Water autoionization and Kw (article) | Khan Academy (2023)

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.

In the following reactions, identify if water is playing the role of an acid, a base, or neither.

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

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$1:11, colon, 1 molar ratio. For any sample of pure water, the molar concentrations of hydronium, $\text{H}_3\text{O}^+$H3​O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, and hydroxide, $\text{OH}^-$OH−start text, O, H, end text, start superscript, minus, end superscript, must be equal:

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 $K_\text{w}$Kw​K, start subscript, start text, w, end text, end subscript.

The expression for the autoionization constant is

Remember that when writing equilibrium expressions, the concentrations of solids and liquids are not included. Therefore, our expression for $K_\text{w}$Kw​K, start subscript, start text, w, end text, end subscript does not include the concentration of water, which is a pure liquid.

We can calculate the value of $K_\text{w}$Kw​K, start subscript, start text, w, end text, end subscript at $25\,^\circ\text{C}$25∘C25, degrees, start text, C, end text using $[\text{H}_3\text{O}^+]$[H3​O+]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, which is related to the $\text{pH}$pHstart text, p, H, end text of water. At $25\,^\circ\text{C}$25∘C25, degrees, start text, C, end text, the $\text{pH}$pHstart text, p, H, end text of pure water is $7$77. Therefore, we can calculate the concentration of hydronium ions in pure water:

In the last section, we saw that hydronium and hydroxide form in a $1:1$1:11, colon, 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}$25∘C25, degrees, start text, C, end text:

This is a little tough to visualize, but $10^{-7}$10−710, start superscript, minus, 7, end superscript 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 $[\text{OH}^-]$[OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket and $[\text{H}_3\text{O}^+]$[H3​O+]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, we can use these values in our equilibrium expression to calculate $K_\text{w}$Kw​K, start subscript, start text, w, end text, end subscript at $25^\circ\text{C}$25∘C25, degrees, start text, C, end text:

Concept check: How many hydroxide and hydronium ions are in one liter of water at $25^\circ\text{C}$25∘C25, degrees, start text, C, end text?

The fact that $K_\text{w}$Kw​K, start subscript, start text, w, end text, end subscript is equal to $10^{-14}$10−1410, start superscript, minus, 14, end superscript at $25\,^\circ\text{C}$25∘C25, degrees, start text, C, end text leads to an interesting and useful new equation. If we take the negative logarithm of both sides of $\text{Eq. 1}$Eq.1start text, E, q, point, space, 1, end text in the previous section, we get the following:

We can abbreviate $-\log{K_\text{w}}$−logKw​minus, log, K, start subscript, start text, w, end text, end subscript as $\text{p}K_\text w$pKw​start text, p, end text, K, start subscript, start text, w, end text, end subscript, which is equal to $14$1414 at $25\,^\circ\text{C}$25∘C25, degrees, start text, C, end text:

Therefore, the sum of $\text{pH}$pHstart text, p, H, end text and $\text{pOH}$pOHstart text, p, O, H, end text will always be $14$1414 for any aqueous solution at $25\,^\circ\text{C}$25∘C25, degrees, start text, C, end text. Keep in mind that this relationship will not hold true at other temperatures, because $K_\text{w}$Kw​K, start subscript, start text, w, end text, end subscript is temperature dependent!

An aqueous solution has a $\text{pH}$pHstart text, p, H, end text of $10$1010 at $25\,^\circ\text{C}$25∘C25, degrees, start text, C, end text.

What is the concentration of hydroxide ions in the solution?

One way to solve this problem is to first find $[\text{H}^+]$[H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket from the $\text{pH}$pHstart text, p, H, end text:

We can then calculate $[\text{OH}^-]$[OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket using Eq. 1:

Another way to calculate $[\text{OH}^-]$[OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket is to calculate it from the $\text{pOH}$pOHstart text, p, O, H, end text of the solution. We can use Eq. 2 to calculate the $\text{pOH}$pOHstart text, p, O, H, end text of our solution from the $\text{pH}$pHstart text, p, H, end text. Rearranging Eq. 2 and solving for the $\text{pOH}$pOHstart text, p, O, H, end text, we get:

We can now use the equation for $\text{pOH}$pOHstart text, p, O, H, end text to solve for $[\text{OH}^-]$[OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket.

Using either method of solving the problem, the hydroxide concentration is $10^{-4}\text{ M}$10−4M10, start superscript, minus, 4, end superscript, start text, space, M, end text for an aqueous solution with a $\text{pH}$pHstart text, p, H, end text of $10$1010 at $25\,^\circ\text{C}$25∘C25, degrees, start text, C, end text.

We have seen that the concentrations of $\text{H}_3\text{O}^+$H3​O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript and $\text{OH}^-$OH−start text, O, H, end text, start superscript, minus, end superscript are equal in pure water, and both have a value of $10^{-7}\text{ M}$10−7M10, start superscript, minus, 7, end superscript, start text, space, M, end text at $25\,^\circ\text{C}$25∘C25, degrees, start text, C, end text. 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 $\text{H}_3\text{O}^+$H3​O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript and $\text{OH}^-$OH−start text, O, H, end text, start superscript, minus, end superscript.

• In a neutral solution, $[\text{H}_3\text{O}^+]=[\text{OH}^-]$[H3​O+]=[OH−]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, equals, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket

• In an acidic solution, $[\text{H}_3\text{O}^+]>[\text{OH}^-]$[H3​O+]>[OH−]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket, is greater than, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket

• In a basic solution, $[\text{OH}^-]>[\text{H}_3\text{O}^+]$[OH−]>[H3​O+]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, is greater than, open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket

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 $[\text{H}_3\text{O}^+]$[H3​O+]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket and $[\text{OH}^-]$[OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket?

The moment we dissolve other acids or bases in water, we change $[\text{H}_3\text{O}^+]$[H3​O+]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket and/or $[\text{OH}^-]$[OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket such that the product of the concentrations is no longer is equal to $K_\text{w}$Kw​K, start subscript, start text, w, end text, end subscript. 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$25∘C25, degrees, start text, C, end text has a hydronium ion concentration of $10^{-7}\,\text M$10−7M10, start superscript, minus, 7, end superscript, start text, M, end text, the added acid increases the concentration of $\text{H}_3\text{O}^+$H3​O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript. In order to get back to equilibrium, the reaction will favor the reverse reaction to use up some of the extra $\text{H}_3\text{O}^+$H3​O+start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript. This causes the concentration of $\text{OH}^-$OH−start text, O, H, end text, start superscript, minus, end superscript to decrease until the product of $[\text{H}_3\text{O}^+]$[H3​O+]open bracket, start text, H, end text, start subscript, 3, end subscript, start text, O, end text, start superscript, plus, end superscript, close bracket and $[\text{OH}^-]$[OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket is once again equal to $10^{-14}$10−1410, start superscript, minus, 14, end superscript.

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

• $[\text H^+]>[\text{OH}^-]$[H+]>[OH−]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket, is greater than, open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket because the added acid increased $[\text H^+]$[H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket. Thus, our solution is acidic!

• $[\text{OH}^-]<10^{-7}\,\text M$[OH−]<10−7Mopen bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket, is less than, 10, start superscript, minus, 7, end superscript, start text, M, end text because favoring the reverse reaction decreased $[\text{OH}^-]$[OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket 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!

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 $[\text H^+]$[H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket and $\text{pH}$pHstart text, p, H, end text 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 $[\text H^+]$[H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket or $[\text{OH}^-]$[OH−]open bracket, start text, O, H, end text, start superscript, minus, end superscript, close bracket 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 $\text H^+$H+start text, H, end text, start superscript, plus, end superscript or $\text{OH}^-$OH−start text, O, H, end text, start superscript, minus, end superscript is within ~$2$22 orders of magnitude (or less than) of $\text{10}^{-7}\,\text M$10−7Mstart text, 10, end text, start superscript, minus, 7, end superscript, start text, M, end text. We will now go through an example of how to calculate the $\text{pH}$pHstart text, p, H, end text of a very dilute acid solution.

Let's calculate the $\text{pH}$pHstart text, p, H, end text of a $6.3 \times 10^{-8}\,\text M$6.3×10−8M6, point, 3, times, 10, start superscript, minus, 8, end superscript, start text, M, end text $\text{HCl}$HClstart text, H, C, l, end text solution. $\text{HCl}$HClstart text, H, C, l, end text completely dissociates in water, so the concentration of hydronium ions in solution due to $\text{HCl}$HClstart text, H, C, l, end text is also $6.3 \times 10^{-8}\,\text M$6.3×10−8M6, point, 3, times, 10, start superscript, minus, 8, end superscript, start text, M, end text.

If we ignore the autoionization of water and simply use the formula for $\text{pH}$pHstart text, p, H, end text, we get:

Easy! We have an aqueous acid solution with a $\text{pH}$pHstart text, p, H, end text that is greater than $7$77. But, wait, wouldn't that make it a basic solution? That can't be right!

Since the concentration of this solution is extremely dilute, the concentration of the hydronium from the hydrochloric acid is close to the $[\text{H}^+]$[H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket contribution from the autoionization of water. That means:

• We have to include the contribution from autoionization to $[\text{H}^+]$[H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket

• Since the autoionization of water is an equilibrium reaction, we must solve for the overall $[\text{H}^+]$[H+]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket using the expression for $K_\text{w}$Kw​K, start subscript, start text, w, end text, end subscript:

If we say that $x$xx is the contribution of autoionization to the equilibrium concentration of $\text H^+$H+start text, H, end text, start superscript, plus, end superscript and $\text{OH}^-$OH−start text, O, H, end text, start superscript, minus, end superscript, the concentrations at equilibrium will be as follows:

Plugging these concentrations into our equilibrium expression, we get:

Rearranging this expression so that everything is equal to $0$00 gives the following quadratic equation:

We can solve for $x$xx using the quadratic formula, which gives the following solutions:

Since the concentration of $\text{OH}^-$OH−start text, O, H, end text, start superscript, minus, end superscript can't be negative, we can eliminate the second solution. If we plug in the first value of $x$xx to get the equilibrium concentration of $\text H^+$H+start text, H, end text, start superscript, plus, end superscript and calculate $\text{pH}$pHstart text, p, H, end text, we get:

Thus we can see that once we include the autoionization of water, our very dilute $\text{HCl}$HClstart text, H, C, l, end text solution has a $\text {pH}$pHstart text, p, H, end text that is weakly acidic. Whew!