrogen ion and causes a shift of the equilibrium to the right.

After the magnesium ammonium phosphate is separated from the solution containing the sodium and ammonium ions, it is dissolved in acetic acid. This treatment with acid causes the following equilibrium to be shifted to the right.

MgNH4PO4 + H+ ⇔ Mg++ + NH4+ + HPO4

2- This results because the phosphate ion in a saturated solution of MgNH4PO4 is removed from the solution by combining with hydrogen ions, making the product of the concentrations of the Mg++, NH4

+, and PO4-3 ions less

than the solubility product constant. Magnesium Reagent and sodium hydroxide are added to the solution containing magnesium ions. Tile magnesium hydroxide which is formed adsorbs the Magnesium Reagent (a dye, p-nitrobenzene-azo-alpha-naphthol) and forms a blue precipitate, which confirms the presence of magnesium. Identification of Potassium

The identification of potassium is made from the solution obtained from the separation of magnesium ammonium phosphate. Ammonium ions must be removed prior to the test for the potassium ion because the precipitant for potassium ions will also precipitate ammonium ions. Concentrated nitric acid is added to the solution, it is evaporated to dryness, and then it is ignited to expel volatile ammonium salts.

NH4C1 → NH3 + HC1


NH4NO3 → NH3 + HNO3 NH3 + HNO3 → N2O + 2 H2O

The potassium ion is precipitated from solution in the form of the yellow complex salt, K2NaCo(NO2)6, upon

the addition of a solution of Na3Co(NO2)6

2 K+ + Na+ + Co(NO2)6-3 ⇔ K2NaCo(NO2)6

Potassium gives a characteristic violet flame test which is masked by the intense yellow color produced by the

sodium ion. By observing the flame through cobalt glass, which filters out the yellow color of the sodium ion, one may detect the potassium ion in the presence of the sodium ion.

Identification of Ammonium Ions

Since ammonium ions are added in the course of the analysis at various points, it becomes necessary to test for this ion on the original solution of a known or unknown. The presence of the ammonium ion in a solution may be detected by the addition of a strong, nonvolatile base such as sodium hydroxide, with subsequent heating of the solution and liberation of gaseous ammonia.

NH4+ + OH- (from NaOH) ⇔ NH3↑ + H2O

The gaseous ammonia may be detected by moist red litmus paper and by its characteristic odor.

NH3 + H2O ⇔ NH4+ + OH-

During the heating of the solution one must not permit the sodium hydroxide solution to come in contact with

the litmus paper because this would void the test for ammonia.


1. Why is Group V often referred to as the soluble group? 2. Explain why the test for the ammonium ion must be made on a portion of the original sample during

the analysis of either a Group V known or unknown or a general unknown involving all of the groups.

3. Why must the ammonium ion be removed before making the chemical test for the potassium ion? 4. Write the equation for the dissolution of MgNH4PO4 in an acid. Explain this reaction in terms of

solubility product and ionic equilibria theory. 5. Why are (NH4)2SO4 and (NH4)2C2O4 added prior to the analysis of Group V? 6. Explain the use of cobalt glass in the flame test for potassium. 7. What is the chemistry of the chemical test for potassium? 8. Why may red litmus turn blue upon prolonged exposure to the air of the laboratory?

PROBLEMS 1. How many ml. of NH3 gas measured at 25" C and 740 mm. pressure will be evolved when a solution

containing 10.0 mg. of NH4Cl is treated with NaOH and heated? Ans. 4.7 mL 2. What weight of MgNH4PO4 can be formed from 5.0 mg. of Mg++? Ans. 28.2 mg.


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