Sophomore Chemistry! Ionization of water! Sophomore Chemistry! Ionization equilibrium of water !!!!!

If it is neutral, the acid-base is completely neutralized, that is, H+ and OH- react completely.

The concentration of H+ in sulfuric acid is:

In NaOH, pH = 12, i.e. the H+ concentration is 10 -12 mol L, so OH- is 1 mol L (the product of the two is equal to 10 -12).

So the volume ratio of sulfuric acid to NaOH is 10:1

1.Neutralization shows that the concentration of hydrogen ions in the solution is equal to the concentration of hydroxides, and it can be approximated that the amount of hydrogen ions in sulfuric acid is equal to the amount of hydroxide ions in Naoh, at this time, there is, x y=10:1

2.On the other hand, there is 10 (-11)x=10 (-12)y, x y= 1:10, but this algorithm is obviously not feasible, and the content is too small, even less than the ionization of water itself.

So the answer can only be the former.

To be honest, I don't really understand your algorithm.

i'm so sorry~

You don't understand that neutral is h+=oh-, just make the two equal, and the answer is 10:1

Here I am:

An important principle in the quantitative calculation of acid-base neutralization is the quantitative comparison of n(h+) and n(oh-).

If it is neutral after mixing, then n(h+) = n(oh-).

Now the question arises: n(h+) n(oh-) in our equation are ionized by acid and base, respectively,!! and not water and electricity ionization!!

Therefore, the correct formula for mixing neutral is c(h+)v strong acid = c(oh-)v strong base.

i.e. your first algorithm is correct. [but not the simplest algorithm].

The best algorithm is H2SO4 with pH = 1, C(H+) = mol L

NaOH at pH = 12, c(OH-) = 1mol L (kW = 10 -12 at this temperature).

Hence v acid = 1x v base.

v acid:v base = 10:1

But if it's the other way around:

Calculated by c(oh-) v strong acid = c(h+) v strong base, which is your second algorithm, the result is 1:10, natural error!

So, why can't you do it?

Because: in acid solution, oh- is ionized by water; In the same way, in alkaline solution, H+ is water-ionized;

Acid-base mixing is not the neutralization of H+ ionized by water and OH- by ionization!!

Therefore, the formula of c(oh-) v strong acid = c(h+) v strong alkali is not valid!

Knowledge supplement: In fact, in any environment, do not consider the "neutralization reaction between H+ and OH- of water ionization"!

NAOH is a strong electrolyte.

In the aqueous solution, it is completely ionized into sodium ions and hydroxide ions.

The ionization equation is: NaOH=NA+OH-H2SO4 is a strong electrolyte, which is completely ionized in an aqueous solution to generate hydrogen ions.

For sulfate ions, the ionization equation is: H2SO4 = 2H + SO4 2-, so the answer is: NaOH = NA2 + OH - H2 SO4 =2H + SO4 2-

First of all, I like it, even though you have a problem with your idea, it's nice to think like this in high school.

The problem is that you ignore the ionization of the water itself. Total H+ concentration in solution = H+ concentration ionized by sulfuric acid + H+ concentration of water ionization.

In very dilute sulfuric acid, the concentration of H+ ionized by sulfuric acid is the H+ concentration of water ionization.

Considering the ionization of water, the concentration of H+ in the solution is still slightly greater than the concentration of OH-, and the solution is slightly acidic.

Specific computing high school does not require mastery.

The concentration of hydrogen ions and hydroxide ions in the water itself are both to the power of 10-7, and the concentration of hydrogen ions and hydroxide ions are equal, so the water is neutral.

In the very small concentration of sulfuric acid, you only consider the hydrogen ions ionized by the sulfuric acid, and do not consider that the water itself will also ionize hydrogen ions.

Water ionizes H+ ions by itself, and the hydrogen ions and hydroxide ions ionized by water are the same.

You ignore the hydrogen ions produced by the ionization of the water itself.

At a constant 25 degrees, the ion product constant of water (in fact, the equilibrium constant of water self-coupling ionization) kw=10 -14 remains unchanged, that is, the product of hydrogen ion concentration and hydroxide ion concentration in the aqueous solution system remains unchanged, which is 10 -14.

For sulfuric acid solution, the concentration of hydrogen ions ionized by sulfuric acid is 1mol l, which is much greater than the concentration of hydrogen ions ionized by water, so the total hydrogen ion concentration in the solution is about equal to 1mol l, and because the above product remains unchanged, the hydroxide ion concentration in the solution = 10 -14mol l is calculated. According to the ionization equation of water: H2O = (reversible sign) = H+ +OH-, it can be seen that the concentration of hydrogen ions ionized by water is the same as the concentration of hydroxide ions ionized by it, so it is also equal to 10 -14mol L.

Water is an extremely weak electrolyte. Normally, 1 liter of water can ionize only 10 -7 molecules. However, water is a highly polar molecule, which can ionize many alkali, acid, and salt molecules to varying degrees.

Water is a very weak electrolyte, so it is difficult to ionize hydrogen ions and hydroxide ions. Because the weak ions ionized by salt are combined with hydrogen ions or hydroxide ions ionized by water, the concentration of hydrogen ions or hydroxide ions is reduced, so the ionization of which is promoted, and weak ions such as: acetate ions or carbonate ions or ammonium ions can promote the ionization of water.

FeCl3 is combined with hydroxide, so that the hydroxide group is reduced, the concentration of the product is reduced, and the ionization equilibrium of water moves in the direction of positive reaction, so the concentration of hydrogen ions is greater than the concentration of hydroxide ions. It's not that you think that hydrochloric acid is produced and promotes the ionization of water. The addition of hydrochloric acid to the water inhibits the ionization of the water.

Because salt hydrolysis is the reverse reaction of acid-base neutralization reaction, the degree of hydrolysis is less. So most of the ferric chloride solution is ferric chloride. Weak acids and weak alkali salts are hydrolyzed, and the situation is very complicated, and we do not require mastery in high school.

Because the concentration of H+ from water ionization is 1 10 13mol L, it means that the ionization of water is suppressed.

Then it could be an acid solution or an alkaline solution.

If it is a strong acid solution, the concentration of H+ and OH- from the ionization of water is 1 10 13mol l 1.

At this time, the OH- in solution is ionized by H2O.

So the concentration of H+ in the solution is 1 10 1mol l, and the concentration of the strong acid solution is , Hno3) or.

If it is a strong alkaline solution, the concentration of H+ and OH- from the ionization of water is 1 10 13mol l 1.

The H+ in the solution is the H+ ionized by H2O, so the concentration of OH- in the solution is 1 10 1mol L

The concentration of the strong alkali solution is, NaOH and so on) or etc.).

As for weak acids, weak bases are partially ionized, and the concentration of acid and base is the concentration of h+ (oh-).

Hope, thank you.

The difference between strong and weak acids lies in the difficulty of ionization of hydrogen ions, strong acids are easy to ionize, and weak acids are not easy to ionize. In aqueous solution, strong acid and strong base are completely ionized (there are individual dibasic acids that are completely ionized by an H ion), and there are hydrogen ions ionized by strong acid itself in the strong acid solution, and hydrogen ions ionized by water, the same is true for strong alkali, there are also hydrogen ions ionized by water, water will be ionized in strong acid, strong alkali and weak acid and weak alkali solution Water will be ionized, but the concentration is not the same and H ion ** is different from the destination, 10 -13 hydrogen ions are inhibited by ionization in water, which can be too many H ions or too many OH ions.

The H+ concentration in H2SO4 with pH=1 is 10 (-1) mol L, and the H+ concentration in pH=12 NaOH is 10 (-12) mol L, then its OH - concentration is 10 (-12) 10 (-12) = 1 mol L, because the solution is neutral after the reaction, so the amount of H+ substances is equal to OH - (that 10 -6 is ionized by H2O, which has nothing to do with acid and base), and the volume ratio is 10:1

In this question, neither hydrogen ions nor hydroxide ions alone can be used. Because hydroxide is ionized in sulfuric acid, it will change with the change of pH, and the same is true for hydrogen ions in Naoh. The correct way is to multiply the concentration of hydrogen ions in sulfuric acid by its volume, and then calculate the concentration of hydroxide in NaOH by its volume according to the ion product of water, so that the two are equal.

I'm sure you already know how to do it, so you don't need to say the answer.

This statement is wrong.

At room temperature, the ion product of water = 1*10 -14

Under normal circumstances, the concentration of hydrogen ions ionized by water is 1 10 -7mol l, and now the concentration of hydrogen ions ionized by water in the solution is 1 10 -3mol l, indicating that the solute promotes the ionization of water, and the ionization of water in the solution of acid or alkali is inhibited, and salt hydrolysis can promote the ionization of water, so the solution is definitely not hydrochloric acid, but only a salt solution.

The hydrogen ions ionized by water in the solution are 10-3, that is to say, the hydroxide ions ionized by water are also 10-3

This means that the ionization of water is moving forward, and there are two ways to move water forward, one is to reduce the concentration of hydrogen ions, and the other is to reduce the concentration of hydroxide ions, which means that there are two possibilities: the solution may be acidic or it may be alkaline.

In addition, just by looking at the 10-3 hydrogen ions ionized by water, it does not mean that there are 10-3 hydrogen ions in the water, and some of them may have been rereacted, and the same is true for hydroxide ions, so just looking at the 10-3 hydrogen ions ionized by water does not tell that the solution is acidic, and the solution may be acidic or alkaline.

First of all, you should understand that acids and bases inhibit the ionization of water, and the salts that can be hydrolyzed promote the ionization of water, and because the concentration of hydrogen ions and hydroxide ions ionized by water at room temperature is 1 10 -7, and now the concentration of hydrogen ions in the solution is greater than 1 10 -7, so there is a substance in the solution that promotes water ionization, which is a salt that can be ionized.

Obviously, the ionization of water is promoted, and hydrochloric acid inhibits the ionization of water, right?

Why hydrochloric acid? There are many kinds of acids

First: the mass mixing of the above substances and other substances, let their concentrations be: c(H2S) = C(Nahs) = A (mol L).

h2s = h+ hs- ka1=

Ignoring the H+ ions produced by water ionization, let the [H+] produced by H2S be x:

There is a-x x a+x on balance

Then: [h+][hs-] [h2s]=ka1

Due to the ionic effect, the ionization degree of H2S is smaller, so a-x a; a+x a is substituted into the above equation:

x*a/a=ka1

x= (mol/l)

So: [h+]=x= (mol l).

Again, find [S2-]: The S2- ion is a product of secondary ionization.

hs- = h+ +s2- (ka2= )

ka2 = [h+][s2-] / [hs-] =

Since the second step ionization is very small, it can be considered [H+] mol L); hs-] a (mol/l)

Then: [s2-] = *a (

*a ) mol/l)

The second question: the concentration of H2S at the mass concentration of the equivalent is A (mol L).

h2s = h+ hs- (ka1=；ka2=

Let [h+] be x (mol l).

At equilibrium, there are: a-x x x ignore secondary ionization due to ka1 ka2 and count as monoacid.

Then: x 2 ( a-x )=ka1 c ka1 , a-x a

x=√( a)

So: [h+]=x= ( a) (mol l).

Again, find [S2-]: In the same way, S2- ions are the product of secondary ionization.

hs- = h+ +s2- (ka2= )

ka2 = [h+][s2-] / [hs-] =

Since the second step ionization is very small, it can be considered [H+] Hs-].

Then: [s2-] ka2= (mol l).

Conclusions: 1. [H+] The former is smaller than the latter. (Due to the homoionic effect, the concentration of hydrogen ions is greatly reduced.) ）

2. [S2-] The former is larger than the latter. (Since the secondary ionization is very small, the concentration of S2- ions is determined by the concentration of Hs- ions, and the former has a large amount of NaHs, which greatly increases its ionization).