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Autodysocjacja wody i Kw

Autojonizacja wody, stała autodysocjacji Kw i związek między [H⁺] i [OH⁻] w wodnych roztworach. Tłumaczenie na język polski: fundacja Edukacja dla Przyszłości, dzięki wsparciu wolontariuszy.

Kluczowe informacje

Woda może ulec autodysocjacji, tworząc jony $H_{3} O^{+}$ ‍ i ${OH}^{-}$ ‍.
W stanie równowagi stała autojodysocjacji wody, $K_{w}$ ‍, wynosi $10^{- 14}$ ‍ w $25^{\circ} C$ ‍.
W obojętnym roztworze $[H_{3} O^{+}] = [{OH}^{-}]$ ‍
W kwasowym roztworze $[H_{3} O^{+}] > [{OH}^{-}]$ ‍
W zasadowym roztworze $[{OH}^{-}] > [H_{3} O^{+}]$ ‍
Dla wodnych roztworów w $25^{\circ} C$ ‍, zawsze prawdziwe jest, że:

$K_{w} = [H_{3} O^{+}] [{OH}^{-}] = 10^{- 14}$ ‍

$pH + pOH = 14$ ‍

Z istotnym wkładem autodysocjacji wody do $[H_{3} O^{+}]$ ‍ oraz $[{OH}^{-}]$ ‍ mamy do czynienia w przypadku bardzo słabych roztworów kwasów i zasad.

Woda jest amfoteryczna

Woda jest jednym z powszechnych rozpuszczalników dla reakcji kwasów z zasadami. W poprzednim artykule o zasadach i kwasach Brønsteda-Lowry'ego omówiliśmy amfoteryczność wody i to, że może reagować zarówno jako kwas, jak i zasada Brønsteda-Lowry'ego.

Zadanie $1$ ‍: Określanie roli wody w reakcji

W poniższych reakcjach ustal, czy woda reaguje, jako kwas, zasada czy żadne z dwóch.

1

Autodysocjacja wody

Since acids and bases react with each other, this implies that water can react with itself! While that might sound strange, it does happen

-

water molecules exchange protons with one another to a very small extent. We call this process the autoionization, or self-ionization, of water.

Wymianę protonów można zapisać w postaci zbilansowanego równania:

H_{2} O (l) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {OH}^{-} (a q)

One water molecule donates a proton (orange sphere) to a neighboring water molecule, which acts as a Bronsted-Lowry base by accepting that proton. The products of the reversible acid-base reaction are hydronium and hydroxide.

One water molecule is donating a proton and acting as a Bronsted-Lowry acid, while another water molecule accepts the proton, acting as a Bronsted-Lowry base. This results in the formation of hydronium and hydroxide ions in a

1 : 1

molar ratio. For any sample of pure water, the molar concentrations of hydronium,

H_{3} O^{+}

, and hydroxide,

{OH}^{-}

, must be equal:

[H_{3} O^{+}] = [{OH}^{-}] w czystej wodzie

Note that this process is readily reversible. Because water is a weak acid and a weak base, the hydronium and hydroxide ions exist in very, very small concentrations relative to that of non-ionized water. Just how small are these concentrations? Let's find out by examining the equilibrium constant for this reaction (also called the autoionization constant), which has the special symbol

K_{w}

The autoionization constant, $K_{w}$ ‍

The expression for the autoionization constant is

K_{w} = [H_{3} O^{+}] [{OH}^{-}] (Eq. 1)

Remember that when writing equilibrium expressions, the concentrations of solids and liquids are not included. Therefore, our expression for

K_{w}

does not include the concentration of water, which is a pure liquid.

We can calculate the value of

K_{w}

25^{\circ} C

using

[H_{3} O^{+}]

, which is related to the

pH

of water. At

25^{\circ} C

, the

pH

of pure water is

7

. Therefore, we can calculate the concentration of hydronium ions in pure water:

[H_{3} O^{+}] = 10^{- pH} = 10^{- 7} M at 25^{\circ} C

In the last section, we saw that hydronium and hydroxide form in a

1 : 1

molar ratio during the autoionization of pure water. We can use that relationship to calculate the concentration of hydroxide in pure water at

25^{\circ} C

[{OH}^{-}] = [H_{3} O^{+}] = 10^{- 7} M at 25^{\circ} C

This is a little tough to visualize, but

10^{- 7}

is an extremely small number! Within a sample of water, only a small fraction of the water molecules will be in the ionized form.

Now that we know

[{OH}^{-}]

and

[H_{3} O^{+}]

, we can use these values in our equilibrium expression to calculate

K_{w}

25^{\circ} C

K_{w} = (10^{- 7}) \times (10^{- 7}) = 10^{- 14} at 25^{\circ} C

Concept check: How many hydroxide and hydronium ions are in one liter of water at $25^{\circ} C$ ‍?

We know that at

25^{\circ} C

, the following relationship is true:

[{OH}^{-}] = [H_{3} O^{+}] = 10^{- 7} \frac{mol}{l}

To oznacza, że 1 litr wody zawiera

10^{- 7} mol

zarówno jonów hydroniowych jak i jonów wodorotlenkowych. Możemy przekonwertować mole na liczbę jonów używając liczby Avogadro:

10^{- 7} mol \times \frac{6.022 \times 10^{23} ions}{1 mol} = 6.022 \times 10^{16} ions

This sounds like a huge number! To put this in perspective, we can compare the number of hydronium and hydroxide to the number of water molecules in the same volume. There are

3.34 \times 10^{25}

molecules of water in one liter of water. That is nearly

10^{9}

larger than the number of

H_{3} O^{+}

and

{OH}^{-}

ions in solution!

Relationship between the autoionization constant, $pH$ ‍, and $pOH$ ‍

The fact that

K_{w}

is equal to

10^{- 14}

25^{\circ} C

leads to an interesting and useful new equation. If we take the negative logarithm of both sides of

Eq. 1

in the previous section, we get the following:

\begin{aligned} - \log K_{w} & = - \log ([H_{3} O^{+}] [{OH}^{-}]) \\ = - (\log [H_{3} O^{+}] + \log [{OH}^{-}]) \\ = - \log [H_{3} O^{+}] + (- \log [{OH}^{-}]) \\ = pH + pOH \end{aligned}

We can abbreviate

- \log K_{w}

p K_{w}

, which is equal to

14

25^{\circ} C

p K_{w} = pH + pOH = 14 at 25^{\circ} C (Eq. 2)

Therefore, the sum of

pH

and

pOH

will always be

14

for any aqueous solution at

25^{\circ} C

. Keep in mind that this relationship will not hold true at other temperatures, because

K_{w}

is temperature dependent!

Example $1$ ‍: Calculating $[{OH}^{-}]$ ‍ from $pH$ ‍

An aqueous solution has a

pH

10

25^{\circ} C

What is the concentration of hydroxide ions in the solution?

Method $1$ ‍: Using Eq. $1$ ‍

One way to solve this problem is to first find

[H^{+}]

from the

pH

\begin{aligned} [H_{3} O^{+}] & = 10^{- pH} \\ = 10^{- 10} M \end{aligned}

We can then calculate

[{OH}^{-}]

using Eq. 1:

\begin{aligned} K_{w} & = [H_{3} O^{+}] [{OH}^{-}] Rearrange to solve for [{OH}^{-}] \\ [{OH}^{-}] & = \frac{K_{w}}{[H_{3} O^{+}]} Plug in values for K_{w} {and [H}_{3} O^{+}] \\ = \frac{10^{- 14}}{10^{- 10}} \\ = 10^{- 4} M \end{aligned}

Method $2$ ‍: Using Eq. $2$ ‍

Another way to calculate

[{OH}^{-}]

is to calculate it from the

pOH

of the solution. We can use Eq. 2 to calculate the

pOH

of our solution from the

pH

. Rearranging Eq. 2 and solving for the

pOH

, we get:

\begin{aligned} pOH & = 14 - pH \\ = 14 - 10 \\ = 4 \end{aligned}

We can now use the equation for

pOH

to solve for

[{OH}^{-}]

\begin{aligned} [{OH}^{-}] & = 10^{- pOH} \\ = 10^{- 4} M \end{aligned}

Using either method of solving the problem, the hydroxide concentration is

10^{- 4} M

for an aqueous solution with a

pH

10

25^{\circ} C

Definitions of acidic, basic, and neutral solutions

We have seen that the concentrations of

H_{3} O^{+}

and

{OH}^{-}

are equal in pure water, and both have a value of

10^{- 7} M

25^{\circ} C

. When the concentrations of hydronium and hydroxide are equal, we say that the solution is neutral. Aqueous solutions can also be acidic or basic depending on the relative concentrations of

H_{3} O^{+}

and

{OH}^{-}

W roztworze neutralnym, $[H_{3} O^{+}] = [{OH}^{-}]$ ‍
W roztworze kwasowym, $[H_{3} O^{+}] > [{OH}^{-}]$ ‍
W roztworze zasadowym, $[{OH}^{-}] > [H_{3} O^{+}]$ ‍

Practice $2$ ‍: Calculating $pH$ ‍ of water at $0^{\circ} C$ ‍

If the $p K_{w}$ ‍ of a sample of pure water at $0^{\circ} C$ ‍ is $14.9$ ‍, what is the $pH$ ‍ of pure water at this temperature?

Practice $3$ ‍: Calculating $p K_{w}$ ‍ at $40^{\circ} C$ ‍

The

pH

of pure water at

40^{\circ} C

is measured to be

6.75

Based on this information, what is the $p K_{w}$ ‍ of water at $40^{\circ} C$ ‍?

Autoionization and Le Chatelier's principle

We also know that in pure water, the concentrations of hydroxide and hydronium are equal. Most of the time, however, we are interested in studying aqueous solutions containing other acids and bases. In that case, what happens to

[H_{3} O^{+}]

and

[{OH}^{-}]

The moment we dissolve other acids or bases in water, we change

[H_{3} O^{+}]

and/or

[{OH}^{-}]

such that the product of the concentrations is no longer is equal to

K_{w}

. That means the reaction is no longer at equilibrium. In response, Le Chatelier's principle tells us that the reaction will shift to counteract the change in concentration and establish a new equilibrium.

For example, what if we add an acid to pure water? While pure water at

25^{\circ} C

has a hydronium ion concentration of

10^{- 7} M

, the added acid increases the concentration of

H_{3} O^{+}

. In order to get back to equilibrium, the reaction will favor the reverse reaction to use up some of the extra

H_{3} O^{+}

. This causes the concentration of

{OH}^{-}

to decrease until the product of

[H_{3} O^{+}]

and

[{OH}^{-}]

is once again equal to

10^{- 14}

Once the reaction reaches its new equilibrium state, we know that:

$[H^{+}] > [{OH}^{-}]$ ‍ because the added acid increased $[H^{+}]$ ‍. Thus, our solution is acidic!
$[{OH}^{-}] < 10^{- 7} M$ ‍ because favoring the reverse reaction decreased $[{OH}^{-}]$ ‍ to get back to equilibrium.

The important thing to remember is that any aqueous acid-base reaction can be described as shifting the equilibrium concentrations for the autoionization of water. This is really useful, because that means we can apply Eq. 1 and Eq. 2 to all aqueous acid-base reactions, not just pure water!

Autoionization matters for very dilute acid and base solutions

The autoionization of water is usually introduced when first learning about acids and bases, and it is used to derive some extremely useful equations that we've discussed in this article. However, we will often calculate

[H^{+}]

and

pH

for aqueous solutions without including the contribution from the autoionization of water. The reason we can do this is because autoionization usually contributes relatively few ions to the overall

[H^{+}]

[{OH}^{-}]

compared to the ions from additional acid or base.

The only situation when we need to remember the autoionization of water is when the concentration of our acid or base is extremely dilute. In practice, this means that we need to include the contribution from autoionization when the concentration of

H^{+}

{OH}^{-}

is within ~

2

orders of magnitude (or less than) of

10^{- 7} M

. We will now go through an example of how to calculate the

pH

of a very dilute acid solution.

Example $2$ ‍: Calculating the $pH$ ‍ of a very dilute acid solution

Let's calculate the

pH

of a

6.3 \times 10^{- 8} M

HCl

solution.

HCl

completely dissociates in water, so the concentration of hydronium ions in solution due to

HCl

is also

6.3 \times 10^{- 8} M

Try 1: Ignoring the autoionization of water

If we ignore the autoionization of water and simply use the formula for

pH

, we get:

\begin{aligned} pH & = - log [H^{+}] \\ = - log [6.3 \times 10^{- 8}] \\ = 7.20 \end{aligned}

Easy! We have an aqueous acid solution with a

pH

that is greater than

7

. But, wait, wouldn't that make it a basic solution? That can't be right!

Try 2: Including the contribution from autoionization to $[H^{+}]$ ‍

Since the concentration of this solution is extremely dilute, the concentration of the hydronium from the hydrochloric acid is close to the

[H^{+}]

contribution from the autoionization of water. That means:

We have to include the contribution from autoionization to $[H^{+}]$ ‍
Since the autoionization of water is an equilibrium reaction, we must solve for the overall $[H^{+}]$ ‍ using the expression for $K_{w}$ ‍:

K_{w} = [H^{+}] [{OH}^{-}] = 1, 0 \times 10^{- 14}

If we say that

x

is the contribution of autoionization to the equilibrium concentration of

H^{+}

and

{OH}^{-}

, the concentrations at equilibrium will be as follows:

[H^{+}] = 6.3 \times 10^{- 8} M + x

[{OH}^{-}] = x

Plugging these concentrations into our equilibrium expression, we get:

\begin{aligned} K_{w} & = (6.3 \times 10^{- 8} M + x) x = 1.0 \times 10^{- 14} \\ = x^{2} + 6.3 \times 10^{- 8} x \end{aligned}

Rearranging this expression so that everything is equal to

0

gives the following quadratic equation:

0 = x^{2} + 6.3 \times 10^{- 8} x - 1.0 \times 10^{- 14}

We can solve for

x

using the quadratic formula, which gives the following solutions:

x = 7, 3 \times 10^{- 8} M, - 1, 4 \times 10^{- 7} M

Since the concentration of

{OH}^{-}

can't be negative, we can eliminate the second solution. If we plug in the first value of

x

to get the equilibrium concentration of

H^{+}

and calculate

pH

, we get:

\begin{aligned} pH & = - log [H^{+}] \\ = - log [6.3 \times 10^{- 8} + x] \\ = - log [6.3 \times 10^{- 8} + 7.3 \times 10^{- 8}] \\ = - log [1.36 \times 10^{- 7}] \\ = 6.87 \end{aligned}

Thus we can see that once we include the autoionization of water, our very dilute

HCl

solution has a

pH

that is weakly acidic. Whew!

Podsumowanie

Water can undergo autoionization to form $H_{3} O^{+}$ ‍ and ${OH}^{-}$ ‍ ions.
W stanie równowagi stała autojodysocjacji wody, $K_{w}$ ‍, wynosi $10^{- 14}$ ‍ w $25^{\circ} C$ ‍.
W roztworze neutralnym, $[H_{3} O^{+}] = [{OH}^{-}]$ ‍
W roztworze kwasowym, $[H_{3} O^{+}] > [{OH}^{-}]$ ‍
W roztworze zasadowym, $[{OH}^{-}] > [H_{3} O^{+}]$ ‍
Dla wodnych roztworów w $25^{\circ} C$ ‍, zawsze prawdziwe jest, że:

$K_{w} = [H_{3} O^{+}] [{OH}^{-}] = 10^{- 14}$ ‍

$pH + pOH = 14$ ‍

Z istotnym wkładem autodysocjacji wody do $[H_{3} O^{+}]$ ‍ oraz $[{OH}^{-}]$ ‍ mamy do czynienia w przypadku bardzo słabych roztworów kwasów i zasad.

Źródła

Ten artykuł został napisany na podstawie następujących źródeł:

"Water Autoionization" from UC Davis ChemWiki, CC BY-NC-SA 3.0
“Temperature dependence of the pH of pure water” from UC Davis ChemWiki, CC BY-NC-SA 3.0

Zmodyfikowany artykuł może być używany zgodnie z licencją CC BY-NC-SA 4.0.

Dodatkowe źródła

Zumdahl, S.S., i Zumdahl S.A. (2003). Atomic Structure and Periodicity. W Chemistry (edycja 6, str. 290-94), Boston, MA: Houghton Mifflin Company.

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Chemia

Kurs: Chemia > Rozdział 13

Kluczowe informacje

Woda jest amfoteryczna

Zadanie 1‍ : Określanie roli wody w reakcji

Autodysocjacja wody

The autoionization constant, Kw‍

Relationship between the autoionization constant, pH‍ , and pOH‍

Example 1‍ : Calculating [OH−]‍ from pH‍

Method 1‍ : Using Eq. 1‍

Method 2‍ : Using Eq. 2‍

Definitions of acidic, basic, and neutral solutions

Practice 2‍ : Calculating pH‍ of water at 0∘C‍

Practice 3‍ : Calculating pKw‍ at 40∘C‍

Autoionization and Le Chatelier's principle

Autoionization matters for very dilute acid and base solutions

Example 2‍ : Calculating the pH‍ of a very dilute acid solution

Try 1: Ignoring the autoionization of water

Try 2: Including the contribution from autoionization to [H+]‍

Podsumowanie

Chcesz dołączyć do dyskusji?

Zadanie $1$ ‍: Określanie roli wody w reakcji

The autoionization constant, $K_{w}$ ‍

Relationship between the autoionization constant, $pH$ ‍, and $pOH$ ‍

Example $1$ ‍: Calculating $[{OH}^{-}]$ ‍ from $pH$ ‍

Method $1$ ‍: Using Eq. $1$ ‍

Method $2$ ‍: Using Eq. $2$ ‍

Practice $2$ ‍: Calculating $pH$ ‍ of water at $0^{\circ} C$ ‍

Practice $3$ ‍: Calculating $p K_{w}$ ‍ at $40^{\circ} C$ ‍

Example $2$ ‍: Calculating the $pH$ ‍ of a very dilute acid solution

Try 2: Including the contribution from autoionization to $[H^{+}]$ ‍