Chemical ion test, a method for the detection of ions

The question requires different cation and anion compositions, which means that they are not reused.

From the stem of the question, it can be seen that H+ is greater than the substance is acid, and from the given anion, it can only be SO4 2 (because Oh-, HCO3- and H+ cannot coexist in large quantities, and H+ solution H+ composed of H+ and Cl-- should be equivalent to the introduction of substance B as H2SO4 accordingly

According to the knowledge that C must contain Cl-, it is known that A can precipitate with other substances, so it must contain Ba2+, then the anion can only be OH-, and the cation in C can be Mg2+.

So A is BACL2, B is H2SO4, C is MGCL2, and D is NH4HCO3. So the conclusion is incorrect

This question is a question in the 2011 Sichuan Provincial High School Entrance Examination, and the answer is D; Analysis: Only Baso4, BaCO3, and Mg(OH)2 can form white precipitates, so A can only be BA(OH)2;C(h) in solution B > indicates that B is sulfuric acid; Acrylic solution.

There is a white precipitate insoluble in dilute Hno3 in the AgNO3 solution, indicating that C is MgCl2, so Ding is NH4HCO3.

Answer: d

The four substances are:

A: Ba(OH)2B: H2SO4C: MGCL2 D: NH4HCO3, so term D is wrong.

The method of testing the ions is as follows:

1. Carbonate ions: dilute hydrochloric acid. When carbonate ions encounter dilute hydrochloric acid, carbon dioxide and water will be generated, and when gas is generated, it proves that the tested substance contains carbonate ions.

2. Chloride ions: silver nitrate and dilute nitric acid. Silver ions will combine with chloride ions to form silver chloride to form a precipitate, but since silver carbonate is also insoluble in water, dilute nitric acid is used, and when no gas is generated, it proves that the tested substance contains chloride ions.

3. Silver ions: sodium chloride. Silver ions combine with chloride ions to form silver chloride to form precipitates.

4. Hydroxide ion: acid-base indicator, colorless phenolphthalein. If phenolphthalein turns red, the substance under test contains hydroxide ions.

5. Ferrous ions: light green in the solution, that is, the substance under test contains ferrous ions.

6. Copper ions: sodium hydroxide. Copper ions combine with hydroxide ions to form copper hydroxide, resulting in a blue precipitate.

7. Ammonium ion: use alkali and moistened red litmus test paper. Ammonium ions react with alkali to produce alkaline ammonia gas, which turns litmus paper blue.

8. Barium ions: dilute sulfuric acid. Barium ions combine with sulfate ions to form barium sulfate, forming a precipitate. <>

9. Ferrous ions: add K Fe(CN) dropwise to the solution to generate a blue precipitate; KSCN solution was added to the solution, which did not appear red, and then new chlorinated water was added, and the solution turned red.

10. Iron ions: KSCN solution is added dropwise to the solution for reaction, and the solution turns blood red; NaOH solution is added dropwise to the solution to generate a reddish-brown precipitate.

The test of cation is a basic knowledge point in chemistry, and if you can't master it well, it is easy to lose points.

Methods for testing cations

1. Flame color reaction: Na+: yellow; k+: purple (seen through blue cobalt glass); ca2+: brick red;

2. H+: H+ acidic. When the purple litmus test solution turns red, when it is moistened, the blue litmus test paper turns red;

3. NH4+: Strengthen alkali (NaOH) heating in the test solution to produce gas that turns the wet red litmus test paper blue;

nh4++oh-nh3↑+h2o;nh3+h2o nh3·h2o nh4++oh-

4. Fe3+: Blood-red KSN or NH4SCN solution: Fe3++SCN-==Fe(SCN)]2+; Passage NaOH solution reddish-brown precipitate:

fe3++3oh-==fe(oh)3↓

5. Fe2+: When encountering NaOH solution, a white precipitate is generated, which is quickly converted into gray-green in the air and finally becomes a reddish-brown precipitate

fe3++2oh-=fe(oh)2↓;4fe(oh)2+o2+2h2o==4fe(oh)3 ;Add a small amount of KSCN to the test solution without obvious change, and then add chlorine water to appear in the blood manuscript red: 2Fe2++Cl2==2Fe3++2Cl-; fe3++scn-==fe(scn)]2+

6. Mg2+: white precipitate is generated in NaOH solution, and excessive NaOH precipitation is insoluble: Mg2++2OH-==mg(OH)2, but the precipitate can be dissolved in NH4Cl solution;

7. Al3+: white precipitate is generated in NaOH solution (appropriate amount), and excessive precipitation and dissolution of NaOH solution: Al3++3OH-==Al(OH)3;

al(oh)3+oh-==alo2-+2h2o

8. Cu2+: blue precipitate is generated in NaOH solution, and the heat becomes black precipitate: Cu2++2OH-==Cu(OH)2; cu(oh)2cuo+h2o

9. Ba2+: white precipitate is generated in dilute H2SO4 or sulfate solution, and dilute Hno3 precipitate is insoluble: Ba2++SO42-==BaSO4

10. AG+: Add NaOH solution to generate a white precipitate, which quickly turns into a brown precipitate and dissolves in ammonia AG++OH-==AGoh;

2agoh==ag2o+h2o;agoh+2nh3·h2o==[ag(no3)2]oh+2h2o

Dilute HCl or soluble oxide solution followed by dilute HNO3 to generate a white precipitate: AG+ +Cl-==AGCL