The chemical formula writing principle of adding sodium bisulfate dropwise to a barium bicarbonate s

This kind of problem is generally discussed with a constant amount of 1mol of substance. The amount of barium bicarbonate in this question remains unchanged, take 1mol of barium bicarbonate, and require the dropwise addition of sodium bisulfate until the precipitation is complete, that is, the amount of barium ions and sulfate ions is 1:1, that is, 1mol Nahso4 needs to be added.

At this time, 1 mole of barium ions precipitates with 1 molar sulfate ions, and 2 moles of bicarbonate only react with 1 molar of hydrogen ions, and the ion equation is Ba2+ +SO4 2- +HCO3 - H+ = BaSO4 + H2O + CO2, and the chemical equation is BA(HCO3)2+NaHSO4=BASO4+NaHCO3+H2O+CO2.

At this time, there are still bicarbonate ions in the solution, and sodium bisulfate is added, and the hydrogen ions react with bicarbonate ions to form water and carbon dioxide.

Count a small amount as one, and then add another reactant as needed. For example, a hydrogen ion needs a bicarbonate, etc. Try.

1) Towards sodium bisulfate.

Add barium hydroxide solution drop by drop to the solution until neutral.

2nahso4+ba(oh)2=na2so4+baso4↓+2h2o2h^+ so4^2- +2oh^- ba^2+ =baso4↓+2h2o

2) Add barium hydroxide solution to sodium bisulfate solution drop by drop.

The ions completely sank into the air, and the smile brightened and rose to the lake.

nahso4+ba(oh)2=baso4↓+h2o+naohh^+ ba^2+ +oh^- so4^2- =baso4↓ +h2o

Categories: Education Academic Exams >> Gaokao.

Problem description: Add barium hydroxide solution (excess) dropwise to sodium bisulfate solution until the barium sulfate precipitate disappears. How do you write a chemical equation? What are the patterns of such reactions?

Analysis: If Qi Heguo no longer precipitates, you should think like this:

The amount of substances (that is, the number of ions) of Ba and SO4 should be the same as the amount of sodium bisulfate reacted with barium hydroxide.

ba(oh)2 + nahso4 = baso4 + h2o + naoh

It is important to see if the precipitation is complete.

PS: Barium sulfate precipitates do not go away.

This is a step-by-step reaction examined in the high and high ion equations.

When thinking, pay attention to the proportional relationship of the reactants themselves, for example, if one mole of barium hydroxide is added dropwise, one mole of barium ions and two moles of hydroxide are added.

Answer(1) 2H SO42- BA2 2OH === 2H2O BASO4

When H+ and OH- are 1:1, they are neutral, but at this point there is an excess of SO42-.

2)h++so42- ＋ba2＋oh- ===baso4↓+h2o

3) Same as 2

What do you mean: the ionic reaction that occurs in the process of making a solution neutral???

To neutral: Ba2++2OH +2H++SO=BaSO4+2H2O

The sulfate reaction is exactly the same as the excess:

ba2++oh－+h++so=baso4+h2o

Actually, this can be written according to the chemical equation:

2nahso₄+ba(oh)₂*****baso₄↓＋na₂so₄＋2h₂o

The equation for conversion into ions is:

2h⁺＋（so₄）²ba⁺＋ 2oh⁻＝＝baso₄↓＋2h₂o

Because although the ratio of sulfate and hydrogen ions in bisulfate is 1:1, it is obvious that the concentration of barium ions here is insufficient to completely react with sulfate, and finally there is still sulfate left, so the concentration of sulfate ions and hydrogen ions in the equation is not 1:1.

Because sulfate is not all involved in the reaction. Assuming that the hydrogen ions and sulfate in the solution are 2m before the reaction, 2m hydroxide should be added to make the solution neutral, and only m barium ions are added at this time, leaving m sulfate that does not participate in the reaction. So the ratio of hydrogen ions to sulfate is not 1:1

The first question, 2 molecules of sodium bicarbonate and 1 molecule of barium hydroxide react to form sodium carbonate, barium carbonate and water, and the second question, sodium carbonate and barium hydroxide react 1:1 to generate barium carbonate and sodium hydroxide, and the first step is to neutralize the reaction and consume hydrogen ions for the purpose.

The second step is the precipitation reaction, which consumes carbonate ions.

A small amount of barium hydroxide, so sodium bicarbonate is excessive, and the ion equation is ba2+2oh-

2hco3-

baco32h2o+co3

2- Chemical equation: 2NaHCO3

ba(oh)2=na2co3+

baco3+2h2o

2h+ +so42- +ba2+ +2oh- =baso4↓+2h2o

Its sub-distribution is.

so42-+ba2+=baso4↓

Thank you.

I know, the scenery is pretty good, and there must be no mistakes

Your ionic equation.

Yes, but only if it is.

In the case of excess Ba(OH)2, the solution is alkaline at this time, and the ion equation written for the condition you give should be.

Ba2+ +2OH- +2H+ +SO42-=BaSO4 (precipitation) +2H2O

Description: Barium hydroxide.

Add sodium bisulfate dropwise to the solution.

The solution happens to be neutral, the alkali excess solution cannot be neutral, so the alkali can not be excessive, when the limbs are acidic, it will not be excessive, I write these two equations, you will know whether the solution is neutral, the stoichiometric number of reactants is not 1:1, that is, the alkali is excessive (you may ask the reaction is obviously a 1:1 reaction, how can the alkali be overdosed?)

This is relative to NaHSO4, in which NAHso4 is insufficient, which in turn is the BA(OH)2 excess) equation

Ba(OH)2+NaHSO4=BASO4(precipitate)+NaOH+H2O

The stoichiometric number of reactants is not 1:2, and the reaction equation:

Ba(OH)2+2NaHSO4=BASO4(precipitate)+Na2SO4+H2O

Obviously, the first solution is alkaline, and the second solution is neutral.

It is right to add dropwise to the base, but you don't pay attention to the topic, and finally the solution is neutral, this is the topic, the title does not let you write the reaction equation in the dropwise addition process, your equation can be regarded as the first step of the reaction! At this time, the solution is still alkaline.