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But did you know that even if you drink 'pure' water, you don't actually get just H2O molecules? You actually also get trace amounts of hydronium (H3O+) and hydroxide (OH−) ions. This is due to the autoionization of water, a cool property of the one liquid that is so vital to our body.
- This article is about the autoionization of water.
- First, we will define what the autoionization of water means.
- Then, we will look at the chemical equation for theautoionization of water.
- After, we will zoom in on the autoionization constant of water, known as Kw, and learn how to write its expression.
- Once we have a foundational understanding of autoionization, we can then consider the relationship between the autoionization of water and pH.
- Finally, we'll practice calculating pH using examples of the autoionization of water. We'll also consider the limitations involved in these sorts of calculations.
Autoionization of Water Definition
In the article Acids and Bases, you learned that water has a pretty special property. Water is amphoteric, which means that it can act as both a Bronstead-Lowry acid (a proton donor) and a Bronstead-Lowry base (a proton acceptor). Take a look:
- Water molecules can act as acids by dissociating into hydroxide ions (OH-) and hydrogen ions (H+). This is equivalent to losing a proton.
- Water molecules can also act as bases by combining with hydrogen ions to form hydronium ions (H3O+). This is equivalent to gaining a proton.
- Both reactions are reversible - the hydronium ion and hydroxide ion can recombine to form water once more.
- This amphoteric nature means that two water molecules can even react with each other. In this case, the first loses a proton, which the second one then gains.
The ability of water to react with itself and form two ions is known as autoionization.
The autoionization of water is the reaction between two water molecules to produce a hydroxide ion (OH-) and a hydronium ion (H3O+).
Autoionization of Water Equation
To better visualize what exactly happens in the autoionization of water, let’s write a chemical equation to show the process.
Firstly, a water molecule splits into H+ and OH-. Then, a second water molecule reacts with the H+ produced to form H3O+:
$$H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$$
$$H_2O(l)+H^+(aq)\rightleftharpoons H_3O^+(aq)$$
If we combine the two steps, we end up with one overall equation for the autoionization of water:
$$2H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)$$
Notice a few things about the equation:
- H3O+ and OH- are produced in an equal ratio. Therefore, in a solution of pure water, [H3O+ ] and [OH-] are the same.
- The reaction is reversible. This means that it eventually forms a state of dynamic equilibrium. Like with all equilibria, we can represent the autoionization of water using an equilibrium constant, which we'll look at in just a second.
Simplifying the equation
The aqueous hydronium ion, H3O+(aq), is really nothing more than a hydrated proton. We can instead represent it using an aqueous hydrogen ion, H+(aq). This allows us to simplify the equation for the autoionization of water:
$$H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)$$
H3O+(aq) is commonly replaced with H+(aq) in most acid-base chemistry. For the rest of this article, we'll use this simplified form of notation.
Autoionization Constant of Water
So, we now know more about autoionization and have an equation to back it up. This equation shows that the autoionization of water is reversible, and so forms a dynamic equilibrium. Therefore, we can represent autoionization with an equilibrium constant.
The equilibrium constant, Keq, is a value that tells us the relative amounts of reactants and products in a system at equilibrium.
The equilibrium constant for the autoionization of water is calculated using concentration, and so is based on Kc. This constant is important enough that it has its own name: Kw. Kw is also known as the ion-product constant of water.
Kw Expression
Kw is expressed by:
$$K_w=[H^+]_{eqm}[OH^-]{eqm}$$
We often ignore the subscript eqm in the Kw expression. Hence, you'll commonly see \(K_w=[H^+][OH^-]\) . Once again, we'll stick to the simpler notation for the rest of the article.
Try deriving the Kw expression yourself from the general expression for Kc.
For the general equation \(aA(g)+bB(g) \rightleftharpoons cC(g)+dD(g)\) , the equilibrium constant Kc has the following expression:
$$K_c=\frac{{[C]_{eqm}}^c\space {[D]_{eqm}}^d}{{[A]_{eqm}}^a\space {[B]_{eqm}}^b}$$
Apply that to the equation for the autoionization of water:
$$K_w=\frac{[H^+]_{eqm}[OH^-]_{eqm}}{[H_2O]_{eqm}}$$
However, can you see that H2O is liquid? Remember that in equilibrium constant expressions, we don't include the concentration of any pure liquids or solids. Therefore, we ignore [H2O] in the Kw expression, leaving us with the familiar form shown earlier:
$$K_w=[H^+]_{eqm}[OH^-]{eqm}$$
Kw Properties
Kw is an equilibrium constant. As a result, it shares common properties with all forms of Keq:
- Kw tells us about the position of the equilibrium reaction. Hence, it tells us about the extent of the autoionization of water.
- The value of Kw is fixed at a particular temperature. However, change the temperature, and you change the value of Kw.
- On the other hand, Kw is unaffected by factors such as concentration and pressure.
For example, at 25 °C (298 K), Kw always equals 1.0 × 10-14. This is a small value and so tells us that the position of the equilibrium reaction lies far to the left - the vast majority of water molecules don't ionize.
But despite its diminutive size, Kw is still significant. This is because it allows us to calculate [H+], [OH-], and pH for aqueous solutions. Let's find out how.
Autoionization of Water and pH
Remember that the value of Kw is fixed at a particular temperature. We say that it is temperature-dependent. At 25 °C, Kw equals 1.0 × 10-14. Therefore, in all aqueous solutions, \( [H^+][OH^-]=1.0\times 10^{-14}\) . Provided we know information about any of [H+], [OH-], or pH, we can use Kw to calculate the other values.
The concentration of hydrogen ions in pure water at 25ºC is 1.0 × 10-7 M. We can use this information to calculate its pH:
$$p\,H=-log_{10}[H^+]$$
$$pH=-\log_{10}(1.0\times 10^{-7})$$
$$pH=7.0$$
Likewise, we can work out the concentration of hydroxide ions, using the fact that Kw equals 1.0 × 10-14 at this temperature:
$$K_w=[H^+][OH^-]$$
$$1.0\times 10^{-14}=(1.0\times 10^{-7})\times [OH^-]$$
$$[OH^-]=1.0\times 10^{-7}\space M$$
Notice that in pure water, the concentration of hydrogen ions equals the concentration of hydroxide ions. We'd expect this result - after all, the balanced chemical equation for the autoionization of water produces an equal ratio of H+ and OH-. This makes pure water neutral.
A neutral substance contains equal concentrations of hydrogen ions and hydroxide ions. Hence, [H+] = [OH-].
At 25 °C, neutral substances (be it water or another aqueous solution) have a pH of exactly 7.0. But as temperature changes, the value of Kw changes, and the pH of neutral substances does too. One thing remains constant: no matter the temperature, neutral substances always have the same concentration of hydrogen ions as hydroxide ions.
Well, that's the pH of neutral substances covered. However, most aqueous solutions aren't neutral. Instead, they are either acidic or basic; in these solutions, [H+] ≠ [OH-]. Let's see how this affects their acidity and pH:
- If [H+] > [OH-], the solution is acidic. At 25 °C, it has a pH below 7.0.
- If [H+] < [OH-], the solution is basic. At 25 °C, it has a pH above 7.0.
Fig.1-A diagram showing how the relative concentrations of hydrogen and hydroxide ions affect the pH of aqueous solutions.
Always remember that for aqueous solutions, \([H^+][OH^-]\equiv K_w\) !
Autoionization of Water Examples
Time to pull together what we have learned by practicing some example calculations involving the autoionization of water. Unless stated otherwise, assume that the temperature is 25 °C in all the following questions. First, have a go at finding pH using Kw and [OH-]. Here are the steps:
- Use the expression for Kw to find [H+].
- Use [H+] to calculate pH.
- Find the pH of an aqueous solution containing 5.6 × 10-4 M OH- ions.
- State whether the solution is acidic, neutral, or basic.
To solve part a, use the expression for Kw to find [H+]. Because the temperature is 25 °C, we know that Kw equals 1.0 × 10-14:
$$K_w=[H^+][OH^-]$$
$$1.0\times 10^{-14}=[H^+]\times (5.6\times 10^{-4})$$
$$[H^+]=1.7857\times 10^{-11}$$
Now, use the calculated value of [H+] to work out pH:
$$pH=-\log{10}[H^+]$$
$$pH=-\log_{10}(1.7857\times 10^{-11})$$
$$pH=10.748$$
pH is given as standard to one decimal place. Hence, our final answer for part a is 10.7.
To find part b, we compare the pH value just calculated to the pH of pure water at this temperature, which is 7.0. Because 10.7 > 7.0, the solution is basic.
You can work out pH differently using the relationship between pKw, pH, and pOH. Let's derive the equation linking the three variables together:
- Start with the expression for Kw: $$K_w=[H^+][OH^-]$$
- Take negative logs of both sides: $$-log_{10}K_w=-log_{10}([H^+][OH^-])$$
- Expand using the laws of logarithms: $$-log_{10}K_w=-log_{10}[H^+]+-log_{10}[OH^-]$$
- Simplify your answer: $$pK_w=pH+pOH$$
So, for our example above:
$$-\log{10}(1.0\times 10^{-14})=pH-\log{10}(5.6\times 10^{-4})$$
$$pH=10.74=10.7$$
Time to move on to a different sort of problem. Here, you need to work backwards and find both [H+] and [OH-] using pH and Kw:
- Use pH to find [H+].
- Use [H+] and the expression for Kw to find [OH-].
Try this problem:
The pH of an aqueous solution is 3.5. Find [OH-] to three significant figures.
First, find [H+]:
$$[H^+]=10^{-pH}$$
$$[H^+]=10^{-3.5}$$
$$[H^+]=6.162\times 10^{-4}\space M$$
Now, use [H+] to find [OH-]:
$$K_w=[H^+][OH^-]$$
$$1.0\times 10^{-14}=(6.162\times 10^{-4})\times [OH^-]$$
$$[OH-]=3.162\times 10^{-11}\space M$$
Rounding to three significant figures, we find that [OH-] = 3.16 × 10-11 M.
Finally, have a go at finding the pH of pure water at a higher temperature. Remember that this changes the value of Kw. The method is similar to the one you used in the first problem.
At 75 °C, pure water has an ion-product constant of 2.0 × 10-13. Find its pH.
First, we find [H+] using the ion-product constant, which is simply another name for Kw. You might think that there is an issue here - we don't know [OH-]. However, we do know that we are working with pure water, which contains equal concentrations of hydrogen and hydroxide ions. Hence, in this case, [H+] = [OH-]:
$$K_w=[H^+][H^+]=[H^+]^2$$
$$2.0\times 10^{-13}=[H^+]^2$$
$$[H^+]=\sqrt{2.0\times 10^{-13}}$$
$$[H^+]=4.472\times 10^{-7}\space M$$
Now, work out pH:
$$pH=-\log_{10}(4.472\times 10^{-7})$$
$$pH=6.349$$
Rounding to one decimal place, the pH of pure water at this temperature equals 6.3.
Notice how at a higher temperature, both Kw and [H+] are larger? This can be explained by applying Le Chatelier's principle to the autoionization of water, looking in particular at the reaction's enthalpy change.
Here's the balanced equation for the autoionization of water:
$$H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)\qquad \Delta ^\circ H= \text{positive}$$
We can see that the forward reaction is endothermic. Le Chatelier's principle tells us that increasing the temperature shifts the position of the equilibrium to the right, counteracting the disturbance to the system. This increases the concentration of both H+ and OH- ions. Thus, Kw increases.
Limitations of the Autoionization of Water
We've seen how we can use the autoionization of water and Kw to find the pH of an aqueous solution. This is handy for all sorts of acid-base calculations. In these problems, we ignore the H+ and OH- ions from the ionization of water itself, and assume that they all come from the other acid or base involved. This typically isn't a problem. Remember that Kw is very low, meaning that only a small proportion of water molecules ionize at equilibrium. The ions that come from the autoionization of water have very little effect on the pH of the solution.
However, water's autoionization becomes an issue if working with extremely dilute acids or bases, such as those with [H+] or [OH-] values less than two orders of magnitude away from 1.0 × 10-7. Because these acids and bases also have such low values of [H+] or [OH-], the H+ and OH- ions from water become much more important, and you can't ignore their effect on pH.
Autoionization of Water - Key takeaways
- The autoionization of water is the reaction between two water molecules to produce a hydroxide ion (OH-) and a hydronium ion (H3O+).
- We represent the autoionization of water with the equation \(2H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)\) . This can be simplified to \(H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)\)
Kw is the equilibrium constant for the autoionization of water. It is also known as the ion-product constant. The expression for Kw is given by \(K_w=[H^+]_{eqm}[OH^-]{eqm}\)
Kw is temperature-dependent, meaning that it is always the same at a particular temperature. At 25 °C, Kw = 1.0 × 10-14.
We can use Kw to calculate [H+], [OH-], and the pH of aqueous solutions. For example, at 25 °C, pure water has a pH of 7.0.
References
- Anne Marie Helmenstine, Ph.D, 'How Much of Your Body Is Water?' ThoughtCo (07/09/2021)
- Shaun K Riebl, Brenda M. Davy, 'The Hydration Equation: Update on Water Balance and Cognitive Performance', 'ACSMs Health Fit', J. 2013 November/December, 17(6): 21–28
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Frequently Asked Questions about Autoionization of Water
What is the autoionization of water?
The autoionization of water is the reaction between two water molecules to produce a hydroxide ion and a hydronium ion.
Is the autoionization of water exothermic or endothermic?
The autoionization of water is an endothermic reaction.
Why is the autoionization of water important?
The autoionization of water is important because it allows us to calculate the pH of aqueous solutions.
How do you calculate pH when considering the autoionization of water?
To calculate pH using the autoionization of water, you need to use the equilibrium constant Kw. You first use the expression for Kw to find [H+]. You then use [H+] to find pH. We provide you with plenty of worked examples in this article, so check them out if you aren't sure about the process.
How do you calculate the standard enthalpy for autoionization of water?
You can use Hess' law and standard enthalpies of formation to calculate the standard enthalpy change for the autoionization of water.
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