Why is the solution of c OH 1x10 14 ionized by water possible acidic or alkaline

Electrolysis of water. h2o=oh-

The product of H+ ions is kw

Take acetic acid, for example.

ch3coo-

h+ch3cooh

The ionization constant is 1 kN

ch3coo-

h2ooh-

ch3cooh

Ionization constant. kh

kh=kw1/kn

Except for the last equal sign.

Everything else is reversible.

Let's give an example

The H+ concentration of HCl solution with a pH of 1 should be 10- if at room temperature.

Aqueous. The ion product constant is 10-

So the concentration of oh- should be 10-

And where did this oh-come from? Of course.

It's water ionization. The h+ of the water is the same as the oh- of the water and electricity ionization, so it is also 10-

Again: The H+ concentration of NaOH solution with pH 13 is 10- Same as above and this H+ is ionized by water.

Therefore, the oh+ ionized by water is also 10-

Aren't these two particles ionized by water at a concentration of 10- and one acidic?

An alkaline dahehe.

At room temperature, in pure water, the ionization of water c(h+)=c(oh-)=1x10 -7

moll is now in this solution, and the water ionized c(oh-) = 1x10-5

mol/l＞1x10^-7

mol l i.e. solute promotes the ionization of water, so the answer is bc

Hydrogen ions and hydroxide ions ionized from neutral water at room temperature.

The concentration is 10-7 mol l

According to the ionization equilibrium.

c（h2o）= c（oh）+c（h）

Therefore, whether the solution is acidic or alkaline, it will inhibit the ionization of water, and the concentration of hydroxide ions ionized by water can be reduced to the minus 14th power of c(oh-)=10.

Yes, it is like this: c(h+)=may be acidic, may be alkaline, and is related to the ion product of water.

The c(h+)=c(oh-) produced by the ionization of water in solution is equal, so.

c(h+)=c（oh-）=

pH = 10 in this solution, i.e. c(H+) = 10 -10; C(OH)- =C(H+) produced by ionization of water does not produce H+ by ionization

C(oh-) = c(h+) = 10 -10 produced by water ionization

pH = 10 in this solution, i.e. c(H+) = 10 -10; C(OH)- =C(H+) produced by ionization of water does not produce H+ by ionization

C(oh-) = c(h+) = 10 -10 produced by water ionization

H2O=OH- +H+ (reversible).

Above is the electrolysis of water, and it can be found that the hydroxide ions ionized by water are the same as the hydrogen ionization naurion. In a neutral solution, hydroxide ions and hydrogen ions are both 10 (-7) mol l, and the reason why they decrease is because the ionization of water with Liang is inhibited, and according to the chemical equilibrium, it can be known that both acid and alkali will inhibit the ionization of water.

So the concentration of hydrogen ions ionized by water is equal to 10 -14, why can the solution become alkaline or acidic.

The concentration of H+ and OH- ionized by water at room temperature is the same, and when the concentration of H+ ionized by water is 10 -12mol L, the ionized OH- concentration is also 10 -12mol LThere are two ** H+ and OH- in the solution blind pin, one is ionized and the other is administered.

Assuming that H+ is all provided by the ionization of water (no external H+), then the H+ concentration is 10 -12, the Oh- concentration is 10 -2, pH = 12, alkaline solution.

Assuming that OH- is all provided by water ionization (no external OH-), then the OH- concentration is 10 -12, and the H+ concentration is 10 -2mol L, pH = 2, and the acidic Qin Shenye solution.

C(H+) = 10-12 mol L produced by water ionization, the first possibility: the solution is acidic, and the C(H+) = 10-12 mol L produced by water ionization, then the concentration of OH produced at the same time C(OH) = 10-12 mol L, according to the ion product constant of water, C(H+) = 10-2 mol L in the solution, pH 2.

The second possibility: the solution is alkaline, and the water ionization produces c(h+) = 10-12 mol l, and according to the ion product constant of water, c(oh) = 10-2 mol l in the solution can be calculated as pH 12.

Both acids and bases inhibit the ionization of water, so the solution may be acidic or alkaline.