The problem of Na2CO3 vs. NaHCO3

First, with only a small amount of H+, the carbonate reacts as follows.

co3(2-)+h(+)=hco3(-)

Whereas, bicarbonate directly reacts with H+ to form CO2+H2O

According to the title, 60g of 10% NAOH is divided into two parts, so there are 6*10% 2=3g NAOH in A and B

1) First look at A, since HCl is added, gas is immediately generated, and only NaHCO3 is present in the solution. Because if you have Na2CO3, you must consume a part of H+ to produce HCO3- first.

Then, there is only NaHCO3 in the solution, which is conserved by the Na elemental mass.

naoh---nahco3

3---x solution gives x=

Answer: Solution A contains only NaHCO3

2) HCl is added to B, and in the AG at the beginning, no gas is produced, so it definitely contains Na2CO3

Because according to the ratio, Na2CO3 + HCl = NaHCO3. Therefore, the Na2CO3 in the solution just reacts with a g HCl to form NaHCO3. However, the gas is released only when 3A g HCl is added, and the NAHCO3 produced by Na2CO3 can only react to A G HCl, so it must also contain NaHCO3.

According to the amount of H+ consumed, there is a Na2CO3 X molecule; NaHCO3 Y molecule.

Na2CO3 consumes 2A g of HCl; NaHCO3 consumes HCl of Ag

So the ratio of Na2CO3 to NaHCO3 is 1:1, and the ratio of Na contained in them is 2:1

Then it is conserved according to the mass of the Na element.

The NA of 2 3 goes into NA2CO3, while the NA of 1 3 is in NAHCO3.

That is, NaOH has 2 3*3=2g to generate Na2CO3, while NaOH has 1g to generate NaHCO3

2naoh---na2co3

2---y contains Na2CO3 as y=

naoh---nahco3

1---z contains NaHCO3 as Z=

Boy, this kind of question is a fart without a picture, it is analyzed according to the image,

First, the common name is different:

1. Na2CO3 is commonly known as soda ash and soda. White powder. It is relatively stable and not easy to decompose when heated.

2. NaHCO3 is commonly known as baking soda.

Second, the degree of stability is different:

1. Na2CO3 white powder. It is relatively stable and not easy to decompose when heated.

2. NaHCO3 white fine crystals. Unstable and easy to decompose when heated.

3. Different molecular weights:

1. Molecular weight of Na2CO3:

2. Molecular weight of NaHCO3: .

Fourth, the use is different:

1. Na2CO3 is an important organic chemical raw material, mainly used in the production of flat glass, glass products and ceramic glaze. It is also widely used in household washing, acid neutralization, and food processing.

2. NaHCO3 is used as a starter in the food industry, a generator of carbon dioxide in soft drinks and cold drinks, and a preservative of butter. It can be directly used as a raw material in the pharmaceutical industry for ** hyperacidity. It can also be used in film production, tanning, mineral processing, smelting, metal heat treatment, as well as in the fiber and rubber industries.

At the same time, it is used as a detergent for wool, a foam fire extinguishing agent, and for agricultural soaking.

Na2CO3 after evaporation and burning: Na2CO3

The equation is: Na2CO3 + H2O ==NaHCO3 + NaOH

NaHCO3 after evaporation and burning: Na2CO3

The equation is: 2NaHCO3===Na2CO3+H2O+CO2

Na2SO3 after evaporation and burning: Na2SO4

The equation is: 2Na2SO3 + O2===2Na2SO4

Problem Solving Ideas: A Nahco 3 solid is unstable and decomposes by heat to form sodium carbonate;

B Ca(OH)2 solution reacts with a sufficient amount of NaHCO3 solution to form calcium carbonate, sodium carbonate and water;

c Judged according to the conditions under which the metathesis reaction occurs;

When NaHCO3 and Na2CO3 react with acid, NaHCO3+2HC2NaCl+H2O+CO2, NaHCO3+HCl+H2O+CO2, respectively, are answered

A NaHCO3 solid is unstable and decomposes by heat by the equation 2NaHCO3

Na2CO3+H2O+CO2, so the NaHCO3 solid in the Na2CO3 solid can be removed by heating, so A is correct;

B Na2CO3 solution and NaHCO3 solution can react with Ca(OH)2 solution to form calcium carbonate precipitate, so B is wrong;

C Na2CO3 reacts with CaCl2 solution to form calcium carbonate and sodium chloride, NaHCO3 and CaCl2 solution do not meet the conditions for ion reaction to occur, and will not react, so C is wrong;

When D NaHCO3 and Na2CO3 react with acid, Na2CO3+2HCl 2NaCl+H2O+CO2 and NaHCO3+HCl+H2O+CO2 occur respectively, and it can be seen from the equation that 106g of sodium carbonate completely reacts to form 1mol of carbon dioxide, and 84g of sodium bicarbonate completely reacts to form 1mol of dioxide, so D is wrong;

Therefore, a 2, the correct statement about na 2co 3 and nahco 3 is ( ).

A In order to remove the NaHCO3 solids from the Na2Co3 solids, a heating method can be employed.

The Bona 2CO3 solution can react with the Ca(OH)2 solution, while the NaHCO3 solution cannot.

Both CNa2Co3 and NaHCO3 can react with CaCl2 solution.

The same mass of Na2CO3 and NaHCO3 produced when reacted with sufficient hydrochloric acid is also of the same mass of CO2.

HCl reacts with Na2CO3 first.

The correct explanation involves the knowledge of hydrolysis of salts.

The action of CO32- and solvent H2O in Na2CO3 solution: CO32- +H2O <=reversible) = >HCO3- +OH-1

The role of HCO3- and solvent H2O in NaHCO3 solution: HCO3- +H2O <=reversible) = >H2CO3 +OH-2

According to the facts, the same concentration of Na2CO3 solution is more alkaline than NaHCO3 solution, so it can be seen that the 1 reaction generates more OH-, so the 1 reaction is carried out to a greater extent, that is, CO32- can bind H+ in H2O more, so CO32- reacts with H+ first.

The explanation of 4L can also be used, if H+ reacts with HCO3- first, the resulting CO2 reacts with CO32- to form HCO3-

co32- +co2+h2o=2hco3-

na2co3+co2+h2o=2nahco3

It reacts with NaHCO3 first, Na2CO3 is a strong base and weak salt, which needs to be hydrolyzed into HCO3- first, and then reacts with HCl.

It reacts with sodium carbonate first. Because if it reacts with sodium bicarbonate first, the CO2 generated will react with sodium carbonate to form sodium bicarbonate.

Na2CO3 is very alkaline, so you should react with him first.

NaHCO3 reacts first, has H ions, and can be used for titration to identify the two substances.

The reason for the formation of crystals is because the atoms of the substance are arranged more regularly, and the reason for the appearance of powder is because the atoms of the substance are arranged in a disorderly manner, and sodium carbonate can be understood to be generated by the thermal decomposition of sodium bicarbonate.

The main knowledge point of this question is the hydrolysis and ionization of salts, carbonate ions and bicarbonate ions will be hydrolyzed, reversible reactions occur, the solution is alkaline, and the alkalinity of carbonate ion hydrolysis is stronger than that of bicarbonate ions, if you know this principle, it is obvious that A is Na2CO3 and B is NaHCO3. A brief analysis is provided below:

co32-h2o

Reversible reaction) HCO3-OH- (hydrolysis of carbonate ions to bicarbonate ions, which will continue to hydrolyze).

hco3-h2o

Reversible reaction) H2CO3OH- (bicarbonate ion hydrolysis).

Obviously, the first addition of dilute hydrochloric acid is equivalent to the addition of H+ ions, which will neutralize with the OH- ions produced by hydrolysis, thereby promoting hydrolysis. Therefore, the most frequent and fast bubbles should be a solution with bicarbonate ions, because carbonate ions are dihydrolyzed, and bicarbonate is generated first and then carbonic acid (CO2 produced by carbonic acid decomposition); The hydrolysis of bicarbonate ions directly produces carbonic acid, that is, bubbles.

As for the second addition of Ca(NO3)2, it is equivalent to an excess of sodium carbonate and sodium bicarbonate, and a small amount of calcium nitrate, ionized from the equation of sodium bicarbonate: NaHCO3=Na+

hco3-na++h+

CO32- It can be seen that only carbonate ions can directly precipitate with calcium ions, obviously the concentration of carbonate ions in sodium carbonate is larger, so the precipitation produced is more, the precipitation in the sodium bicarbonate solution is produced by the hydrolysis of calcium bicarbonate, and there are still a lot of calcium bicarbonate in the solution, so the precipitation is more sodium carbonate.

nahco3

with CA(NO3)2

In fact, it is not reacting, in fact, it is just that they have calcium bicarbonate in their solution, which is easy to understand and can be regarded as the following reaction: 2NaHCO3

ca(no3)2

ca(hco3)2+2nano3

Note that Ca(HCO3)2 is a water-soluble salt, and the CO32- ions produced by the hydrolysis of Ca2+ ions and HCO3- ions form CaCO3 precipitates, which is easy to understand, and can be regarded as Ca(HCO3)2 hydrolysis to produce CaCO3 precipitate.

After talking so much nonsense, I don't know if the landlord can understand, it's been 6 years since I graduated from high school, and I don't remember much about the knowledge I may have, so I suggest that the landlord focus on understanding the hydrolysis and ionization of salts, and hydrolysis is a reversible reaction, which is the key. This kind of test center will be tested in the college entrance examination every year, and it must be mastered.

[Specific steps].

By heating. NaHCO3 heating can change to Na2CO3:

2NaHCO3 = Na2CO3 + H2O + CO2 by adding other substances like carbon dioxide. CO2 is introduced into Na2CO3 solution, and Na2CO3 can be converted to NaHCO3

Na2CO3 + CO2 + H2O = 2NaHCO3 sodium carbonate plus a small amount of hydrochloric acid to sodium bicarbonate Sodium bicarbonate plus sodium hydroxide to sodium carbonate or heat sodium bicarbonate to become water, carbon dioxide and sodium carbonate.

Depending on the acidity of the salts (acids) formed, the "weak ions" are hydrolyzed the weaker they are. The acid corresponding to Na2CO3 is HCO3-; The acid corresponding to NaHCO3 is H2CO3; Acidic apparently H2CO3 HCO3-. The weaker the more hydrolyzed, so the degree of hydrolysis of Na2CO3 is greater, i.e.:

It is more alkaline and has a greater pH.

Do experiments, directly prepare the same concentration of the two solutions, with a pH test paper to know.

(1) Na2CO3 solution is introduced into CO2

Na2CO3 + CO2 + H2O = 2NaHCO3 (2) sodium bicarbonate solid is decomposed into sodium carbonate by heating.

2NaHCO3 = (heated) Na2CO3 + CO2 (rising arrow) + H2O

1) By heating. NaHCO3 heating can change to Na2CO3:

2NaHCO3 = Na2CO3 + H2O + CO22) by adding other substances such as carbon dioxide. CO2 is introduced into Na2CO3 solution, and Na2CO3 can be converted to NaHCO3

na2co3 + co2 + h2o = 2nahco3

1.With a sufficient amount of carbon dioxide: CO2 + Na2CO3 + H2O = = 2NaHCO3

The degree of hydrolysis of sodium carbonate is greater than that of sodium bicarbonate, which is judged by the alkalinity of the two at the amount and concentration of the same substance.

For strong alkalis and weak salts, hydrolysis is greater than ionization, indicating that the solution is alkaline.

It can be explained by the degree of hydrolysis (primary hydrolysis and secondary water) and ionization.

1) For NaHCO3, bicarbonate is hydrolyzed and ionized.

Hydrolysis equation: HCO3-+H2O=H2CO3+OH-, OH- will be released, making the solution alkaline;

The ionization equation: HCO3-=(CO3)2-+H+, H+ will be released, making the solution acidic;

However, since the hydrolysis of bicarbonate is greater than ionization, C(OH-)>C(H+), so the sodium bicarbonate solution is alkaline.

Note: Hydrolysis or ionization here generally refers to the hydrolysis or ionization of the corresponding weakly acidic acid. )

2) For Na2CO3, hydrolysis of carbonate occurs.

CO3)2-+H2O=HCO3-+OH-, which will release OH- and make the solution alkaline;

HCO3-+H2O==H2CO3+OH-, which also releases OH-, but the degree of secondary hydrolysis is much smaller than that of primary hydrolysis (negligible);

Of course, at this time, the bicarbonate obtained by first-stage hydrolysis will also be ionized to obtain H+, but the first hydrolysis reaction is the main one. Secondary hydrolysis and especially ionization of bicarbonate can be ignored. Therefore, the Na2CO3 solution is alkaline.

It can be seen from the above that the degree of hydrolysis of sodium carbonate and sodium bicarbonate can be reflected by their primary hydrolysis, and the amount and concentration of the hydrolyzed oh- substance can reflect the degree of hydrolysis, that is, the alkalinity of the two can be judged by the amount concentration of the same substance, because the alkalinity of sodium carbonate is much greater than that of sodium bicarbonate at the amount concentration of the same substance, so the degree of hydrolysis of sodium carbonate is greater than that of sodium bicarbonate.