# Water autoionization and Kw

Autoionization of water, the autoionization constant Kw, and the relationship between [H⁺] and [OH⁻] in aqueous solutions.

## Key points

• Water can undergo autoionization to form $\text{H}_3\text{O}^+$ and $\text{OH}^-$ ions.
• The equilibrium constant for the autoionization of water, $K_\text{w}$, is $10^{-14}$ at $25\,^\circ\text{C}$.
• In a neutral solution, $[\text{H}_3\text{O}^+]=[\text{OH}^-]$
• In an acidic solution, $[\text{H}_3\text{O}^+]>[\text{OH}^-]$
• In a basic solution, $[\text{OH}^-]>[\text{H}_3\text{O}^+]$
• For aqueous solutions at $25\,^\circ\text{C}$, the following relationships are always true:
$K_\text{w}=[\text{H}_3\text{O}^+][\text{OH}^-]=10^{-14}$
$\text{pH}+\text{pOH}=14$
• The contribution of the autoionization of water to $[\text{H}_3\text{O}^+]$ and $[\text{OH}^-]$ becomes significant for extremely dilute acid and base solutions.

## Water is amphoteric

Water is one of the most common solvents for acid-base reactions. As we discussed in a previous article on Brønsted-Lowry acids and bases, water is also amphoteric, capable of acting as either a Brønsted-Lowry acid or base.

## Practice $1$: Identifying the role of water in a reaction

In the following reactions, identify if water is playing the role of an acid, a base, or neither.
Reaction
Role of water
• $\text{CH}_3\text{NH}_2(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{CH}_3\text{NH}_3^+(aq)+\text{OH}^-(aq)$
• $\text{HSO}_4^-(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{SO}_4^{2-}(aq)+\text{H}_3\text{O}^+(aq)$
• $\text{Cu}^{2+}(aq)+6\text{ H}_2\text{O}(l)\rightarrow\text{Cu(H}_2\text{O)}_6^{2+}(aq)$
• $\text{base }$
• $\text{acid}$
• $\text{neither}$

When water donates $\text H^+$ to form $\text{OH}^-$, it is acting as a Bronsted-Lowry acid. When water accepts a proton for form $\text{H}_3 \text O^+$, it is acting as a Bronsted-Lowry base.

## Autoionization of water

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.
The proton exchange can be written as the following balanced equation:
$\qquad\qquad\text{ H}_2\text{O}(l)+\text{H}_2\text{O}(l)\rightleftharpoons\text{H}_3\text{O}^+(aq)+\text{OH}^-(aq)$
space filling models to show two water molecules, where each water molecule is represented as a large red sphere (oxygen) stuck to two small grey sphere (hydrogen). The products are hydronium ion, which has 3 hydrogens and a positive charge, and hydroxide, which has one hydrogen and a negative charge.
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, $\text{H}_3\text{O}^+$, and hydroxide, $\text{OH}^-$, must be equal:
$[\text{H}_3\text{O}^+]=[\text{OH}^-]~~\text{in pure water}$
Excellent question! Pure water means water with no other compounds dissolved in the water. Pure water must also be degassed to remove dissolved gases such as carbon dioxide, $\text{CO}_2$. This is because when carbon dioxide dissolves in water, it can form small amounts of the weak acid carbonic acid, $\text H_2 \text{CO}_3$.
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 $K_\text{w}$.

## The autoionization constant, $K_\text{w}$

The expression for the autoionization constant is
$K_\text{w}=[\text{H}_3\text{O}^+][\text{OH}^-]\quad\quad\text{(Eq. 1)}$
Remember that when writing equilibrium expressions, the concentrations of solids and liquids are not included. Therefore, our expression for $K_\text{w}$ does not include the concentration of water, which is a pure liquid.
We can calculate the value of $K_\text{w}$ at $25\,^\circ\text{C}$ using $[\text{H}_3\text{O}^+]$, which is related to the $\text{pH}$ of water. At $25\,^\circ\text{C}$, the $\text{pH}$ of pure water is $7$. Therefore, we can calculate the concentration of hydronium ions in pure water:
$[\text{H}_3\text{O}^+]=10^{-\text{pH}}=10^{-7}\text{ M}~~\text{at }25\,^\circ\text{C}$
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 $25^\circ\text{C}$:
$[\text{OH}^-]=[\text{H}_3\text{O}^+]=10^{-7}\text{ M}~~\text{at }25\,^\circ\text{C}$
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 $[\text{OH}^-]$ and $[\text{H}_3\text{O}^+]$, we can use these values in our equilibrium expression to calculate $K_\text{w}$ at $25^\circ\text{C}$:
$K_\text{w}=(10^{-7})\times(10^{-7})=10^{-14}~~\text{at }25\,^\circ\text{C}$
Concept check: How many hydroxide and hydronium ions are in one liter of water at $25^\circ\text{C}$?
We know that at $25^\circ\text{C}$, the following relationship is true:
$[\text{OH}^-]=[\text{H}_3\text{O}^+]=10^{-7}\,\dfrac{\text{mol}}{\text L}$
That means one liter of water contains $10^{-7}\,\text{mol}$ each of hydronium ions and hydroxide ions. We can convert from moles to the number of ions using Avogadro's number:
$10^{-7}\,\cancel{\text{mol}}\times \dfrac{6.022\times 10^{23} \,\text {ions}}{1\,\cancel{\text{mol}}}=6.022\times 10^{16} \,\text{ions}$
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 $\text{H}_3\text{O}^+$ and $\text{OH}^-$ ions in solution!

## Relationship between the autoionization constant, $\text{pH}$, and $\text{pOH}$

The fact that $K_\text{w}$ is equal to $10^{-14}$ at $25\,^\circ\text{C}$ leads to an interesting and useful new equation. If we take the negative logarithm of both sides of $\text{Eq. 1}$ in the previous section, we get the following:
\begin{aligned}-\log{K_\text{w}}&=-\log({[\text{H}_3\text{O}^+}][\text{OH}^-])\\ \\ &=-\big(\log[\text{H}_3\text{O}^+]+\log[\text{OH}^-]\big)\\ \\ &=-\log[\text{H}_3\text{O}^+]+(-\log[\text{OH}^-])\\ \\ &=\text{pH}+\text{pOH}\end{aligned}
No worries! There are a just few rules you will want to remember for problems in chemistry involving concepts like $\text {pH}$. For a review of useful rules for logarithm calculations, see Sal's video on properties of logarithms.
We can abbreviate $-\log{K_\text{w}}$ as $\text{p}K_\text w$, which is equal to $14$ at $25\,^\circ\text{C}$:
$\text{p}K_\text{w}=\text{pH}+\text{pOH}=14~~\text{at }25\,^\circ \text C\quad\quad\text{(Eq. 2})$
Therefore, the sum of $\text{pH}$ and $\text{pOH}$ will always be $14$ for any aqueous solution at $25\,^\circ\text{C}$. Keep in mind that this relationship will not hold true at other temperatures, because $K_\text{w}$ is temperature dependent!

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

An aqueous solution has a $\text{pH}$ of $10$ at $25\,^\circ\text{C}$.
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 $[\text{H}^+]$ from the $\text{pH}$:
\begin{aligned}[\text{H}_3\text{O}^+]&=10^{-\text{pH}}\\ \\ &=10^{-10}\,\text M\\\end{aligned}
We can then calculate $[\text{OH}^-]$ using Eq. 1:
\begin{aligned}K_\text{w}&=[\text{H}_3\text{O}^+][\text{OH}^-]~~~\quad\quad\text{Rearrange to solve for }[\text{OH}^-]\\ \\ [\text{OH}^-]&=\dfrac{K_\text{w}}{[\text{H}_3\text{O}^+]}\qquad\quad\qquad\text{Plug in values for }K_\text w \,\text{and [H}_3 \text O^+]\\ \\ &=\dfrac{10^{-14}}{10^{-10}}\\ \\ &=10^{-4}\text{ M}\end{aligned}

### Method $2$: Using Eq. $2$

Another way to calculate $[\text{OH}^-]$ is to calculate it from the $\text{pOH}$ of the solution. We can use Eq. 2 to calculate the $\text{pOH}$ of our solution from the $\text{pH}$. Rearranging Eq. 2 and solving for the $\text{pOH}$, we get:
\begin{aligned}\text{pOH}&=14-\text{pH}\\ \\ &=14-10\\ \\ &=4\end{aligned}
We can now use the equation for $\text{pOH}$ to solve for $[\text{OH}^-]$.
\begin{aligned}[\text{OH}^-]&=10^{-\text{pOH}}\\ \\ &=10^{-4}\text{ M}\end{aligned}
Using either method of solving the problem, the hydroxide concentration is $10^{-4}\text{ M}$ for an aqueous solution with a $\text{pH}$ of $10$ at $25\,^\circ\text{C}$.

## Definitions of acidic, basic, and neutral solutions

We have seen that the concentrations of $\text{H}_3\text{O}^+$ and $\text{OH}^-$ are equal in pure water, and both have a value of $10^{-7}\text{ M}$ at $25\,^\circ\text{C}$. 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 $\text{H}_3\text{O}^+$ and $\text{OH}^-$.
• In a neutral solution, $[\text{H}_3\text{O}^+]=[\text{OH}^-]$
• In an acidic solution, $[\text{H}_3\text{O}^+]>[\text{OH}^-]$
• In a basic solution, $[\text{OH}^-]>[\text{H}_3\text{O}^+]$

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

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

For aqueous solutions at $0\,^\circ \text C$
$\text{p}K_\text{w}=14.9=\text{pH}+\text{pOH}$
Since pure water at $0\,^\circ \text C$ is neutral, we know that $[\text{H}_3\text{O}^+]=[\text{OH}^-]$ and $\text{pH}=\text{pOH}$.
If we plug this relationship into the equation for $\text{p}K_\text{w}$, we get:
\begin{aligned}\text{pH}+\text{pH}&=14.9=2 \times \text{pH}\\ \\ \text{pH}&=\dfrac{14.9}{2}=7.45\end{aligned}
The $\text{pH}$ of pure water at at $0\,^\circ \text C$ is $7.45$.

## Practice $3$: Calculating $\text pK_{\text w}$ at $40\,^\circ\text{C}$

The $\text{pH}$ of pure water at $40\,^\circ\text{C}$ is measured to be $6.75$.
Based on this information, what is the $\text pK_{\text w}$ of water at $40\,^\circ\text{C}$?

Pure water is neutral, so we know that $[\text{H}_3\text{O}^+]=[\text{OH}^-]$.
If we take the $-\text{log}$ of both sides of the equation, we get
$\text{pH}=\text{pOH}$
We can find the relationship between $\text{pH}$ and $\text{p}K_\text{w}$ for pure water at $40\,^\circ\text{C}$ using Eq. 2:
\begin{aligned}\text{p}K_\text{w}&=\text{pH}+\text{pOH}\\ \\ &=2 \times \text{pH}\\ \\ &=2\times 6.75 \\ \\ &=13.5\end{aligned}
The $\text{p}K_\text{w}$ for pure water at $40\,^\circ\text{C}$ is $13.5$.

## 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 $[\text{H}_3\text{O}^+]$ and $[\text{OH}^-]$?
The moment we dissolve other acids or bases in water, we change $[\text{H}_3\text{O}^+]$ and/or $[\text{OH}^-]$ such that the product of the concentrations is no longer is equal to $K_\text{w}$. 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 $25\,^\circ \text C$ has a hydronium ion concentration of $10^{-7}\,\text M$, the added acid increases the concentration of $\text{H}_3\text{O}^+$. In order to get back to equilibrium, the reaction will favor the reverse reaction to use up some of the extra $\text{H}_3\text{O}^+$. This causes the concentration of $\text{OH}^-$ to decrease until the product of $[\text{H}_3\text{O}^+]$ and $[\text{OH}^-]$ is once again equal to $10^{-14}$.
Once the reaction reaches its new equilibrium state, we know that:
• $[\text H^+]>[\text{OH}^-]$ because the added acid increased $[\text H^+]$. Thus, our solution is acidic!
• $[\text{OH}^-]<10^{-7}\,\text M$ because favoring the reverse reaction decreased $[\text{OH}^-]$ 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 $[\text H^+]$ and $\text{pH}$ 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 $[\text H^+]$ or $[\text{OH}^-]$ 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 $\text H^+$ or $\text{OH}^-$ is within ~$2$ orders of magnitude (or less than) of $\text{10}^{-7}\,\text M$. We will now go through an example of how to calculate the $\text{pH}$ of a very dilute acid solution.

## Example $2$: Calculating the $\text{pH}$ of a very dilute acid solution

Let's calculate the $\text{pH}$ of a $6.3 \times 10^{-8}\,\text M$ $\text{HCl}$ solution. $\text{HCl}$ completely dissociates in water, so the concentration of hydronium ions in solution due to $\text{HCl}$ is also $6.3 \times 10^{-8}\,\text M$.

### Try 1: Ignoring the autoionization of water

If we ignore the autoionization of water and simply use the formula for $\text{pH}$, we get:
\begin{aligned}\text{pH}&=-\text{log}[\text H^+]\\ \\ &=-\text{log}[6.3 \times 10^{-8}]\\ \\ &=7.20\end{aligned}
Easy! We have an aqueous acid solution with a $\text{pH}$ 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 $[\text{H}^+]$

Since the concentration of this solution is extremely dilute, the concentration of the hydronium from the hydrochloric acid is close to the $[\text{H}^+]$ contribution from the autoionization of water. That means:
• We have to include the contribution from autoionization to $[\text{H}^+]$
• Since the autoionization of water is an equilibrium reaction, we must solve for the overall $[\text{H}^+]$ using the expression for $K_\text{w}$:
$K_\text{w} =[\text H^+][\text{OH}^-]=1.0\times10^{-14}$
If we say that $x$ is the contribution of autoionization to the equilibrium concentration of $\text H^+$ and $\text{OH}^-$, the concentrations at equilibrium will be as follows:
$[\text H^+]=6.3 \times 10^{-8}\,\text M+x$
$[\text {OH}^-]=x$
Plugging these concentrations into our equilibrium expression, we get:
\begin{aligned}K_\text{w} &=(6.3 \times 10^{-8}\,\text M+x)x=1.0\times10^{-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:
$0=x^2+6.3 \times 10^{-8}x-1.0\times10^{-14}$
We can solve for $x$ using the quadratic formula, which gives the following solutions:
$x=7.3 \times 10^{-8}\,\text M, -1.4 \times 10^{-7}\,\text M$
Since the concentration of $\text{OH}^-$ 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 $\text H^+$ and calculate $\text{pH}$, we get:
\begin{aligned}\text{pH}&=-\text{log}[\text H^+]\\ \\ &=-\text{log}[6.3 \times 10^{-8}+x]\\ \\ &=-\text{log}[6.3 \times 10^{-8}+7.3 \times 10^{-8}]\\ \\ &=-\text{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 $\text{HCl}$ solution has a $\text {pH}$ that is weakly acidic. Whew!

## Summary

• Water can undergo autoionization to form $\text{H}_3\text{O}^+$ and $\text{OH}^-$ ions.
• The equilibrium constant for the autoionization of water, $K_\text{w}$, is $10^{-14}$ at $25\,^\circ\text{C}$.
• In a neutral solution, $[\text{H}_3\text{O}^+]=[\text{OH}^-]$
• In an acidic solution, $[\text{H}_3\text{O}^+]>[\text{OH}^-]$
• In a basic solution, $[\text{OH}^-]>[\text{H}_3\text{O}^+]$
• For aqueous solutions at $25\,^\circ\text{C}$, the following relationships are always true:
$K_\text{w}=[\text{H}_3\text{O}^+][\text{OH}^-]=10^{-14}$
$\text{pH}+\text{pOH}=14$
• The contribution of the autoionization of water to $[\text{H}_3\text{O}^+]$ and $[\text{OH}^-]$ becomes significant for extremely dilute acid and base solutions.