Ion Concentration Comparison, High School Chemistry, High School Chemistry Ion Concentration Size Co

Except for 9, which is a weak base, the rest are weak alkali salts, so NH4+ +OH- = + H+ This is the NH4+ hydrolysis equation, that is, the pH and NH4+ concentration of the solution are related to the degree of hydrolysis, and the greater the degree, the greater the concentration of NH4+ and H+, the greater the pH.

NH4+ is the same, then the degree of hydrolysis of the salt depends on the acid ions behind these 8 salts, there are weak salts, weak acids and weak alkali salts are mutually reinforcing in hydrolysis, and the acid roots are regarded as x, then there is x- +H2O = Hx +oh- in the solution, which is the hydrolysis equation of acid roots.

The hydrolysis equation can be seen as the inverse of the ionization equation, and then the problem becomes to rank the degree of ionization of the raw acid.

Involves multi-step hydrolysis Level 1 > Level 2.

HSO4-OK NH4+ concentration from large to small: 1 6 3 8 7 2 5 4 9PH from small to large: 8 7 6 1 2 3 4 5 6 9 There are still questions.

NH4+ from large to small: 6, 1, 3, 8, 7, 2, 4, 5, 9

pH from small to large: 8, 7, 6, 1, 2, 4, 3, 5, 9

What's this 5 of yours? I am considered to be ch3coonh4 processing! If not, look at it yourself!

c(NH4+), from large to small:

pH: From acid to alkali: Satisfied

The solution is electrically neutral and must satisfy the conservation of charge, i.e., the total number of positive charges of all cations is equal to the total number of negative charges of all anions. According to the conservation of charge, there is: c(k+).

c(h＋)c(oh―)

c(C6H5O-) if pH

c(oh―)

then c(k+).

At pH >7, c(h).

c(c6h5o-)

At pH >7, c(h).

c(oh―)

then c(k+).

c(c6h5o-)

Select ahcl==h+

cl-nh3*h2o==nh4+

oh-h2o==h+

oh - the symbol in the middle, which is a reversible symbol).

At the beginning of the addition, the concentration of chloride ions is the same as the concentration of ammonia ions, because ammonia ions will combine with hydroxide ions in water to form NH3*H2O, so the concentration of chloride ions is greater than that of ammonia ions, and because ammonia ions are only a small part of hydrolysis, and the solution itself is acidic (because ammonia and hydrochloric acid react to form strong acid and weak alkali salts), so the concentration of ammonia ions is greater than that of hydrogen ions, and the concentration of hydrogen ions is greater than that of hydroxide ions

i.e. C(Cl-)>C(NH4+)>C(H+)>C(OH-).

According to the conservation of charge, there is: c(k+).

c(h＋)c(oh―)

c(c6h5o-)

Therefore, the total charge of the cation is greater than the total charge of the anion, and the charge is not conserved.

phc(oh―)

Hence c(k+).

pH>7, due to c(h).

c(c6h5o-)

pH>7, due to c(h).

c(oh―)

Hence c(k+).

c(c6h5o-)

The number of ions of acetic acid is greater than that of hydrochloric acid, because the concentration of acetic acid in the two solutions of the same pH is large, so the original number of ions is the same, after dilution, it promotes the continued ionization of acetic acid, so the amount of ionic substances increases, and the amount of ionic substances in hydrochloric acid does not change, according to.

The amount of matter = the number of particles Avogadro's constant!!

Mainly consider acetic acid ionization, weak electrolyte, after dilution, ionization equilibrium is carried out positively, weakening the ion concentration of the original solution is reduced, the answer is almost the same, and then considering the weak ionization of water, all the order comes out.

na2s:c（na+）>c(s2-)>

c(oh-)>c(hs-)>c(h+)

Explanation: This substance is a salt, completely ionized, and then.

s2-h2o

hs-oh-

hs-h2o

h2soh-

The generation of 1molHS- produces 1moloh-, so c(OH-) > c(hs-).

H2S: C(H+)>C(Hs-)>C(S2-)>C(OH-) Explanation: H2S is a moderately strong acid, which ionizes, and [H+] is relatively large, so [OH-] is very small.

nahs:c(na+)>c(hs-)>c(oh-)>c(h+)>c(s2-)

Explanation: Hydrolysis ionization is a small fraction.

hs-+h2o=h2s+oh-

hs-=h+

S2-HS- Hydrolysis is greater than ionization.

c（h2s）＞c（s2-）

co2+2naoh=na2co3+h2o

According to the question. co2

And because NaOH+CO2=NaHCO3

The remaining reacts with NaOH and the resulting NaHCO3 to produce Na2CO3 NaOH + NaHCO3 = Na2CO3 + H2OC (HCO3-) slightly larger than C (CO3

I'll try to explain it.

In fact, the main reliance on high school to judge the concentration of examples.

1.Column ion equation.

2.Determine acidity and alkalinity.

3.If the first two steps cannot be completely judged, the column reaction equilibrium, and the ions ** are judged.

I will only give the example of a weakly salted Naa solution, the first one is relatively simple, and I write all the concentrations in parentheses.

1.Column ion equation.

There are two first, charge balancing.

na+】+h+】==【a-】+oh-】

Note the number of charges charged by the ions).

Secondly, the mass balance.

na+】=【a-】+ha】

Be careful not to miss ions here.

2.Determine acidity and alkalinity.

Without going into detail here, the most basic situation is that the weak acid solution is acidic, and the weak alkaline is alkaline. Weak acids and strong alkaline salt solutions are alkaline.

The Naa solution is alkaline due to A-hydrolysis. 【oh-】>h+】

You combine the above three equations to get [na+]>a-] and [oh-] = [ha] + [h+].

There is an important assumption that must be made at the initial concentration of NAA, for example. If the initial concentration of this salt is too low (say 10 (-8)), then the ionization of the water dominates in the water, and the conclusions are inconsistent with what you wrote.

If the initial concentration of NaA is relatively large, then the ionization of water has little effect, mainly the hydrolysis of weak acid group A-. In general, the degree of hydrolysis of weak acid roots is very low, for example, only 5% of A- hydrolysis produces OH-, but the OH- produced in this way is still more than that of water ionization. Therefore, in a salt solution with a large concentration, [oh-] and [h+] are considered to be small amounts, much smaller than the unhydrolyzed a-.

So [na+] >ac-] >oh-] >ha].

To determine [ha] and [h+], we use the third step.

3.Column balanced.

In order to judge [ha] we list the equilibrium formula in the solution, which is also not heavy or leaky:

Hydrolysis of A-H2O

HA+OH- water ionization.

h2o====h++

oh- According to the previous assumptions, these two reactions, the hydrolysis reaction is dominant, and the water ionization is very little, so [ha] >

H+] produces oh- from these two equilibriums, the first equilibrium produces an equal ha

and oh-, the second equilibrium additionally produces oh- (but rarely, as has already been assumed.) From here we also get that the total amount of oh- is slightly greater than ha. 【oh-】>ha】

You can write the first HA solution yourself.

Hehe, I've been confused when I was learning this.

I've just finished your question. 、。In fact.

You can see it this way.

1.Salt is a substance that is completely ionized when put into water.

Whether it is a strong acid and a weak alkali salt or a strong alkali and a weak salt.

Let's say ammonium bicarbonate.

Put it in water to ionize the ionized ions and bicarbonate ions.

It should be noted that:

Both carbonic acid and acidic salts of carbonic acid are mainly hydrolyzed.

However, both sulphuric acid and the acid salts of sulphuric acid are dominated by electricity.

Because it is a strong acid.

Ammonium bicarbonate is ionized in water and the bicarbonate ions are hydrolyzed after the ionization of the root ions and bicarbonate ions.

Bicarbonate ions are hydrolyzed to produce carbonic acid.

However, the degree of hydrolysis is small.

So all exist in the form of carbonic acid molecules.

2.You look at what the hydrolysis of carbonate ions produces.

Bicarbonate ions, which are then rehydrolyzed by bicarbonate ions.

Carbonic acid is generated. Hydrolysis is mainly based on the first step, so the degree of hydrolysis in the first step is much greater than that in the second step, so the degree of hydrolysis of bicarbonate ions is not as large as that of carbonate ions.

There is also double hydrolysis, for example, ammonium bicarbonate, bicarbonate ion hydrolysis, the generation of hydroxide ions, ammonium ions are also hydrolyzed, to generate hydrogen ions, but the bicarbonate ions are combined with the hydrogen ions in the water, and the ammonium ions are combined with the hydroxide ions in the water, if they react like this, promote the ionization of water, so that the concentration of hydroxide ions and hydrogen ions in the water increases, according to the principle of Le Chartrete, the degree of hydrolysis of bicarbonate ions and ammonium ions increases.

And there's more. To deal with this kind of problem, you also need to look at the amount of weak electrolyte substances in the solution, ammonium sulfate solution has 2mol ammonium ions, and ammonium chloride solution has 1moi ammonium ions, and their hydrolysis ion concentrations are different.

The degree of ionization of weak electrolytes is very small, so most of them are still molecules, and then there are hydrogen ions, because it is ionized by weak acid and water, which is definitely more than acid radicals, and hydroxide radicals are the least and can be conserved with electric charge! In NAA, the hydrolysis is alkaline, and it can also be conserved with charge, and the HA molecule is still ranked behind the hydroxide or due to the ionization of water.

Both questions are because it is a weak acid, and the weak acid root will be hydrolyzed in solution,,, if you can hydrolyze, then you will understand.

1. Acidity: The concentration of hydrogen ions is greater than the concentration of hydroxides, because the hydrogen ions have a positive charge, the hydroxide has a negative charge, and the number of positive and negative charges in the solution is equal, so the concentration of sodium ions is less than the concentration of a-ions. That is: a- is greater than na+ greater than h + greater than oh-

Alkaline: The concentration of hydrogen ions is less than the concentration of hydroxides, because the hydrogen ions have a positive charge, the hydroxide is negatively charged, and the number of positive and negative charges in the solution is equal, so the concentration of sodium ions is greater than the concentration of a-ions. That is: Na+ is greater than A - Greater than OH - Greater than H+

Neutral: The concentration of hydrogen ions is equal to the concentration of hydroxides, and the same as above obtains: the concentration of sodium ions is equal to the concentration of a-ions. That is: na+ is equal to a - greater than oh - is equal to h+

2. D This substance cannot exist, and I ignore it here.

The more acidic the solution, the smaller the degree of ammonium hydrolysis, and the higher the concentration at the same concentration. At the same volume and concentration, if not hydrolyzed, the ammonium ion concentration of C and E is twice that of A and B, while the hydrolysis is a small part, so it can be judged that the concentration of C and E is much greater than that of A and B. Comparing C and E again, because the acidity of sulfuric acid is much greater than that of carbonic acid, the degree of hydrolysis of ammonium sulfate is much less than that of ammonium carbonate, so the ammonium ion concentration:

e is greater than c; Comparing A and B, because hydrochloric acid is more acidic than acetic acid, the hydrolysis of ammonium chloride is weaker than that of ammonium acetate, so the ammonium ion concentration of ammonium chloride is greater than that of ammonium acetate. That is, the final result is: E is greater than C and A is greater than B.

If the final ammonium ions are required to be the same, then you only need to change all the greater than signs of the answer to less than signs. Because the ammonium ions that can be produced are less, and the final ammonium ions are the same, it can only be satisfied if the previous concentration is large.

I don't know if it's clear, you can continue to ask questions ......

NH4HCO3 ionization and hydrolysis how to deal with the two of them when comparing ion concentrations (especially mixed solutions) hehe I have been confused when I was also learning this I just finished this question. 、。In fact.

Alas, I forgot about it all and returned it to the teacher again.

1.First of all, we need to analyze the HA with the NAA solution, which we call a buffer solution. In the solution there is HA ionization and Naa hydrolysis and water ionization, and for the solution of weak acid and weak salt, the ionization of water cannot be ignored.

Analysis: H2O==H+ +OH-, HA==H+ +A- ionization is greater than hydrolysis, A- +H+==HA (sodium salt) hydrolysis is greater than ionization. When the solution is neutral, that is, pH = 7, C (OH-) = C (H+), that is, H+ ionized by HA is hydrolyzed by the weak acid group A- in NAA, then it means that the amount of NaA added to the solution is more than that added by HA.

Being neutral means that c(a-) is less than the ionic product of water. Then na+>h+=oh->a-. In the same way, when acidic, it means that H+>OH-, H+ ionized by HA is greater than that of A-, a weak acid group hydrolyzed by NAA.

Then there is a->na+>h+>oh-. When alkaline, Na+>A->OH->H+.

The 2D option is a complex, also known as a coordination compound. The correct writing is [Fe(NH4)2]SO4. The [Fe(NH4)2] here is 2+ valence, which is a stable complex ion and cannot be ionized, so there will be no ammonium ions in this solution.

Detailed analysis, equal volume and equal concentration description of the amount of such substances. A ammonium chloride is a strong acid and weak alkali salt, ammonium ion hydrolysis, combined ammonium acetate is a weak acid and weak alkali salt, acetate hydrolysis, ammonium hydrolysis, promote each other, so that there is less ammonium. c Ammonium carbonate is also a weak acid and weak alkali salt, the difference is that after the hydrolysis of carbonate, H2CO3 will be precipitated by CO2, and the carbonate becomes a little less, and the ammonium is more.

dAs mentioned earlier, there is no ionization of ammonium ions. 1mol of E(NH4)2SO4 can ionize 2mol NH4+. So the arrangement is e, c, a, b, d.