Ions in the following groups can coexist in large quantities in solution

H2CO3 in DA decomposes to produce water, so it cannot coexist.

There is a BASO4 precipitate in B, which also does not meet the conditions for coexistence.

H+ and OH in c will produce water.

Therefore, only d can coexist under the condition that there is no water generation, and no gas precipitation is generated, and sometimes it will be indicated that it is an acid condition, that is, there is h in the implied solution, and if it is alkaline, it is oh, so this is also to be noted

There will also be a mark on whether there is a color, such as K mammoanate and so on can not be commonly used, such as h and oh and h2CO3 and nitrate and so on can not coexist!

a;H+ and CO3 2- cannot coexist in large quantities (H2CO3 is unstable and easily decomposed) b; Ba2+ and SO2- cannot coexist in large quantities (BaSO4 is a precipitate)c; H+ and OH- cannot coexist in large quantities (acid-base neutralization) d; OK.

Da: H+ and CO3 form carbonic acid, which is unstable and decomposes into water and carbon dioxide.

B: Ba2+ and SO42- cannot coexist in large quantities and will form barium sulfate precipitation.

C: H+ and OH- cannot coexist in large quantities, they combine to form water.

The common situation that cannot coexist in large quantities in solution is: ion exchange reaction occurs to generate 1 precipitate 2 volatile substances 3 weak electrolytes that are difficult to ionize So the answer is d

Choose a, [al(oh)4)]-is alo2-.

Double hydrolysis of Al(OH)3 precipitates occurs with AlO2-.

h2o+alo2-=al(oh)3↓

4oh-=alo2- +2h2o

What is this aluminium hydroxide ion? Is metaaluminic acid heel? If yes.

Answer: a

I think it's d, because acetate and bicarbonate are hydrolyzed in solution, making the solution alkaline. Whereas a more alkaline solution can make alo2- exist, because there is no possibility of reaction, it will only react with hydrogen ions.

In A, the stable valence states of chlorine are -1 and +7 valence, Clo- is unstable, and SO32- is the same as B ammonium ions and bicarbonate ions react and do not coexist.

C phosphoric acid is a moderately strong acid, so HPO42- must be hydrolyzed, and the solution is alkaline, so HCO32- cannot coexist.

A, NH4+, OH-, do not coexist, will react to form NH3+H2OB, Ca2+, Al3+, NO3-, Cl-, can coexist in large quantities C, AG+, Br-, will react to form AGBR precipitate D, Fe2+, H+, NO3-, Fe2+ will undergo redox reaction with Hno3 so answer B

A is an alkaline solution of OH- ions and Fe3+ will form Fe(OH)3 precipitate.

The solution in B is an acidic solution, and the NO3- ions show strong oxidation with the slow opening, which will oxidize Fe2+ to Fe3+

c can coexist in large numbers.

In d, HCO3- ions and H+ could not coexist in large quantities.

A The phenolphthalein test solution turns red, indicating that the solution is alkaline and cannot coexist with Fe3+ (iron hydroxide precipitation is generated);

b The purple litmus test solution turns red, indicating that the solution is acidic, and in an acidic environment, ions can coexist in the option;

c Colorless solution, (mNO4)2- colored;

dBicarbonate and OH do not coexist.

a. Do not coexist. The solution that turns the phenolphthalein solution red is alkaline, and Fe3+ will precipitate B and coexist. The solution that turns the purple litmus solution red, acidic, can coexist c, not coexist.

MNO4- is purplish-red, not a colorless solution D, and does not coexist. OH- can react with NaHCO3 to obtain Na2CO3 option B

The answer is b,..Fe3+ in A cannot exist in large quantities in alkaline solution, Mno4 in C is a colored ion, and Oh- in D is not.

Choosing ab i- will reduce Fe3+ to Fe2+, and cannot coexist in large quantities;

C Po4 3- will react with H2Po4- to form Hpo4 2-, and the Voltson collapse cannot coexist in large quantities.

a Can coexist.

A correct. b.Fe(scn)3 blood-red complexes are generated.

d.A solution that reacts with aluminum to form H2 may be.

2Al+6HCl==2AlCl3+3H2 2AL+2NaOH+2H2O=2Naalo2+3H2 Ions must be able to coexist in large quantities, because if it is an alkaline solution, Cu(Oh)2 precipitate will be generated, so it is wrong.

Iron ions in selection A and B are complexed with thiocyanide to form iron thiocyanide; In C, bicarbonate and hydroxide can not coexist in large quantities, and hydrated carbon dioxide will be generated, and D reacts with aluminum to form hydrogen gas as an alkaline solution, and copper ions cannot exist in large quantities.

Select b carbonate ion reacts with hydrogen ion to release carbon dioxide gas, and the two ions cannot coexist.

Silver ions react with chloride ions to form silver chloride precipitates, and the two ions cannot coexist.

Calcium ions react with carbonate ions to form calcium carbonate precipitates, and the two ions cannot coexist.

Nitrate ions, sodium ions, and hydrogen ions can neither precipitate nor gas, nor water, and the three can coexist.

Coexistence, i.e., free ions do not interact and influence each other, that is, no reaction occurs.

There are three main types of ionic interactions at the high school level:

1.Metathesis reactions, in which ions interact to form precipitates, gases, or weak electrolytes (usually water), require a solubility table at the end of the book.

2.Redox reaction, there are more redox reactions between ions, but there are fewer reducing ions, mainly sulfur ion S2-, sulfite SO32-, iodine ion I-, ferrous ion Fe2+, when these ions appear, special attention should be paid to redox reaction.

3.The complexation reaction is mainly reflected in the complexation between some subgroup metal ions and some anions or small molecules, such as silver ammonia ions (that is, silver ions and ammonia molecules cannot coexist in large quantities), and the products produced when iron ions are tested by KSCN (that is, iron ions and SCN- cannot coexist in large quantities).

The answer is b

When there is no reaction, for example, CO3- and H+ in A will produce bubbles, Ag+ and Cl- in C will precipitate, and Ca+ and CO3- in D.