A solution may contain several of H, NH4, Mg2, Al3, Fe3, CO32, SO42, and NO3.

The answer is b. Sodium hydroxide is added, there is no precipitation at the beginning, and the solution contains H+, neutralizing sodium hydroxide, it must not contain nitrate ions, because nitric acid will be generated and nitric oxide will be produced. When the generated precipitate is more sodium hydroxide, it is partially dissolved, indicating that the precipitate contains aluminum hydroxide, so the solution contains Al3 and does not contain carbonate ions, because complete double hydrolysis will occur, and the precipitate is partially dissolved and white, so there is Mg2+ and no Fe3+.

The intermediate precipitation is the reaction of NH4+ and sodium hydroxide, so it must also contain NH4+. NH4+ +OH- = NH3H2O, consumes NaOH but does not produce precipitation, consumes sodium hydroxide mol, then ammonia ions have mol. The anion in the solution is only SO42-, so it must also contain SO42-.

From the image, the ratio of the quantities of the three ionic species can be calculated to be 2:2:1. Hope.

I didn't see the figure, in response to your second question, metaaluminate can't be generated in one step, even if there is, it will react with Al3+ to form Al(Oh)3, so as a whole, it is still regarded as all Al(Oh)3 is generated first, and then excessive OH- will generate metaaluminate and water.

I think this question is c

Well, first of all, the front of the image is flat, then it is oh-+h+=h2o, there is h+

Later, there was precipitation, Al3+, Fe3+, Mg2+, it is possible, and then it was precipitated a little, this is the gender of Al(OH), there must be Al3+

Look at the number of grids, 6 grids in the previous growth process, 1 grid in the precipitation reduction process, al(oh)3+oh-=[al(oh)4]-

It can be seen that 6 parts of OH- were used in the previous part, and the final precipitation was exactly half of the maximum, so the amount of Al3+ and the substance of another ion is equal. Now we know that the ion is Fe3+, which consumes 6 parts of OH- together with Al3+

Later, I saw that there must be anions, so with AL3+ and H+, don't think about CO32-, it's SO42-.

Oh yes, seeing that the amount of precipitation is still flat, then at this time it is NH4++OH-=NH3·H2O, and there is NH4+

Then absolutely none is CO32-, and what is uncertain is NA+

So look at the options, A says that there is Mg2+, but in fact there is Fe3+.

b does not necessarily have na+

C is correct and D and NH3·H2O.

Hope it helps.

D must have Na2SO4 in the end, Na comes from Naoh, SO42 - there is already there.

I'm not sure if it's right or wrong, but I tried my best.

Hope it helps!!

The correct answer is c

The reason for this is that the following reactions occurred in stages:

1. There is no precipitation when NaOH is added at the beginning, indicating that the solution is acidic, and the reaction is: H+ +OH- = H2O

2. Precipitation begins to occur (combined and then partially dissolved), and the reaction is: Al3+ +3Oh- = Al(OH)3

fe3+ +3oh- = fe(oh)3

3. The precipitation does not dissolve and does not increase, indicating that the ammonium ion participates in the reaction, and the reaction is: NH4+ +OH- = NH3 + H2O (or ammonia monohydrate NH3·H2O).

4. The precipitate is partially dissolved, indicating that Al(OH)3 and NaOH react, and the remaining precipitate is Fe(Oh)3, and the reaction is: Al(OH)3 + NaOH = NaALO2 + 2H2O

1。And because there are only two kinds of anions, and CO32 and H+, Al3+ and Fe3+ and other ions can not coexist in large quantities, there must be SO42, and the solution formed by the reaction contains Na2SO4, but due to the excess of NAOH, it should be a mixture of the two.

Al3+, SO42, CO32 plasma. When a certain amount of concentration of NaOH is added to the solution, firstly, by precipitation finally NaOH disappears there must be Al3+, then CO32- is excluded, because both will be double hydrolyzed.

First of all, there must be Al3+ after the disappearance of NaOH by precipitation, then CO32- is excluded, because both will be double hydrolyzed; For your question, you have to remember that 4molNaOH will dissolve 1molAl(OH)3, so add a block of NaOH at the end, which must correspond to the previous generation of Al(OH)3 when Al3+ will consume 3 blocks of NaOH, then there are 3 cells to generate MG(OH)2 or Fe(OH)3, there is no precipitation at the beginning to generate proof that there must be H+ in the solution, and if it does not disappear after the point, there must be NH4+, otherwise it will disappear immediately; From the data you provided, the ordinate should represent the number of moles of precipitation, what do you think the cation and OH- combine to produce the same as Al3+ to generate AL(OH)3 moles? It must be a 3-valent ion, so it must not be a mixture of Mg2+ or Mg2+ and Fe3+, otherwise it will definitely not be the case on the ordinate, the problem you ignore is probably here, think about it yourself, just remind this. Therefore, the anion in the solution must be SO42-, so the solute composition of the solution basically comes out.

Select C According to the image, the precipitation does not appear immediately, it can be known that there must be ions that react with hydroxide and do not produce precipitation, it may be ammonium or hydrogen ions, and then dissolve after the precipitation must have aluminum ions, and the precipitation is not dissolved immediately, indicating that there are ions and hydroxide reactions without precipitating, here it can be determined that the front is hydrogen ions, and the reaction here is ammonium ions. Because of the acidic conditions, the hydroxide group is neutralized first, and there is no possibility of aluminum hydroxide formation in the presence of a large number of hydrogen ions. According to the coordinates, 6mol of hydroxide is used to generate the precipitate, half of which is aluminum hydroxide, and the other half must be +3 valence.

It's iron. Positive and negative ions are needed in the solution, so there must be anions, but it is not certain what they are.

As for why it is not magnesium ions but iron ions: As can be seen from the graph, the amount of precipitation is the same for each block of sodium hydroxide added, and if it is magnesium ions, there is no guarantee that the rate of precipitation is the same as that of aluminum ions.

Be careful that the amount of precipitated material is equal. Therefore, it can only be Fe3+ and Al3+

1. Add AGNO3 to produce precipitation, indicating that it may contain one or more of CO32-, Cl-, and SO42-.

2. Add NaOH to obtain precipitated gas, indicating that it contains mol Mg2+ and mol NH4+, and it is impossible to determine whether it contains Al3+

3. Add BaCl2 acidic solution to obtain a precipitate, indicating that it contains mol SO42-

The cations mol Mg2+, mol NH4+, and the anion mol SO42-

The positive charge is less mol than the negative charge, indicating that there is also an anion, which can only be mol Cl- (because of MgCO3 precipitation).

So a, wrong. If it does not contain Al3+, it contains mol Cl-. If it contains Al3+, it is greater than mol

b. Error. It is not possible to know for sure if AL3+ exists.

c. Correct. d. Error. mol SO42-, concentration mol L option C

(1) Must have magnesium ions, aluminum ions, SO42 - there are no Ba2+ ions, ferrous ions, CO32 - there may be Cl - I -

2) Barium + + SO4 2 - = BASO4.

3) After causing a small amount of chlorine, add the starch solution to see if the presence or absence of blue is produced if it contains i-instead.

Add excess Ba(NO3)2 and add silver nitrate to the supernatant if the Cl-opposite of the white precipitate.

The solution evaporates until dry, burning, and looks with or without the chromogenic reaction yellow, if there are Na+ ions present and not.

(1) Adding agno3 solution to the first solution produces precipitation

There may be CO3-, Cl-SO42-

2) After adding a sufficient amount of NaOH solution to the second solution for a full reaction, the final precipitate is filtered, washed, and dried, the same below), and the gas is collected at the same time, and all the gases are escaped from the solution).

There must be: NH4+ Mg2+ There may be: Al3+ (3) Add a sufficient amount of hydrochloric acidified BaCl2 solution to the third solution, and obtain a precipitate after a full reaction.

There must be: SO42-

From this, it can be seen that the following statement about the composition of the original solution is correct: (c) I hope it helps you!

Select C, from (1) know that there is at least one carbonate, sulfate, and chloride ions, and from (2) know that magnesium ions must exist, because it is an excess of sodium hydroxide, even if there are aluminum ions, it will only generate metachlororate, so there must be magnesium ions. From (3), it is known that there must be sulfate ions, because there are enough hydrogen ions in the hydrochloric barium chloride solution to generate barium carbonate precipitate, and barium sulfate precipitation is, the amount of sulfate can be obtained, divided by 100ml to obtain the concentration, not moles per liter.

Answer C, add human silver nitrate has a precipitate, containing chloride ions or carbonate silver sulfate is a slightly soluble, add a sufficient amount of sodium hydroxide to obtain a precipitate, containing magnesium ions, can not be determined to contain aluminum ions, there is gas, indicating that it contains ammonium ions, adding hydrochloric acid acidified barium chloride has a precipitation, indicating that the solution has no carbonate, containing sulfate is three moles per liter, here can not be determined to contain aluminum ions, aluminum ions and carbonate can not coexist, there is double hydrolysis.

The solution must contain cations or: H+, NH4+, Al3+, and the ratio of the amount of each ion is: H+: NH4+: Al3+ = 2:3:1

Certainly do not contain more than the cuffy cations are covered: Mg2+, Fe3+,

Because there are three kinds of anions that exist in large quantities in the solution given by the title: SO42-, I-, and S2-, and it has been determined that there are no two of them: I-, S2-; The solution itself as a whole is electrically neutral (the total number of positive charges of cations in the solution is equal to the total number of negative charges of anions), and it is impossible to have a solution that only contains cations and does not contain anions, so in order to ensure that the solution as a whole is electrically neutral, it must contain anion SO42-; Ba2+ cannot coexist with SO42-, so it must not contain Ba2+.

Hope it helps!

There must be anions in the solution, since there is no I and S2, it can only contain SO42-, because it contains SO42-, it does not contain BA2+, otherwise BASO4 precipitates.

If there are none of SO42- and Ba2+, then what is the anion?

It must contain sulfate ions because: ammonia ions are cations, so there must be anions. There may be three anions in the solution, and both are excluded so it can only be sulfate ions.

Because these two ions will combine to form barium sulfate, it is a precipitate. The title says that there is a certain solution, so it must not contain precipitation.

There are three anions, after excluding I- and S2-, there is only SO42-, and the anion and cation in a solution must have anions to balance, so it must contain SO42-, and if it contains Ba2+, it will form barium sulfate precipitation, so there can be no Ba2+.

The solution is electrically neutral, and since there is none, there must be a sucate root.

1. No AG+, CO32-.

2. If there is Fe2+, then there is no OH-, then there must be SO42-, because there is only this kind of anion left. Then there is no ba2+.

3. There is NH4+.

There must be Fe2+, NH4+, SO42-, and there must be no BA2+, AG+, OH-, CO32-.

**Is there mg2+?

Because the ferrous hydroxide precipitation will be oxidized by the oxygen in the air to form Fe(OH)2 on the surface as soon as it is formed, the precipitation is not white.

White precipitate. If Fe(OH)2 is used, it will quickly transform into a reddish-brown precipitate.

Because the rotten hunger is Fe(Oh)2 is unstable, it will change from white to gray-green, and finally the hunger will return to Fe(Oh)3, which is the final product of the most beautiful child, so it cannot be Fe(Oh)2

Hope it helps, hope, thank you.