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 H, start subscript, 3, end subscript, O, start superscript, plus, end superscript and O, H, start superscript, minus, end superscript ions.
  • The equilibrium constant for the autoionization of water, K, start subscript, w, end subscript, is 10, start superscript, minus, 14, end superscript at 25, space, degree, C.
  • In a neutral solution, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, equals, open bracket, O, H, start superscript, minus, end superscript, close bracket
  • In an acidic solution, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, is greater than, open bracket, O, H, start superscript, minus, end superscript, close bracket
  • In a basic solution, open bracket, O, H, start superscript, minus, end superscript, close bracket, is greater than, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket
  • For aqueous solutions at 25, space, degree, C, the following relationships are always true:
K, start subscript, w, end subscript, equals, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, open bracket, O, H, start superscript, minus, end superscript, close bracket, equals, 10, start superscript, minus, 14, end superscript
p, H, plus, p, O, H, equals, 14
  • The contribution of the autoionization of water to open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket and open bracket, O, H, start superscript, minus, end superscript, close bracket 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.
რეაქცია
Role of water
  • CH3NH2(aq)+H2O(l)CH3NH3+(aq)+OH(aq)\text{CH}_3\text{NH}_2(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{CH}_3\text{NH}_3^+(aq)+\text{OH}^-(aq)
  • HSO4(aq)+H2O(l)SO42(aq)+H3O+(aq)\text{HSO}_4^-(aq)+\text{H}_2\text{O}(l)\rightleftharpoons\text{SO}_4^{2-}(aq)+\text{H}_3\text{O}^+(aq)
  • C, u, start superscript, 2, plus, end superscript, left parenthesis, a, q, right parenthesis, plus, 6, space, H, start subscript, 2, end subscript, O, left parenthesis, l, right parenthesis, right arrow, C, u, left parenthesis, H, start subscript, 2, end subscript, O, right parenthesis, start subscript, 6, end subscript, start superscript, 2, plus, end superscript, left parenthesis, a, q, right parenthesis
  • b, a, s, e
  • a, c, i, d
  • n, e, i, t, h, e, r

When water donates H, start superscript, plus, end superscript to form O, H, start superscript, minus, end superscript, it is acting as a Bronsted-Lowry acid. When water accepts a proton for form H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, it is acting as a Bronsted-Lowry base.
სწორი პასუხებია:

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 happenminuswater 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:
 H2O(l)+H2O(l)H3O+(aq)+OH(aq)\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, colon, 1 molar ratio. For any sample of pure water, the molar concentrations of hydronium, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, and hydroxide, O, H, start superscript, minus, end superscript, must be equal:
open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, equals, open bracket, O, H, start superscript, minus, end superscript, close bracket, space, space, i, n, space, p, u, r, e, space, w, a, t, e, r
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, C, O, start subscript, 2, end subscript. This is because when carbon dioxide dissolves in water, it can form small amounts of the weak acid carbonic acid, H, start subscript, 2, end subscript, C, O, start subscript, 3, end subscript.
Note that this is 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, start subscript, w, end subscript.

The autoionization constant, K, start subscript, w, end subscript

The expression for the autoionization constant is
K, start subscript, w, end subscript, equals, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, open bracket, O, H, start superscript, minus, end superscript, close bracket, space, space, left parenthesis, E, q, point, space, 1, right parenthesis
Remember that when writing equilibrium expressions, the concentrations of solids and liquids are not included. Therefore, our expression for K, start subscript, w, end subscript does not include the concentration of water, which is a pure liquid.
We can calculate the value of K, start subscript, w, end subscript at 25, space, degree, C using open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, which is related to the p, H of water. At 25, space, degree, C, the p, H of pure water is 7. Therefore, we can calculate the concentration of hydronium ions in pure water:
open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, equals, 10, start superscript, minus, p, H, end superscript, equals, 10, start superscript, minus, 7, end superscript, space, M, space, space, a, t, space, 25, space, degree, C
In the last section, we saw that hydronium and hydroxide form in a 1, 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, degree, C:
open bracket, O, H, start superscript, minus, end superscript, close bracket, equals, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, equals, 10, start superscript, minus, 7, end superscript, space, M, space, space, a, t, space, 25, space, degree, C
This is a little tough to visualize, but 10, 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 open bracket, O, H, start superscript, minus, end superscript, close bracket and open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, we can use these values in our equilibrium expression to calculate K, start subscript, w, end subscript at 25, degree, C:
K, start subscript, w, end subscript, equals, left parenthesis, 10, start superscript, minus, 7, end superscript, right parenthesis, times, left parenthesis, 10, start superscript, minus, 7, end superscript, right parenthesis, equals, 10, start superscript, minus, 14, end superscript, space, space, a, t, space, 25, space, degree, C
Concept check: How many hydroxide and hydronium ions are in one liter of water at 25, degree, C?
We know that at 25, degree, C, the following relationship is true:
open bracket, O, H, start superscript, minus, end superscript, close bracket, equals, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, equals, 10, start superscript, minus, 7, end superscript, space, start fraction, m, o, l, divided by, L, end fraction
That means one liter of water contains 10, start superscript, minus, 7, end superscript, space, m, o, l each of hydronium ions and hydroxide ions. We can convert from moles to the number of ions using Avogadro's number:
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, point, 34, times, 10, start superscript, 25, end superscript molecules of water in one liter of water. That is nearly 10, start superscript, 9, end superscript larger than the number of H, start subscript, 3, end subscript, O, start superscript, plus, end superscript and O, H, start superscript, minus, end superscript ions in solution!

Relationship between the autoionization constant, p, H, and p, O, H

The fact that K, start subscript, w, end subscript is equal to 10, start superscript, minus, 14, end superscript at 25, space, degree, C leads to an interesting and useful new equation. If we take the negative logarithm of both sides of E, q, point, space, 1 in the previous section, we get the following:
logKw=log([H3O+][OH])=(log[H3O+]+log[OH])=log[H3O+]+(log[OH])=pH+pOH\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 p, H. For a review of useful rules for logarithm calculations, see Sal's video on properties of logarithms.
We can abbreviate minus, log, K, start subscript, w, end subscript as p, K, start subscript, w, end subscript, which is equal to 14 at 25, space, degree, C:
p, K, start subscript, w, end subscript, equals, p, H, plus, p, O, H, equals, 14, space, space, a, t, space, 25, space, degree, C, space, space, left parenthesis, E, q, point, space, 2, right parenthesis
Therefore, the sum of p, H and p, O, H will always be 14 for any aqueous solution at 25, space, degree, C. Keep in mind that this relationship will not hold true at other temperatures, because K, start subscript, w, end subscript is temperature dependent!

Example 1: Calculating open bracket, O, H, start superscript, minus, end superscript, close bracket from p, H

An aqueous solution has a p, H of 10 at 25, space, degree, 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 open bracket, H, start superscript, plus, end superscript, close bracket from the p, H:
[H3O+]=10pH=1010M\begin{aligned}[\text{H}_3\text{O}^+]&=10^{-\text{pH}}\\ \\ &=10^{-10}\,\text M\\\end{aligned}
We can then calculate open bracket, O, H, start superscript, minus, end superscript, close bracket using Eq. 1:

Method 2: Using Eq. 2

Another way to calculate open bracket, O, H, start superscript, minus, end superscript, close bracket is to calculate it from the p, O, H of the solution. We can use Eq. 2 to calculate the p, O, H of our solution from the p, H. Rearranging Eq. 2 and solving for the p, O, H, we get:
pOH=14pH=1410=4\begin{aligned}\text{pOH}&=14-\text{pH}\\ \\ &=14-10\\ \\ &=4\end{aligned}
We can now use the equation for p, O, H to solve for open bracket, O, H, start superscript, minus, end superscript, close bracket.
[OH]=10pOH=104 M\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, start superscript, minus, 4, end superscript, space, M for an aqueous solution with a p, H of 10 at 25, space, degree, C.

Definitions of acidic, basic, and neutral solutions

We have seen that the concentrations of H, start subscript, 3, end subscript, O, start superscript, plus, end superscript and O, H, start superscript, minus, end superscript are equal in pure water, and both have a value of 10, start superscript, minus, 7, end superscript, space, M at 25, space, degree, 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, start subscript, 3, end subscript, O, start superscript, plus, end superscript and O, H, start superscript, minus, end superscript.
  • In a neutral solution, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, equals, open bracket, O, H, start superscript, minus, end superscript, close bracket
  • In an acidic solution, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, is greater than, open bracket, O, H, start superscript, minus, end superscript, close bracket
  • In a basic solution, open bracket, O, H, start superscript, minus, end superscript, close bracket, is greater than, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket

Practice 2: Calculating p, H of water at 0, space, degree, C

If the p, K, start subscript, w, end subscript of a sample of pure water at 0, space, degree, C is 14, point, 9, what is the p, H of pure water at this temperature?
გთხოვთ, აირჩიეთ მოცემული ვარიანტებიდან ერთი.

For aqueous solutions at 0, space, degree, C
p, K, start subscript, w, end subscript, equals, 14, point, 9, equals, p, H, plus, p, O, H
Since pure water at 0, space, degree, C is neutral, we know that open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, equals, open bracket, O, H, start superscript, minus, end superscript, close bracket and p, H, equals, p, O, H.
If we plug this relationship into the equation for p, K, start subscript, w, end subscript, we get:
pH+pH=14.9=2×pHpH=14.92=7.45\begin{aligned}\text{pH}+\text{pH}&=14.9=2 \times \text{pH}\\ \\ \text{pH}&=\dfrac{14.9}{2}=7.45\end{aligned}
The p, H of pure water at at 0, space, degree, C is 7, point, 45.

Practice 3: Calculating p, K, start subscript, w, end subscript at 40, space, degree, C

The p, H of pure water at 40, space, degree, C is measured to be 6, point, 75.
Based on this information, what is the p, K, start subscript, w, end subscript of water at 40, space, degree, C?
გთხოვთ, აირჩიეთ მოცემული ვარიანტებიდან ერთი.

Pure water is neutral, so we know that open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, equals, open bracket, O, H, start superscript, minus, end superscript, close bracket.
If we take the minus, l, o, g of both sides of the equation, we get
p, H, equals, p, O, H
We can find the relationship between p, H and p, K, start subscript, w, end subscript for pure water at 40, space, degree, C using Eq. 2:
pKw=pH+pOH=2×pH=2×6.75=13.5\begin{aligned}\text{p}K_\text{w}&=\text{pH}+\text{pOH}\\ \\ &=2 \times \text{pH}\\ \\ &=2\times 6.75 \\ \\ &=13.5\end{aligned}
The p, K, start subscript, w, end subscript for pure water at 40, space, degree, C is 13, point, 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 open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket and open bracket, O, H, start superscript, minus, end superscript, close bracket?
The moment we dissolve other acids or bases in water, we change open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket and/or open bracket, O, H, start superscript, minus, end superscript, close bracket such that the product of the concentrations is no longer is equal to K, start subscript, w, 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, space, degree, C has a hydronium ion concentration of 10, start superscript, minus, 7, end superscript, space, M, the added acid increases the concentration of H, start subscript, 3, end subscript, O, 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 H, start subscript, 3, end subscript, O, start superscript, plus, end superscript. This causes the concentration of O, H, start superscript, minus, end superscript to decrease until the product of open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket and open bracket, O, H, start superscript, minus, end superscript, close bracket is once again equal to 10, start superscript, minus, 14, end superscript.
Once the reaction reaches its new equilibrium state, we know that:
  • open bracket, H, start superscript, plus, end superscript, close bracket, is greater than, open bracket, O, H, start superscript, minus, end superscript, close bracket because the added acid increased open bracket, H, start superscript, plus, end superscript, close bracket. Thus, our solution is acidic!
  • open bracket, O, H, start superscript, minus, end superscript, close bracket, is less than, 10, start superscript, minus, 7, end superscript, space, M because favoring the reverse reaction decreased open bracket, O, H, 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!

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 open bracket, H, start superscript, plus, end superscript, close bracket and p, H 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 open bracket, H, start superscript, plus, end superscript, close bracket or open bracket, O, H, 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 H, start superscript, plus, end superscript or O, H, start superscript, minus, end superscript is within ~2 orders of magnitude (or less than) of 10, start superscript, minus, 7, end superscript, space, M. We will now go through an example of how to calculate the p, H of a very dilute acid solution.

Example 2: Calculating the p, H of a very dilute acid solution

Let's calculate the p, H of a H, C, l solution with a hydronium ion concentration of 6, point, 3, times, 10, start superscript, minus, 8, end superscript, space, M.

Try 1: Ignoring the autoionization of water

If we ignore the autoionization of water and simply use the formula for p, H, we get:
pH=log[H+]=log[6.3×108]=7.20\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 p, H 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 open bracket, H, start superscript, plus, end superscript, close bracket

Since the concentration of this solution is extremely dilute, the concentration of the hydronium from the hydrochloric acid is close to the open bracket, H, start superscript, plus, end superscript, close bracket contribution from the autoionization of water. That means:
  • We have to include the contribution from autoionization to open bracket, H, start superscript, plus, end superscript, close bracket
  • Since the autoionization of water is an equilibrium reaction, we must solve for the overall open bracket, H, start superscript, plus, end superscript, close bracket using the expression for K, start subscript, w, end subscript:
K, start subscript, w, end subscript, equals, open bracket, H, start superscript, plus, end superscript, close bracket, open bracket, O, H, start superscript, minus, end superscript, close bracket, equals, 1, point, 0, times, 10, start superscript, minus, 14, end superscript
If we say that x is the contribution of autoionization to the equilibrium concentration of H, start superscript, plus, end superscript and O, H, start superscript, minus, end superscript, the concentrations at equilibrium will be as follows:
open bracket, H, start superscript, plus, end superscript, close bracket, equals, 6, point, 3, times, 10, start superscript, minus, 8, end superscript, space, M, plus, x
open bracket, O, H, start superscript, minus, end superscript, close bracket, equals, x
Plugging these concentrations into our equilibrium expression, we get:
Kw=(6.3×108M+x)x=1.0×1014=x2+6.3×108x\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, equals, x, start superscript, 2, end superscript, plus, 6, point, 3, times, 10, start superscript, minus, 8, end superscript, x, minus, 1, point, 0, times, 10, start superscript, minus, 14, end superscript
We can solve for x using the quadratic formula, which gives the following solutions:
x, equals, 7, point, 3, times, 10, start superscript, minus, 8, end superscript, space, M, comma, minus, 1, point, 4, times, 10, start superscript, minus, 7, end superscript, space, M
Since the concentration of O, H, start superscript, minus, end superscript 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, start superscript, plus, end superscript and calculate p, H, we get:
pH=log[H+]=log[6.3×108+x]=log[6.3×108+7.3×108]=log[1.36×107]=6.87\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 H, C, l solution has a p, H that is weakly acidic. Whew!

შინაარსი

  • Water can undergo autoionization to form H, start subscript, 3, end subscript, O, start superscript, plus, end superscript and O, H, start superscript, minus, end superscript ions.
  • The equilibrium constant for the autoionization of water, K, start subscript, w, end subscript, is 10, start superscript, minus, 14, end superscript at 25, space, degree, C.
  • In a neutral solution, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, equals, open bracket, O, H, start superscript, minus, end superscript, close bracket
  • In an acidic solution, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, is greater than, open bracket, O, H, start superscript, minus, end superscript, close bracket
  • In a basic solution, open bracket, O, H, start superscript, minus, end superscript, close bracket, is greater than, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket
  • For aqueous solutions at 25, space, degree, C, the following relationships are always true:
K, start subscript, w, end subscript, equals, open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket, open bracket, O, H, start superscript, minus, end superscript, close bracket, equals, 10, start superscript, minus, 14, end superscript
p, H, plus, p, O, H, equals, 14
  • The contribution of the autoionization of water to open bracket, H, start subscript, 3, end subscript, O, start superscript, plus, end superscript, close bracket and open bracket, O, H, start superscript, minus, end superscript, close bracket becomes significant for extremely dilute acid and base solutions.

Attributions

This article was adapted from the following articles:
The modified article is licensed under a CC-BY-NC-SA 4.0 license.

Additional References

Zumdahl, S.S., and Zumdahl S.A. (2003). Atomic Structure and Periodicity. In Chemistry (6th ed., pp. 290-94), Boston, MA: Houghton Mifflin Company.