Preparation of Sodium Carbonate extract - Identify Sulfates, Nitrites, Nitrates and Phosphates by using the prepared Sodium carbonate solution

Preparation of Sodium Carbonate extract

200mg of the unknown sample, 1g of Na₂CO₃ and 50ml of distilled water was added into a clean dry flask later boiled using a tripod stand. This was done by clamping a funnel above the flask to reduce evaporation.

It was boiled for about 5 minutes.

Afterwards solution was cooled and filtered using a filter paper.

A colorless filtrate was obtained.

Identify Sulfates, Nitrites, Nitrates and Phosphates by using the prepared Sodium carbonate solution

SO₄^2-

Procedure	Observation
5 drops of Na2CO3 extract was acidified with dil. HCl in boiling tube. Acidity was checked with litmus. It was kept in a hot water bath for about 3 minutes and added 3 drops of BaCl2	When the extract was acidified with HCl, it turned into brownish red colour. White colour precipitate was formed.

NO₂^-

Procedure	Observation
1ml of Na2CO3 extract was heated with dil. H2SO4 in boiling tube. The evolved gas was checked with starch.iodide paper.	With H2SO4 the extract turned into reddish brown colour. Brown fume was evolved when heating. It turned the starch-iodide paper into blue colour.

NO₃^-

Procedure	Observation
1ml of Na2CO3 extract was heated with 100mg Sulphamic acid in a boiling tube. Later dil.NaOH was added and heated until the evolution of NH3 is ceased. It was again checked with Devarda’s alloy.	In the presence of Sulphamic acid the solution turned into brownish red. Pungent smelling brownish colour gas was formed and the it continued for awhile in Devarda’s alloy.

PO₄^3-

Procedure	Observation
1ml of Na2CO3 extract, 1ml of dil. HNO3 and 10ml ammonium molybdate was heated gently in a boiling tube.	With HNO3 the extract turned into brownish red colour. Later a yellow colour precipitate was obtained.

What’s the purpose of preparing Na₂CO₃ extract?

It gives an expedient way of bringing all the anions of the sample into solution. Otherwise they might be insoluble and couldn’t carry out the experiment properly. Also it prevents the cation in interfering.

What is the importance of Na₂CO₃ extract to be clear and colourless?

This provides a better setting to observe formations of precipitates, colour changes which take place during the experiment.

Is the Na₂CO₃ extract basic or acidic?

It is slightly basic.

Test for phenols:- Phenol, 2-napthol, Catechol, Resorcinol

Azo-dye formation

Procedure	Observation
2drops of aniline were dissolved in 1ml of con.HCl and 3ml of water were shaken in a test tube. It was kept in ice. Also 20% sodium nitrite solution and phenol dissolved in NaOH were kept in ice. Few drops of sodium nitrite were added to the cold aniline. Maintaining the temperature around 0-4°C this prepared diazonium solution was added to the cold phenol.	Phenol: intense orange colour dye 2-napthol: red colour dye Catechol: dark brownish dye Resorcinol: red colour dye

Test for primary aromatic amines (Benzoylation):- Anilene, O.toluidene, P.toluidene, M.toluidene

Benzoylation

Procedure	Observation
1ml of the compound, 20ml of 10% NaOH solution and 1.5ml of Benzoyl chloride were mixed together in Stoppered conical flask. Mixture was vigorously shaken.	Aniline: White semi-solid O.toluidene: Pale pink semi-solid M.toluidene: Pale brown semi-solid P.toluidene: Dark brown semi-solid

Tests for Phenols - Discussion

Phenol

Catechol

Resorcinol

2-napthol

Solubility

Some phenols (Ex: Phenol,catechol,resorcinol) are soluble in water and others are sparingly soluble or insoluble in water because of the increasing molecular weight and hydrophobic nature. But all the phenols are capable of reacting with NaOH solution and dissolve.

C₆H₅OH + NaOH ----> C₆H₅ONa + H₂O

Reaction with Sodium carbonate

Phenols give no reaction with sodium carbonate but as the aqueous solution of it was used,phenols dissolved in it giving the same colourations given when testing the solubility in water.As Phenols do not produce gas from sodium carbonate it should be a very weak acid.

Benzoylation

To react with benzoyl chloride, phenol should be modified to make the reaction possible.the -COCl group is bounded directly to the benzene ring. Therefore it's less reactive.phenol is first dissolved in NaOH in order to get the phenoxide. Later it's reacted with the Benzoyl chloride.

Tests for Carboxylic acid - Discussion

Acetic acid

Benzoic acid

Salicylic acid

 Cinnamic acid

Solubility
Acetic acid is soluble in water but Benzoic acid,salicylic acid and cinnamic acid are insoluble due to their heavy molecular weight and increased hydrophobic nature. Carboxylic acids are acidic therefore they can dissolve in bases such as ethanol and sodium hydroxide forming the carboxylate ion/salt.


Reaction with Sodium carbonate
Reacts with sodium carbonate liberating carbon dioxide

Esterification
Acetic acid----> Medicinal or strong odour of apples
Benzoic acid----> Fruity odour
Salicylic acid----> Medicinal odour
Reaction with Ferric chloride
Higher acidity can interfere the final results. Therefore it's important to use neutral ferric chloride neutralized by sodium hydroxide.
FeCl₃ + C₆H₅COO^-   -------->     Fe ( OOC₆H₅)₃
FeCl₃ + CH₃COO^-   -------->     Fe ( OOCH₃)₃
Cinnamic acid with KMnO4
This will give out Benzaldehyde and acetic acid. Almond odour was given due to the formation of benzaldehyde.

The Chemistry of the Group II Analysis

A. The addition of sodium hydroxide does two things: (a) It precipitates the hydroxides of manganese,
nickel, and iron. (b) It forms the hydroxide complexes of aluminum and zinc, since these hydroxides are
amphoteric.
Mn2+(aq) + 2OH–(aq) → Mn(OH)2(s)
Ni2+(aq) + 2OH–(aq) → Ni(OH)2(s)
Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s)
Al3+(aq) + 4OH–(aq) → [Al(OH)4]–(aq) (aluminate ion)
Zn2+(aq) + 4OH–(aq) → [Zn(OH)4]2–(aq) (zincate ion)

B. The addition of nitric acid dissolves the hydroxide precipitate. When sodium bismuthate, a strong
oxidizing agent, is then added, the Mn2+ ion is oxidized to the permanganate ion.
Mn(OH)2(s) → Mn2+(aq) + 2OH–(aq)
Ni(OH)2(s) → Ni2+(aq) + 2OH–(aq)
Fe(OH)3(s) → Fe3+(aq) + 3OH–(aq)
14H+(aq) + 2Mn2+(aq) + 5BiO3–(aq) → 2MnO4–(aq) + 5Bi3+(aq) + 7H2O(l)
                                                                   purple

C. The addition of ammonia forms a precipitate of iron(III) hydroxide and the hexaamminenickel(II)
complex ion. Upon the addition of hydrochloric acid to the precipitate, it dissolves giving Fe3+ ions in
solution. Then treating this solution with ammonium thiocyanate forms a blood-red precipitate of the
hexaisothiocyanatoferrate(III) complex ion.
Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s)
Ni2+(aq) + 6NH3(aq) → [Ni(NH3)6]2+(aq)
Fe(OH)3(s) + 3H+(aq) → Fe3+(aq) + 3HOH(l)
Fe3+(aq) + 6SCN–(aq) → [Fe(NCS)6]3–(aq)
                                            blood-red

D. Adding dimethylglyoxime, H2DMG, to the solution of hexaamminenickel(II) ion gives another
complex ion commonly called nickel dimethylglyoxime. [Note: Dimethylglyoxime (H2DMG, 2,3-
butanedionedioxime, C4H8N2O2) is an organic complexing agent that causes a colorful precipitation of
nickel ions.]
[Ni(NH3)6]2+(aq) + 2H2DMG → [Ni(HDMG)2](s) + 2NH4+(aq) + 4NH3(aq)
                                                                                      pink-red

E. The addition of nitric acid will break up the hydroxide complexes of aluminum and zinc. Then the
addition of ammonia to the resulting solution of Al3+ and Zn2+ ions will do two things: (a) the aluminum
ion will precipitate as aluminum hydroxide and (b) the zinc ion will form the tetraamminezinc(II)
complex ion. Now nitric acid is added to the aluminum hydroxide precipitate causing it to dissolve.
Then the addition of ammonia and aluminon reagent will cause a reddish-pink precipitate of
[Al(OH)3•aluminon] complex to form. (Note: The aluminon reagent is the ammonium salt of aurin
tricarboxylic acid, C22H23N3O9, a red dye.)
[Al(OH)4]–(aq) + 4H+(aq) → Al3+(aq) + 4HOH(l)
[Zn(OH)4]2–(aq) + 4H+(aq) → Zn2+(aq) + 4HOH(l)
Al3+(aq) + 3OH–(aq) → Al(OH)3(s)
Zn2+(aq) + 4NH3(aq) → [Zn(NH3)4]2+(aq)
Al(OH)3(s) + 3H+(aq) → Al3+(aq) + 3HOH(l)
Al3+(aq) + 3NH3(g) + 3H2O(l) + aluminon → [Al(OH)3•aluminon](s) + 3NH4+(aq)
                                                                              reddish-pink

F. The addition of HCl breaks up the tetraamminezinc(II) complex ion. Then the addition of potassium
hexacyanoferrate(II) forms a precipitate of K2Zn3[Fe(CN)6]. These two reactions are shown as one step
below.
[Zn(NH3)4]2+(aq) + 2K4[Fe(CN)6](aq) + 4H+(aq) → K2Zn3[Fe(CN)6]2(s) + 4NH4+(aq) + 6K+(aq)
                                                                                      cream to blue-grey

Separation and Confirmation of Group II Ions: Mn2+, Ni2+, Fe3+, Al3+, and Zn2+

This procedure can be used for the known or unknown solution.

A. Separation of Mn2+, Ni2+, and Fe3+ from Al3+ and Zn2+
1. Place 20 drops of the Group II (known or unknown) solution into a clean centrifuge test tube and add
20 drops of 3M NaOH. Heat in a boiling water bath for 10 min. While heating, be sure to stir about
every minute or so to mix the reagents or the aluminum and zinc won’t precipitate properly. Stir,
centrifuge, and decant the supernatant into a clean tube.
2. To reduce problems with the analysis for zinc and aluminum ions it is important that you not transfer
any precipitate to the supernatant. If you have any question about the purity of the supernatant, centrifuge
and decant it a second time. If you can’t remove traces of precipitate by centrifuging, see your instructor
who will use two disposable polyethylene pipets to filter the solution.
3. Save the precipitate and the supernatant. Label the supernatant "Aluminum and Zinc."
The equations for step 1 are:
Mn2+(aq) + 2OH–(aq) → Mn(OH)2(s)
Ni2+(aq) + 2OH–(aq) → Ni(OH)2(s)
Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s)
Al3+(aq) + 4OH–(aq) → [Al(OH)4]–(aq) (aluminate ion)
Zn2+(aq) + 4OH–(aq) → [Zn(OH)4]2–(aq) (zincate ion)

B. Separation and Confirmation of Mn2+
1. To the tube containing the precipitate from part A, add 12 drops of 6M HNO3. Stir the solution until
the precipitate dissolves.
2. Place 5 drops of this solution into another clean centrifuge test tube and add a small amount of solid
sodium bismuthate (NaBiO3) using a wooden spatula. Use the amount you can fit on the tip of the
spatula.
3. Add 20 drops of distilled water, stir, and centrifuge. A purple-colored solution confirms the presence
of Mn2+.
The equations for step 1 are:
Mn(OH)2(s) → Mn2+(aq) + 2OH–(aq)
Ni(OH)2(s) → Ni2+(aq) + 2OH–(aq)
Fe(OH)3(s) → Fe3+(aq) + 3OH–(aq)
The equation for step 3 is:
14H+(aq) + 2Mn2+(aq) + 5BiO3(aq) → 2MnO4-(aq) + 5Bi3+(aq) + 7H2O(l)
                                                                  purple

C. Separation and Confirmation of Fe3+
1. To the remaining solution from part B above, add 15 drops of 6M NH3·H2O and stir. Centrifuge and
decant the supernatant into a clean centrifuge test tube. Label the tube "Nickel" and set it aside for future
use.
8
2. To the precipitate, add 10 drops of 6M hydrochloric acid, stir, and then add 3 drops of 0.1 M NH4SCN.
A “blood-red” color confirms the presence of Fe3+.
The equations for step 1 are:
Fe3+(aq) + 3OH–(aq) → Fe(OH)3(s)
Ni2+(aq) + 6NH3(aq) → [Ni(NH3)6]2+(aq)
The equations for step 2 are:
Fe(OH)3(s) + 3H+(aq) → Fe3+(aq) + 3HOH(l)
Fe3+(aq) + 6SCN–(aq) → [Fe(NCS)6]3–(aq)
                                            blood-red

D. The Confirmation of Ni2+
1. Add 4 drops of dimethylglyoxime (DMG) to the tube labeled "Nickel" in part C and stir. There is no
need to centrifuge, as the nature of the precipitate (sometimes called a “lake”), is more obvious before
centrifuging. A strawberry-colored ppt confirms the presence of Ni2+.
The equation for step 1 is:
[Ni(NH3)6]2+(aq) + 2H2DMG → [Ni(HDMG)2](s) + 2NH4+(aq) + 4NH3(aq)

                                                          pink-red

E. Separation on Confirmation of Al3+
1. Add 10 drops of 6 M HNO3 to the solution from part A above labeled "Al & Zn" and stir. This will
breakup the hydroxide complexes.
2. Now add 10 drops of 6M NH3•H2O to the solution and stir. Check the solution with universal
indicator paper to be sure it is basic (dark blue).
3. If the solution is acidic, add 10 more drops of 6M NH3 ·H2O. Stir and re-check the acidity of the
solution. Continue adding 6M NH3 ·H2O until the solution is basic. You should get a clear gelatinous
precipitate. It may be hard to see until the tube is centrifuged. Even then, some people have difficulty
seeing the precipitate.
4. Centrifuge and decant the supernatant into a clean tube. Label this tube "Zinc."
5. Add 5 drops of 6M HNO3 to the precipitate that supposedly contains aluminum(III) ions and stir until
it dissolves. If it looks like a precipitate is still present, test the solution for acidity with pH paper. If the
solution is not acidic, add 6M HNO3 one drop at a time, with stirring, until the solution is acidic.
6. Now add 4 drops of aluminon reagent and 10 drops of 6 M NH3•H2O. Stir and centrifuge. A cherrycolored
precipitate confirms the presence of Al3+.
The equations for step 1 are:
[Al(OH)4]–(aq) + 4H+(aq) → Al3+(aq) + 4HOH(l)
[Zn(OH)4]2–(aq) + 4H+(aq) → Zn2+(aq) + 4HOH(l)
The equations for step 2 are:
Al3+(aq) + 3OH–(aq) → Al(OH)3(s)
Zn2+(aq) + 4NH3(aq) → [Zn(NH3)4]2+(aq)
The equation for step 5 is:
Al(OH)3(s) + 3H+(aq) → Al3+(aq) + 3HOH(l)
The equation for step 6 is:
Al3+(aq) + 3NH3(g) + 3H2O(l) + aluminon → [Al(OH)3•aluminon](s) + 3NH4+(aq)
                                                                               reddish-pink

F. Confirmation of Zn2+
1. Add 10 drops of 6 M hydrochloric acid to the supernatant from part E above labeled "Zinc." Stir the
solution.
2. Add 4 drops of 0.1M K4[Fe(CN)6]. Stir and centrifuge the solution. A cream-colored precipitate
confirms the presence of Zn2+. Note: This precipitate may be contaminated with iron from part A and
may be various shades of blue depending on how much contamination is present.
The equation for step 2 is:
[Zn(NH3)4]2+(aq) + 2K4[Fe(CN)6](aq) + 4H+(aq) → K2Zn3[Fe(CN)6]2(s) + 4NH4+(aq) + 6K+(aq)
                                                                                       cream to blue-grey