Ionic Product of Water

You’ve probably been told your entire life that a pH of 7 is neutral. This isn’t actually true. A neutral solution has equal concentrations of hydrogen and hydroxide ions, regardless of its pH. We only think of 7 as the magic number because it is the pH value of water at room temperature. However, water’s pH can vary depending on the temperature. This is all to do with a value known as ${\mathrm{K}}_{\mathrm{w}}$, which is the ionic product of water.

Let’s explore that term in more detail.

How does water dissociate?

You should be familiar with the actions of acids and bases in water. They both dissociate.

• Acids dissociate to form positive hydrogen ions, ${\mathrm{H}}^{+}$, alongside a negative ion.
• Bases dissociate to form negative hydroxide ions, ${\mathrm{OH}}^{-}$, alongside a positive ion.

Water itself dissociates. However, it does something a little different - it behaves amphoterically.

Regardless of whether it is pure or not, water always partially dissociates into hydronium ions, ${\mathrm{H}}_3{\mathrm{O}}^{+}$, and hydroxide ions, ${\mathrm{OH}}^{-}$. One molecule of water acts as an acid by donating a proton, ${\mathrm{H}}^{+}$, whilst a second water molecule acts as a base by accepting the proton. This is a reversible reaction and sets up the equilibrium shown below:

$2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\leftrightharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

The hydronium ion is a conjugate acid. You might remember from Brønsted-Lowry Acids and Bases that this is an acid formed when a base gains a proton. Likewise, the hydroxide ion is a conjugate base. This is a base formed when an acid loses a proton. Conjugate acids and bases behave just like standard acids and bases, and in fact, these two species are both quite strong. This means that once formed, they rapidly react with each other in the reverse reaction of our equation above, forming water again. Therefore, at any one time there are barely any hydronium ions and hydroxide ions in solution - most of the equilibrium consists of water molecules.

You can also represent the hydronium ion, ${\mathrm{H}}_3{\mathrm{O}}^{+}$ as just a proton, ${\mathrm{H}}^{+}$. This simplifies our equation:

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\leftrightharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

Fig. 1 - The dissociation of water

The equilibrium constant of water

You’ll notice that the dissociation of water is an equilibrium reaction, as we mentioned above. (We’d recommend checking Equilibria out.)

You should remember that we can write equilibrium constants for reaction, known as ${\mathrm{K}}_{\mathrm{c}}$. These relate the concentration of products to the concentration of reactants in a closed system at equilibrium. The formula is as follows:

For the reaction $\mathrm{aA}+\mathrm{bB}\leftrightharpoons \mathrm{cC}+\mathrm{dD}$, ${\mathrm{K}}_{\mathrm{c}}=\frac{{\left[\mathrm{C}\right]}^{\mathrm{c}}{\left[\mathrm{D}\right]}^{\mathrm{d}}}{{\left[\mathrm{A}\right]}^{\mathrm{a}}{\left[\mathrm{B}\right]}^{\mathrm{b}}}$

In our dissociation reaction with water, the reactant is water and the products are the hydronium ion and the hydroxide ion. There is exactly one mole of each in the reaction equation. This gives us a formula for ${\mathrm{K}}_{\mathrm{c}}$:

${\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{H}}_2\mathrm{O}\right]}$

We can also simplify it further by replacing the hydronium ion with the hydrogen ion:

${\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{H}}_2\mathrm{O}\right]}$

However, you might remember that we said that only a tiny proportion of water molecules dissociate at any one time. This means that the concentration of water molecules, $\left[{\mathrm{H}}_2\mathrm{O}\right]$, is so large that it is essentially a constant. To make things a little simpler, we can omit it entirely to create a modified equilibrium constant, ${\mathrm{K}}_{\mathrm{w}}$:

${\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

This can be simplified to the following:

${\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

For the rest of this article, we’ll use the second simplified version.

The units of Kw

To work out the units of ${\mathrm{K}}_{\mathrm{w}}$, we multiply the units for $\left[{\mathrm{H}}^{+}\right]$ and $\left[{\mathrm{OH}}^{-}\right]$ together. This gives us the following:

$\mathrm{mol}{\mathrm{dm}}^{-3}\mathrm{x}\mathrm{mol}{\mathrm{dm}}^{-3}={\mathrm{mol}}^2{\mathrm{dm}}^{-6}$

Kw and pKw

Instead of ${\mathrm{K}}_{\mathrm{w}}$, you might be given a value for ${\mathrm{pK}}_{\mathrm{w}}$. Just as how pH is the negative log of $\left[{\mathrm{H}}^{+}\right]$, ${\mathrm{pK}}_{\mathrm{w}}$ is the negative log of ${\mathrm{K}}_{\mathrm{w}}$:

${\mathrm{pK}}_{\mathrm{w}}=-\log \left({\mathrm{K}}_{\mathrm{w}}\right)$

${\mathrm{K}}_{\mathrm{w}}={10}^{-{\mathrm{pK}}_{\mathrm{w}}}$

The effect of temperature on Kw

Like any equilibrium constant, ${\mathrm{K}}_{\mathrm{w}}$ is affected by temperature. Let’s visit our equilibrium equations again, this time including the enthalpy change:

${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\leftrightharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\text{'}\mathrm{\Delta H}=+57.3\mathrm{kJ}{\mathrm{mol}}^{-1}$

What happens when we increase the temperature? The forward reaction is endothermic, meaning that it takes in energy. According to 'Le Chatelier’s Principle', changing the conditions of a reaction shifts the position of the equilibrium to oppose the change. Increasing the temperature means the forward reaction will be favoured to absorb the extra heat, and the equilibrium will shift to the right. The concentrations of the products will increase. This in turn means that ${\mathrm{K}}_{\mathrm{w}}$ increases.

At room temperature, roughly 25℃, ${\mathrm{K}}_{\mathrm{w}}=1.00\mathrm{x}{10}^{-14}$.

Fig. 2 - The effect of heat on the equilibrium reaction of the dissociation of water

How do we find the pH of water?

It’s all very well and good knowing what ${\mathrm{K}}_{\mathrm{w}}$ is and how it changes with temperature. Now we are going to learn how to use it to calculate the pH of water.

At the start of this article, we mentioned that water is neutral, but this doesn’t necessarily mean that it has a pH of 7.

Remember, pH is a measure of hydrogen ion concentration in solution. As temperature increases, ${\mathrm{K}}_{\mathrm{w}}$ increases as the forward reaction in water’s equilibrium dissociation reaction is favoured. This means that the concentration of hydrogen ions increases. Therefore, as temperature increases, the pH of water decreases.

We can use ${\mathrm{K}}_{\mathrm{w}}$ to find out pH, and vice versa. Let’s look at some examples.

We know that ${\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$, and at this temperature it equals $2.09\mathrm{x}{10}^{-14}$. We also know that water is neutral - it has equal concentrations of hydrogen and hydroxide ions. This means that $\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{OH}}^{-}\right]$. We can replace $\left[{\mathrm{OH}}^{-}\right]$ with $\left[{\mathrm{H}}^{+}\right]$ in our equation, as follows:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{H}}^{+}\right]={\left[{\mathrm{H}}^{+}\right]}^2 \\ {\left[{\mathrm{H}}^{+}\right]}^2=2.09\mathrm{x}{10}^{-14} \end{gathered}$$

To find pH, we need to know $\left[{\mathrm{H}}^{+}\right]$. To do this, we can just square root both sides of the equation:

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\sqrt{2.09\mathrm{x}{10}^{-14}} \\ \left[{\mathrm{H}}^{+}\right]=1.45\mathrm{x}{10}^{-7} \end{gathered}$$

Now we can substitute this value into the equation for pH:

$$\begin{gathered} \mathrm{pH}=-\log \left(\right[{\mathrm{H}}^{+}\left]\right) \\ \mathrm{pH}=-\log (1.45\mathrm{x}{10}^{-7})=6.84 \end{gathered}$$

How do we find the pH of bases?

In pH, we explored how we can use ${\mathrm{K}}_{\mathrm{w}}$ to find the concentration of hydrogen ions in solution, and thus the solution’s pH. We’ll recap that now before exploring an alternative method using ${\mathrm{pK}}_{\mathrm{w}}$.

Potassium hydroxide is a strong base and so dissociates fully in solution. This means that the concentration of hydroxide ions is also $0.05\mathrm{mol}{\mathrm{dm}}^{-3}$.

In pH, we used the water dissociation constant to calculate pH. In aqueous solution, ${\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$. At this temperature, its value is $1.00\mathrm{x}{10}^{-14}$. $\left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{OH}}^{-}\right]}$, which we can then use in our equation for pH:

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\frac{1.00\mathrm{x}{10}^{-14}}{0.05}=2\mathrm{x}{10}^{-13} \\ \mathrm{pH}=-\log \left(\right[{\mathrm{H}}^{+}\left]\right) \\ \mathrm{pH}=-\log (2\mathrm{x}{10}^{-13})=12.70 \end{gathered}$$

However, we can also use an alternative method based on a relationship between pH, pOH and ${\mathrm{pK}}_{\mathrm{w}}$:

$\mathrm{pH}+\mathrm{pOH}={\mathrm{pK}}_{\mathrm{w}}$

If we rearrange this, we get the following equation:

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{w}}-\mathrm{pOH}$

Similar to pH, pOH is a measure of hydroxide ion concentration in solution. We calculate it in the same way, taking the negative log of the hydroxide ion concentration. Let’s give it a go using the example above to see if we get the same answer:

$$\begin{gathered} \mathrm{pOH}=-\log \left(\right[\mathrm{OH}\left]\right) \\ \mathrm{pH}=-\log (0.05)=1.30 \end{gathered}$$

We can now put this into the equation relating pH, pOH and pKw. First we need to find pKw:

$$\begin{gathered} {\mathrm{pK}}_{\mathrm{w}}=-\log \left({\mathrm{K}}_{\mathrm{w}}\right) \\ {\mathrm{pK}}_{\mathrm{w}}=-\log (1.00\mathrm{x}{10}^{-14})=14 \\ \mathrm{pH}={\mathrm{pK}}_{\mathrm{w}}-\mathrm{pOH} \\ \mathrm{pH}=14-1.30=12.70 \end{gathered}$$

This is the same answer as the one that we got before.

The following flow charts summarise how you find the pH of water and strong bases:

Fig. 3 - Finding the pH of water and strong bases