The problem of ion coexistence, the question of ion coexistence

A in ALO2- HCO3- promotes the non-coexistence of electricity.

NH4+ ALO2- double hydrolysis in B.

HS-double hydrolysis. Pick D

a strong acid to weak acid.

AlO2- +HCO3- +H2O = CO32- +Al(OH)3 (precipitate).

b Double hydrolysis.

NH4+ +ALO2- +2H2O = NH3·H2O + Al(OH)3 (precipitation).

c Double hydrolysis.

Al3+ +3Hs- +3H2O = Al(OH)3 (precipitation) + 3H2S (gas).

d Can coexist.

The answer is D

A: Dihydrolyzed AlO2- +HCO3- +H2O=Al(OH)3 +2CO32-

Reason: HCO3- is more acidic than metaaluminic acid.

b: 2nh4+ +alo2- +h2o= 2nh3↑+ al(oh)3↓

Cause: Gas, precipitation promotes the right shift of the reaction.

C: Al3+ 3Hs- +3H2O= Al(OH)3 + H2S Reason: Ammonium bicarbonate solution exists with the same 2, so ions in D can coexist in large quantities.

In A, ALO2 wants to take the proton of HCO3 (in high school terms, it is double hydrolysis) and does not coexist.

In B, it is also the same problem as A, NH4 is acidic, ALO2 is alkaline, and it is necessary to double hydrolyze.

In C, aluminum ions are strongly acidic, and HS- can coexist with hydrogen sulfide (also double hydrolysis) in acidic conditions, and NH4 is not acidic enough to make HCO3 double hydrolyzed. Thank you.

ABC can not coexist, in detail, D NH4+ will not be double hydrolyzed with acid root ions, this is a special case of ion coexistence, so choose D

d, the rest undergo a double hydrolysis reaction.

Choose C because Na will react with water in the solution first, and Na+ and OH- will be generated in the solution, and in A, H+ will react with OH- to form H2O, and the solution containing Mno4- is purple; In B, HCO3- reacts with OH- to form CO3- and H2O; In D, Ca2+ reacts with CO32- to form CaCO3 precipitate, and Ca2+ and OH- cannot coexist in large quantities.

Therefore--- choose C

Hey, hey......ky

When a small piece of sodium is added, sodium hydroxide is generated, 2Na+2H2O==2NaOH+H2 (gas mark).

To make the solution alkaline, the bicarbonate in b will react with hydroxide to form carbonate and water, ion equation:

HCO3-+OH-==CO3)2-+H2O, so choose C

，mno4-,so42-,h+

There is mno4-

Purple in color. ,na+.hco3-,no3-

When sodium grains are added, there will be NaOH, then there will be OH-, and OH- can react with HCO3-.

no3-，k+,cl-

Yes, it fits the topic.

no3-,cl-,co32-

CO32- reacts with Ca2+ to form C2CO3 precipitate.

The addition of OH- is selected to have no effect on the solution.

In B, after adding a small sodium granule, a large amount of OH- can be combined with HCO3- to form CO3-, and can be combined with BA2+ in the solution to form barium carbonate precipitate. Therefore they cannot coexist.

You can't remember this by rote. Exams will still be forgotten. You have to analyze it yourself. The above three have strong oxidizing properties. For example, ferrous ions. Halogen ions and so on can't be together, metal ions can.

And then there are the implicit conditions:

For example, 1, it is said that it is a solution of pH=3 or pH=13, which means that it is an acid solution or a alkali solution 2, the concentration of hydrogen ions ionized by water, and the concentration of hydroxide ions ionized by water, are all acidic and alkaline solutions.

3. In the following colorless solution, colored ions are excluded.

Mainly reducible ions, such as Fe2+, S2-, I-, Br-SO32-, etc.

There is also the effect of acidity, such as CO32-, CH3COO- and other weak acid ions do not coexist with H+.

The ones you listed are all strong oxidizing ions that cannot coexist with reducing ions, Fe2+, NO3-

Can not be combined with acidic potassium permanganate (H+ and Mno4-), acidic potassium dichromate (H+ and Cr2

Solution: Only when a large amount of Fe2+ solution has just been injected into the existing strong alkali solution, it is possible that the strong alkali solution contains a large amount of Fe2+, and it is impossible to contain a large amount of Fe2+ in the strong alkali solution. For the conditions given by the title, I can barely understand it this way.

The ions that cannot coexist in large quantities in the given solution are:

a Sulfate ion: (Ba2+) SO4 2-) = BaSO4 B Iron ion: (Fe3+) 3(Oh-) = Fe(Oh)3 D Copper ion:

cu2+) 2(oh-) = cu(oh)2↓

A strong alkali solution contains a large amount of Fe2+ BbA2+ K+, and it is impossible for a strong alkali to contain a large amount of Fe2+.

Among them, ferrous ions are more special, it is positive 2-valent, so it has reducibility, so when nitrate exists, hydrogen ions cannot exist at the same time: ferrous ions will also react with carbonate until the final reddish-brown precipitate of iron hydroxide will be formed: there are aluminum ions will react with carbonate, and finally aluminum hydroxide precipitation and carbon dioxide will be generated, which is a double hydrolysis reaction. ......

Carbonate and hydrogen ions in a.

B metaaluminate (also known as tetrahydroxyalaluminate Al(OH)4-) and hydrogen ion reaction.

ab is the ionization equilibrium of weak acids.

The ferric ions in C cannot coexist with hydroxide, otherwise a brownish-red precipitate (Fe(OH)3) will be produced

Therefore, D is chosen