Moe Kare!!

Moe Kare!! (萌えカレ!!)

 

Categories:	comedy, mature, romance, shoujo
Author:	Ikeyamada Go
Artist:	Ikeyamada Go
Original Publisher:	Shogakukan
Magazine:	Sho-Comi
Start Date:	2005

       Volume:               7

Sypnosis

         Wakamiya Hikaru is a 15 year old teenager who read Shoujo mangas all the time. She wants a boyfriend that'll protect her and be her prince, just like in the mangas. One day... she gets into a tough situation and she was saved by a handsome boy. In return for saving her, she gets kissed by the boy who she calls the "Perverted Prince." Later on, she finds a guy who looks exactly like him, but has different eyes and a complete different personality.

Characters

Wakamiya Hikaru: 15 years old at the beginning of the story. Wakamiya likes reading Shoujo mangas and she often daydreams about her Prince Charming. Throughout the story she is usually labelled as "idiotic" but "nice and thoughtful" by Arata. Wakamiya first developed love towards Takara, an upperclassman. Initially, she detested Arata for his "perverted mind" and also for his immense hatred for Takara. Nevertheless, she gradually learned that Arata is actually a lonely yet gentle person, and got along well with him. After her break-up with Takara, Hikaru underwent a severe depression, at which time Arata was always by her side. Step by step, Wakamiya found herself harboring feelings for Arata as well. At the end of the story, Wakamiya chose not to return to Takara but to stay by Arata's side, as she finally realized that her heart had belonged to Arata now.

Ichikawa Arata: 15 years old at the beginning of the story. Arata is popular with girls and is also called "Perverted Prince" by Hikaru. He is also the half-brother of Ichikawa Takara. Arata harbored feelings of hostility towards Takara for his dad had abandoned Arata's mother, leaving her bearing their child alone. Arata also developed feelings for Hikaru, who at that time was Takara's girlfriend. After Hikaru's break-up with Takara, Arata stayed by her side, cheering her up with his gentleness and ardent love, thus healing Hikaru's broken heart. When Takara returned, Arata decided to give up on Hikaru for the sake of her happiness. Hence, when he ascertained that Hikaru chose him but not Takara, he was overcome with joy and felicity.

Honda Takara: 16 years old at the beginning of the story. Takara may have the same face as Arata, but his personality is completely different: gentle, gallant and considerate. Takara also loves Hikaru but he himself admits that his feelings for her are contrary from those of Arata: whereas his feelings are selfish, Arata's are altruistic. After his childhood friend - Kobayakawa's accident, Takara broke up with Hikaru, thinking that it would be best if Takara stayed by Kobayakawa's side to compensate for her rescuing him. Afterwards, Takara began to repent on his doings and asked if Hikaru could return to his side. However, Takara himself knew that Hikaru's heart had now belonged to Arata's and expressed doubt on whether he could win. At the end of the story, upon hearing Hikaru's rejection, he bid goodbye to her for the very last time, resigning to the fact that he had lost her forever.

credits: http://en.wikipedia.org/Moe_Kare!!

Permanganometric Analysis

Definition Permanganometric

     Permanganometric is one of the quantitative volumetric analysis based on the reaction of permanganate ion oxidation.
Standard solution used is KMnO4. Before used for titration, a solution of KMnO4 must be standardized in advance because not a primary standard solution. Additionally KMnO4 has the following characteristics:
1. Can not be obtained in pure
2. Containing oxides MnO and Mn2O3
3. The solution is not stable (if any organic substance) 

Reaction:

4     MnO₄^- + 2 H₂O → 4 MnO₂ + 3 O₂ + 4 OH^-

4. It should not be filtered with filter paper (organic matter) →with glass wool
5. Should be stored in brown bottles
6. Standardized with primary standard solution. Primary standard substance which is used such as: As₂O₃, Na₂C₂O₄, H₂C₂O₄, Fe(NH₄)₂(SO₄)₂, K₄Fe(CN)₆, logam Fe, KHC₂O₄H₂C₂O₄2H₂O

      Permanganate ion oxidation can take place under acidic, neutral and alkaline. 

In acidic conditions, pH ± 1
Reaction: MnO ^-+ 8 H⁺+ 5 e → Mn²⁺+ 4 H₂O 

       Potassium permanganate can act as an indicator, and titration is generally done in the sour atmosphere because it will be easier to observe the end point titration. But there are some compounds more easily oxidized in an atmosphere of neutral or alkaline example hydrazine, sulfite, sulfide, sulfide and thiosulfate. 

      Reaction in the neutral atmosphere:

      MnO ^- + 4 H⁺+ 3 e → MnO₂ + 2 H₂O

      Reactions in the atmosphere of alkalis or bases :

MnO ^-+ 3e → MnO₄^2-

MnO ^2-+ 2H₂O + 2e → MnO₂ + 4OH^-

MnO ^-+ 2H₂O + 3e → MnO₂ + 4OH^-

Advantages and Disadvantages Analysis with Permanganometri

- The advantages
1. Standard solution, namely KMnO4 easily available and cheap.
2. No need for indicators to TAT. That's because KMnO4 can act as an indicator.
3. The reaction was fast with lots of reagents.
            - Disadvantages
1. There should be standardization of the initial advance.
2. Can take better if done in acidic conditions.
3. The time required for analysis is quite long.

credits: Book of PDTK 1 2011 Chemical Engineering of Diponegoro University

Iodo-Iodimetric Analysis

        Chemical reactions involving the oxidation reduction is widely used by titrimetric analysis. Ions of various elements may be present in conditions of varying oxidation, resulting in the possibility of many redox reactions. Many of these reactions are eligible to be used in titrimetric analysis and its applications quite a lot. 
The titrimetric iodometric analysis which indirectly to substances that are like iron oxidizing III, copper II, where this substance is added to oxidize iodide to form iodine. Iodine is formed will be determined through the use of thiosulfate standard solution.

       Definition of Reduction - Oxidation process of reduction - oxidation (redox) is a process involving transfer of electrons from one reactant to another reactant.
reduction
    Reduction is the capture of one or more electrons by an atom, ion or molecule.
oxidation
     Oxidation is the release of one or more electrons from an atom, ion or molecule. There are no free electrons in a chemical system, and the release of electrons by a chemical substance is always accompanied by electron capture by the other, in other words always followed by oxidation reduction reaction. In the oxidation reduction (redox) changes the valence of the substances that hold the reaction. Here occurred the transfer of electrons from a reducing pair to pair oxidizing. 
The second half reaction of a redox reaction can generally be written as follows: red → Ox + n é where red indicates the reduced form (also called a reductant or reducing agent), Ox is the oxidized form (oxidant or oxidizing agent), n is the number of electrons transferred and é are electrons. Redox reactions are widely used in titrimetric analysis of inorganic substances or organic. To set the end point on the redox titration can be done with the help of potentiometric or indicator.
Examples of redox reactions:

5Fe²⁺ + MnO₄ + 8H⁺ → 5Fe³⁺ + Mn²⁺ +4H₂O

Where: 5Fe²⁺ → 5Fe³⁺ + 5e → an oxidation reaction
           MnO₄ + 8H⁺ + 5e → Mn²⁺ + 4H₂O is a reduction reaction
Is the titrimetric iodometric analysis which indirectly to substances that are like iron oxidizing III, copper II, where this substance is added to oxidize iodide to form iodine. Iodine is formed will be determined through the use of thiosulfate standard solution.

Oxidator + MnO₄ + 8H⁺ + 5e → Mn²⁺ + 4H₂O 

KI → I₂ + 2e

I₂ + Na₂S₂O₃ → NaI + Na₂S₄O₆

  Iodimetri is a titrimetric analysis that directly used for the reducing agents or sodium thiosulfate with iodine solution or by addition of excess standard solution. Excess iodine is titrated back with thiosulfate solution.
Reductant + I2 → 2I-
Na2S2O3 + I2 → NaI + Na2S4O6

Theory of Indicators Amylum
  The indicators used in this method is an indicator of starch or amylum. Amylum is a strong indicator of iodine, which is blue when a substance containing iodine positive. Reason wearing amylum as indicators, including:
- The price is cheap
- Easily obtained
- Changes color when TAT is clear
- Reaction is spontaneous (without heating)
- Can be used at once in the iodo-iodimetri
 
While the weakness of this indicator are:
- Unstable (easily hydrolyzed)
- Easily damaged (infected bacteria)
- It is difficult soluble in water

How to make amylum indicators:
Prepare a 250ml glass beaker, fill with 100ml distilled water. Then enter the 3 grams of starch in 250 ml glass beaker. Heat while stirring until the temperature of 400C, 400C stop after stirring but still heated to a temperature of 600C. Enter the glass beaker into a black plastic bag and then put in a drawer. Let stand about 5 minutes to settle and form three layers, after taking the center of the cold solution as indicator. The reaction mechanism of the reaction mechanism is the reaction stages that describe a whole set of chemical reactions.
Iodo-iodimetri reaction mechanism: 

 2 Cu²⁺ + 4 I^- → 2 CuI + I₂

I₂ + 2 S₂O₃^2- → 2 I- + S₄O ^2-

I₂ + I^- → I₃ ^-

Amylum + I₃^- → AmylumI₃ ^- (blue)

Things Must Be Considered:

1. Titration should be carried out in the cold, in erlenmeyer without catalyst in order to reduce the oxidation of I-by O2 from the air into I2.
2. Na2S2O3 is a secondary standard solution should be standardized first.
3. The addition of the indicator at the end of titration (just before TAT).
4. Titration can not be done in a medium strong acid hydrolysis amylum because it will happen.
5. Titration can not be done in strong alkaline medium because of I2 will oxidize thiosulfate to sulfate.
6. Na2S2O3 solution must be protected from light because light helps bacterial activity that interferes thioparus. 

 
 credits: Book of Basic Praticum of Chemical Engineering 2011

Anions - Qualitative Analysis

Anion Identification Tests

§  Carbonate Ions: The most characteristic reaction of carbonate is the formation of carbon dioxide upon treatment with acid:

CO₃^2-(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l)

The colorless, odorless carbon dioxide can be identified by bubbling it through a saturated solution of barium hydroxide, with which it forms a white precipitate of barium carbonate.

CO₂(g) + Ba²⁺(aq) + 2OH^-(aq) → BaCO₃(s) + H₂O(l)

Assemble a gas-liberation apparatus from a small test tube and a section of bent tubing. Dissolve or suspend a portion of your compound in a small amount of water and place it in the small test tube. Add about 0.5 mL of 6 M HCl and quickly fit the tube into the small test tube, allowing the gas liberated to bubble into a 6" test tube of saturated Ba(OH)₂ solution. The formation of a white precipitate in the large test tube (if the gas liberated is odorless) is a positive test for carbonate. It is imperative to test the gas-liberation apparatus by adding HCl to Na₂CO₃.

§  Sulfide Ions: When treated with nonoxidizing acids (HCl, CH₃COOH) sulfides react to liberate H₂S gas (rotten-egg odor). If the sulfide is very insoluble liberation of the gas may require concentrated acids (indeed some sulfides, HgS, CuS, are so insoluble that dissolution requires special treatment). The gas is generally identified by its odor and its precipitation of colored sulfides of various metal ions. Sulfides or hydrogen sulfide also are oxidized to elemental sulfur and sulfate by oxidizing agents such as permanganate, nitric acid, sulfuric acid, Fe(III), etc.

3H₂S(aq) + 2H⁺(aq) + 2NO₃^-(aq) → 2NO(g) + 4H₂O + 3S(s)

Acidify a sample with 6 M hydrochloric acid and warm. Cautiously smell the gas evolved and also test the gas with a piece of filter paper moistened with lead acetate solution. A foul smelling gas which turns lead acetate paper black constitutes a positive sulfide test.

§  Sulfate Ions: Sulfate is conveniently identified by precipitation of BaSO₄. Other insoluble barium salts contain anions of weak acids (CO₃^2-, SO₃^2-and PO₄^3-). Precipitation of these anions is prevented by acidifying the solution.
Acidify the test solution with 6 M HCl, and add a few drops of 0.2 M BaCl₂₄^2-. solution. A white precipitate indicates the presence of SO

§  Nitrate Ions: The most notable feature of the chemistry of the nitrate ion is its oxidizing ability as illustrated by the following reactions:

3Fe²⁺(aq) + 4H⁺(aq) + NO₃^-(aq) → NO(g) + 2H₂O + 3Fe³⁺(aq)

In the last reaction the nitrogen oxide reacts with excess Fe²⁺ to give the brown complex ion Fe(NO)²⁺. It is the formation of this brown complex that is used to identify NO₃^- (called the brown ring test).
Acidify about 2 mL of the test solution with 3 M H₂SO₄ and then dissolve one-half spatula full of solid FeSO₄.7H₂O in the acidified solution. Cool the solution and then carefully introduce about 0.5 mL of concentrated H₂SO₄ by allowing it to flow down the side of the tilted test tube. Allow the solution to sit undisturbed so that the sulfuric acid forms a definite layer. The formation of a brown color at the interface of the layer constitutes a positive test for nitrate.

§  Phosphate Ions: The precipitation usually used to identify phosphate is the formation of yellow ammonium molybdophosphate from ammonium molybdate in acidic solution.

12MoO₄^2- + 3NH₄⁺ + PO₄^3- + 24H⁺ → (NH₄)₃[P(Mo₁₂O₄₀)] + 12H₂O

Acidify the sample with concentrated nitric acid and add several drops in excess. Then treat the solution with ammonium molybdate reagent and warm. The formation of a yellow crystalline precipitate confirms the presence of phosphate.

§  Chloride, Bromide, and Iodide Ions: All three of these anions form insoluble silver salts. Although the precipitates are of different colors (AgCl white, AgBr cream, AgI yellow) the colors are difficult to distinguish, and confirmatory tests are necessary.
Silver chloride, the most soluble of the three, dissolves readily in 6 M NH₃₃, a much higher concentration of NH₃ being required to form the complex. solution because of formation of the ammonia complex. Furthermore, when the solution of the ammonia complex is acidified, AgCl reprecipitates. Neither AgBr nor AgI will dissolve readily in 6 M NH

Cl^-(aq) + Ag⁺(aq) → AgCl(s)
AgCl(s) + 2NH₃(aq) → Ag(NH₃)²⁺(aq) + Cl^-(aq)
Ag(NH₃)²⁺(aq) + Cl^-(aq) + 2H⁺(aq) → AgCl(s) + 2NH₄⁺(aq)

Bromide and iodide are usually identified by oxidation to the free elements with chlorine. The elements thus formed are extracted into carbon tetrachloride and identified by their color.

2Br^-(aq) + Cl₂(g) → Br₂(g) + 2Cl^-(aq)
2I^- (aq) + Cl₂(g) → I₂(s) + 2Cl^-(aq)

§  Chloride: Acidify the test solution with 3 M HNO₃. Then add several drops of 0.1 M AgNO₃. If a white precipitate forms, centrifuge and remove the supernatant. To the precipitate add 6 M NH₃ with stirring. If the precipitate dissolves, add 6 M HNO₃ to the solution. A white precipitate will form if the original test solution contained Cl^-.

§  Bromide and Iodide: Acidify the sample with several drops of 6 M HCl and add 4-5 drops of carbon tetrachloride. Then add about 0.5 mL of chlorine water and shake. Appearance of an orange-brown carbon tetrachloride layer indicates the presence of bromide. Formation of a purple layer indicates iodide.
 

credits: http://www.wiredchemist.com/chemistry/instructional/laboratory-tutorials/qualitative-analysis

Cations - Qualitative Analysis

Cation Identification Tests

• Generally Soluble Cations

§  Ammonium Ions: Take a small amount of the material to be tested and place it in a 50-mL beaker. Add 6 M NaOH and smell cautiously. The odor of ammonia indicates the presence of ammonium ions. If you do not smell ammonia, warm the beaker and again smell the emitted vapors. The liberated ammonia will also change the color of a moistened strip of red litmus paper held at the entrance of the test tube.

NH₄⁺(aq) + OH^-(aq) → NH₃(g) + H₂O

This test is very reliable. It should be performed whenever the generally soluble cations, NH₄⁺, Na⁺, and K⁺, are suspected.

§  Sodium Ions: The most common method of identification of Na⁺ is the flame test. Sodium imparts a brilliant, long lasting, yellow flame that masks colors from other ions. The test may be performed on a small sample of the unknown treated with concentrated HCl or a few drops of solution unknown treated with concentrated HCl. The flame should be bright and it should last as long as that of 0.1 M NaCl. Sodium is a common impurity and traces will be found in almost any unknown. You must learn to distinguish between an unknown that has sodium ion as the cation and an unknown that has sodium ion as an impurity.

§  Potassium Ions: The most common method of identification of K⁺ is the flame test. The test may be performed on a small sample of the unknown treated with concentrated HCl or a few drops of solution unknown treated with concentrated HCl. The violet flame is not intense but it is clearly visible in the absence of sodium ions. Cobalt glass filters yellow light from sodium impurities and allows the violet flame to be seen. Do not confuse the glowing wire for the potassium flame.

2.      Cations That Form Insoluble Chlorides

§  Silver Ions: Although Ag⁺, Pb²⁺, and Hg₂²⁺ all form insoluble white chlorides, Ag⁺ is the only one of these cations that forms an ammonia complex. Therefore, AgCl dissolves readily in aqueous NH₃. When the resulting solution is acidified with HNO₃, AgCl reprecipitates.

AgCl(s) + 2NH₃(aq) → Ag(NH₃)²⁺(aq) + Cl^-(aq)
Ag(NH₃)²⁺(aq) + Cl^-(aq) + 2H₃O⁺ → AgCl(s) + 2NH₄⁺ + 2H₂O

Add 3 M HCl dropwise to the solution being tested. If a white precipitate is formed, centrifuge and remove the supernatant liquid. Add 6 M NH₃ solution to the precipitate. If the precipitate dissolves, add 6 M HNO₃. Formation of a white precipitate indicates Ag⁺.

§  Lead Ions: Although PbCl₂ is insoluble at room temperature, its solubility is increased dramatically at higher temperatures; it dissolves readily in boiling water. Pb²⁺₃COO)₂. The addition of chromate ion to this lead acetate solution yields a precipitate of yellow lead chromate. also forms an insoluble white sulfate, which dissolves in a solution containing acetate ion due to the formation of the weak electrolyte, Pb(CH

Pb²⁺(aq) + SO₄^2-(aq) → PbSO₄(s)
PbSO₄(s) + 2CH₃COO^-(aq) → Pb(CH₃COO)₂(aq) + SO₄^2-(aq)
Pb(CH₃COO)₂(aq) + CrO₄^2-(aq) → PbCrO₄(s) + 2CH₃COO^-(aq)

To the solution to be tested add 3 M HCl dropwise. (A large excess of HCl must be avoided because of the formation of the soluble chloro complex, PbCl₄^2-.) Centrifuge and remove the supernatant from the white precipitate (PbCl₂). Add hot water to the precipitate and stir. If the precipitate dissolves, Pb²⁺ is indicated. Add 3 M H₂SO₄ to the hot solution. Centrifuge and remove the supernatant liquid from the white precipitate (PbSO₄). To the precipitate add 3 M NH₄(CH₃COO) and stir. If the white precipitate was PbSO₄, it will dissolve. To confirm, add a few drops of 0.5 M K₂CrO₄ to the resulting solution. A yellow precipitate of PbCrO₄ indicates the presence of Pb²⁺.

§  Mercury(I) Ions: When Hg₂Cl₂ is treated with aqueous NH₃ a reaction occurs in which free mercury and amidochloromercury(II) are formed.

Hg₂Cl₂(s) + NH₃(aq) → Hg(l) + HgNH₂Cl(s) + HCl(aq)

The HgNH₂Cl is a white solid, while the Hg in a finely divided state appears black. The resultant mixture is gray to black.
Add 3 M HCl to the solution to be tested for Hg₂²⁺. If a white precipitate forms, centrifuge and remove the supernatant liquid. To the precipitate, add 6 M NH₃ and stir. The appearance of a gray to black precipitate is positive for Hg₂²⁺.

3.      Cations That Form Insoluble Sulfates Identification tests for Pb²⁺ and Ag⁺ (Ag₂SO₄ is moderately soluble) are described above (Cations that form Insoluble Chlorides). Ba²⁺, Sr²⁺, and Ca²⁺ form moderately soluble sulfates.
The alkaline earth ions Mg²⁺, Ca²⁺, Sr²⁺, and Ba²⁺ are one of the best examples of a periodic relationship among the elements of a family. Solubilities of their compounds are graduated nicely and the separations (qualitatively) can be accomplished readily. Flame tests are very important.

§  Barium Ions: Barium ions can be identified by precipitation of its insoluble yellow BaCrO₄²⁺ or Sr²⁺ are present they will also precipitate in the presence of high concentrations of CrO₄^2-. However, the chromates of Ca²⁺ and Sr²⁺ are moderately soluble; their precipitation can be prevented by addition of acetic acid. This weak acid provides sufficient hydronium ions to lower the CrO₄^2-₄ and SrCrO₄ in solution but to allow the BaCrO₄ to precipitate. salt. If Ca concentratiion enough to keep CaCrO

2CrO₄^2-(aq) + 2H⁺(aq) → Cr₂O₇^2-(aq) + H₂O

The flame test on the solid chromate is important for confirmation.
To about 1 mL of solution add 10 drops of 6 M CH₃COOH. Then add a few drops of 0.5 M K₂CrO₄ solution. The appearance of a yellow precipitate indicates the presence of Ba²⁺. To confirm, dissolve the precipitate in concentrated HCl and flame test.

§  Strontium Ions: Strontium can be identified, in the absence of calcium, by precipitating its sulfate. To the solution add 0.1 M H₂SO₄ dropwise. The formation of a finely-divided, crystalline, white precipitate indicates the presence of Sr²⁺. (Ba²⁺ must be absent, of course.) To confirm, dissolve the precipitate in concentrated HCl and flame test.

§  Calcium Ions: If Ba²⁺ and Sr²⁺ are absent, Ca²⁺ may be precipitated as the oxalate from neutral or alkaline solutions. Test the acidity of the solution with litmus paper. If it is acidic, add 3 M NH₃ until basic. Then add 0.2 M (NH₄)₂C₂O₄²⁺. Confirm by adding a few drops of concentrated HCl and flame testing. solution. The formation of a white precipitate indicates the presence of Ca

4.      Cations That Form Ammonia Complexes

§  Cadmium Ions: Cadmium forms a yellow precipitate with sulfide ion either from a neutral solution containing free Cd²⁺ or from an ammoniacal solution of Cd(NH₃)₄²⁺. Since most sulfides are insoluble, and many of them are black, the presence of other metal ions may make it difficult to detect the yellow color of CdS. Therefore, separations must be as complete as possible before testing for Cd²⁺.

Cd²⁺(aq) + S^2-(aq) → CdS(s)
Cd(NH₃)₄²⁺(aq) + S^2-(aq) → CdS(s) + 4NH₃

To a solution of Cd²⁺ or to a solution thought to contain Cd(NH₃)₄²⁺ add 0.1 M Na₂S solution dropwise. The formation of a yellow precipitate confirms the presence of Cd²⁺.

§  Copper(II) Ions: The very distinct deep blue color of the copper ammonia complex can be used to identify Cu²⁺. This identification can be carried out in the presence of other cations which form either colorless ammonia complexes or white precipitates. Thus, Zn²⁺, Cd²⁺, Al³⁺, among others, will not interfere.
In relatively dilute solutions the color of the ammonia complex may not be intense enough to give an unqualified identification, and some other test for confirmation must be used. Cu²⁺ forms a very insoluble reddish-brown hexacyanoferrate(II).

2Cu²⁺(aq) + Fe(CN)₆^4-(aq) → Cu₂Fe(CN)₆(s)

Other cations that react with this reagent to form highly colored precipitates must be absent (Co²⁺ and Fe³⁺ for example). Acidify the test solution with acetic acid. Then add a few drops of 0.1 M potassium hexacyanoferrate(II) solution (K₄Fe(CN)₆). A red-brown precipitate confirms the presence of Cu²⁺.

§  Nickel(II) Ions: Nickel(II) is one of the easiest cations to identify. Ni²⁺ forms a red precipitate with dimethylglyoxime in a buffered acid solution. Palladium(II) is the only other cation which forms a precipitate with this reagent. However, a few other cations can interfere. Cobalt(II) preferentially forms a dark brown solution with dimethylglyoxime, and excess reagent must be used in its presence.
Acidify the solution to be tested with 6 M CH₃COOH. Then add about one mL of 0.2 M NaOOCCH₃ solution. Add dimethylglyoxime solution dropwise. A bright red precipitate is positive for Ni²⁺.

§  Zinc Ions: Zinc forms one of the few insoluble white sulfides. It is precipitated from a solution of the ammonia complex. Small traces of cations that form dark colored sulfides will obviously interfere.
Add an excess of 3 M NH₃ to the test solution, so that any zinc present is in the form of Zn(NH₃)₄²⁺. Then add a few drops of 0.1 M Na₂S solution. A white precipitate indicates the presence of Zn²⁺.

5.      Cations That Form Amphoteric Hydroxides

§  Aluminum Ions: Aluminum is generally identified by making use of the amphoteric property of its hydroxide and the red color of the "lake" AlOH₃ forms with the reagent, aluminon. Aluminon is a dye (an organic molecule, usually fairly large, that absorbs visible light). As the Al(OH)₃ precipitates the dye is adsorbed on the Al(OH)₃ particles. The adsorption of the dye is called "laking." Aluminum is a fairly common impurity and care must be taken that trace quantities are not reported. Since most laboratory manipulations are carried out in glass containers, silica gel, which physically resembles aluminum hydroxide, is also a common impurity.
Adjust the pH of about 1 mL of the test solution (with 3 M NaOH and 3 M HNO₃) to precipitate the hydroxide. Centrifuge the mixture. Remove the mother liquor with a capillary pipet and wash the precipitate with distilled water. Centrifuge the mixture. Remove the mother liquor with a capillary pipet and wash the precipitate with distilled water. Centrifuge the mixture and remove the mother liquor with a capillary pipet. These repeated washings remove other ions from the precipitate. Dissolve the precipitate in 3 M HNO₃. If any precipitate does not dissolve in the nitric acid, remove the supernatant to a clean test tube and discard the residue. Add two drops of aluminon reagent (avoid any excess). Add 3 M NH₃(aq) until the solution is basic. Centrifuge. A red, gelatinous precipitate (sometimes called a red lake) indicates Al³⁺.
Any precipitate that remains after the addition of the nitric acid is probably silica gel, SiO₂•xH₂O. Silica gel is present in many solutions; it is leached from glass containers. Any silica gel present must be removed before the addition of the aluminon and the ammonia because silica gel will also give a red lake.
Do not confuse traces of red-brown ferric hydroxide for the red lake. Other precipitates will also form colors with the reagent. The supernatant liquid will be an intense blue-purple color if too much reagent has been added. This color has nothing to do with the presence of aluminum. The color of the reagent is sensitive to changes in pH, (the reagent is an acid-base indicator).

§  Chromium(III) Ions: Chromium can be taken through a series of colored tests which leaves no doubt as to its identity. Chromium(III) forms a steel green hydroxide which dissolves in excess strong base to give a deeply green colored solution of the hydroxy complex. Treating this complex with 3% hydrogen peroxide gives the yellow solution of the chromate ion, which upon acidification with dilute nitric acid gives the orange color of dichromate. Treatment of the cold solution of dichromate with 3% hydrogen peroxide gives the intense blue color of a peroxide of chromium. (The actual composition of this peroxide is not known, but it is believed to have the empirical formula CrO₅.) This peroxide readily decomposes to the pale violet color of the original hydrated chromium(III) ion. In low concentrations of dichromate the blue color is fleeting, and attention must be focused on the test tube during the addition of the hydrogen peroxide to avoid missing the color change.

Cr(OH)₄^- (green) --(H₂O₂)--(OH^-)→ CrO₄^2- (yellow)
CrO₄^2- --(H⁺)→ Cr₂O₇^2- (orange)
Cr₂O₇^2- --(H₂O₂)--(HNO₃)→ CrO₅ (blue)→ Cr(H₂O)₆³⁺ (violet)

The following color changes are all indicative of Cr³⁺. Add an excess of 6 M NaOH to about one mL of test solution. To this green solution add 10 drops of 3% H₂O₂. Heat the test tube in the water bath until the excess H₂O₂ is destroyed as indicated by the cessation of bubbles. Acidify the yellow solution with 3 M HNO₃. Cool the resulting orange solution in an ice bath. To the cooled solution add a drop or two of 3% H₂O₂ and observe the immediate fleeting blue color.

§  Tin(IV) Ions: Sn⁴⁺ is most conveniently identified by reduction of Sn⁴⁺ to Sn²⁺ with iron. The Sn²⁺ solution is treated with HgCl₂ solution, whereupon Sn²⁺ is oxidized to Sn⁴⁺ and, simultaneously, HgCl₂ is reduced to Hg₂Cl₂ (a silky, white precipitate). The Hg₂Cl₂ is further reduced by Sn²⁺ to Hg, which appears black.

Sn⁴⁺(aq) + Fe(s) → Sn²⁺(aq) + Fe²⁺(aq)
Sn²⁺(aq) + 2HgCl₂(aq) → Sn⁴⁺(aq) + Hg₂Cl₂(s) + 2Cl^-(aq)
Sn²⁺(aq) + Hg₂Cl₂(s) → Sn⁴⁺(aq) + 2Hg(l) + 2Cl^-(aq)

Add some concentrated HCl to the solution to be tested for Sn⁴⁺. Place an iron brad (or small iron wire) in this solution and heat in a water bath for 5 minutes. Take the clear solution (filter if necessary) and add HgCl₂ solution dropwise. The appearance of a silky, white precipitate, which then turns black, confirms the presence of tin.

6.      Other Cations

§  Manganese(II) Ions: Manganese is easily identified by oxidation of Mn²⁺ to purple MnO₄^- using sodium bismuthate (NaBiO₃). Heat must be avoided to prevent the decomposition of permanganate ion to brown, insoluble manganese dioxide. Chloride ion must be absent, because it reduces permanganate ion to either manganese dioxide or manganese(II) depending upon the conditions.
Acidify the test solution with 3 M HNO₃. Add solid NaBiO₃ and stir. Centrifuge. If the supernatant has the characteristic purple color of MnO₄^-, Mn²⁺ was present.

2Mn²⁺ + 14H⁺ + 5NaBiO₃ → 5Bi³⁺ + 5Na⁺ + 7H₂O + 2MnO₄

§  Bismuth(III) Ions: Bismuth(III) forms a highly insoluble hydroxide which upon treatment with the hydroxy complex of tin(II) is immediately converted to free bismuth, a black precipitate.

3Sn(OH)₄^2-(aq) + 2Bi(OH)₃(s) → 2Bi(s) + 3Sn(OH)₆^2-(aq)

Precipitate Bi³⁺ from the test solution with 3 M NaOH and centrifuge the precipitate. Then, to a solution of tin(II) chloride add with stirring 6 M sodium hydroxide until the precipitate of tin(II) hydroxide which first forms just redissolves. This solution is then added dropwise to the precipitate of bismuth(III) hydroxide. The rapid formation of a black color confirms bismuth.

§  Iron(III) Ions: The Fe³⁺ ion is readily identified in a dilute nitric acid solution through the blood red color of its thiocyanate complex. A large excess of reagent should be avoided.

Fe³⁺(aq) + SCN^-(aq) → Fe(SCN)²⁺(aq)

Acidify the solution with 3 M HNO₃. Then add a few drops of 0.1 M NH₄SCN solution. The solution turns red if Fe³⁺ is present.

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