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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**.

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Jetzt kostenlos anmeldenYou’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**.

The ionic product of water, ${\mathrm{K}}_{\mathrm{w}}$, is a modified equilibrium constant for the dissociation of water.

Let’s explore that term in more detail.

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**.

An amphoteric substance behaves as both an acid and a base.

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:

$2{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)\leftrightharpoons {\mathrm{H}}_{3}{\mathrm{O}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

Note the state symbols used. Water is liquid, whilst the hydronium ions and hydroxide ions are aqueous - meaning they are dissolved in water.

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:

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

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

An equilibrium is a state of reaction where the rates of the forward and backward reactions are the same, and the concentrations of reactants and products remain constant.

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

$\left[\mathrm{A}\right]$ means the concentration of A. ${\left[\mathrm{A}\right]}^{\mathrm{a}}$means you take the number of moles of A given in the equation and raise the concentration of A to this power. For example, if one of your reactants has a concentration of $0.5\mathrm{mol}{\mathrm{dm}}^{-3}$ and there are 2 moles of it in the equation, ${\left[\mathrm{A}\right]}^{\mathrm{a}}=0.52$.

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.

${\mathrm{K}}_{\mathrm{w}}$ actually equals ${\mathrm{K}}_{\mathrm{c}}\mathrm{x}\left[{\mathrm{H}}_{2}\mathrm{O}\right]$. The water molecules on the top and bottom of the equation cancel out to leave just $\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$.

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:

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

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 ${\mathrm{K}}_{\mathrm{w}}$:

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

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

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℃, ${\mathrm{K}}_{\mathrm{w}}=1.00\mathrm{x}{10}^{-14}$.

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.

A neutral solution has equal concentrations of hydrogen and hydroxide ions.

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.

At 40℃, the ${\mathrm{K}}_{\mathrm{w}}$ of water is $2.09\mathrm{x}{10}^{-14}$. Calculate its pH.

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:

${\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{H}}^{+}\right]={\left[{\mathrm{H}}^{+}\right]}^{2}\phantom{\rule{0ex}{0ex}}{\left[{\mathrm{H}}^{+}\right]}^{2}=2.09\mathrm{x}{10}^{-14}$

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

$\left[{\mathrm{H}}^{+}\right]=\sqrt{2.09\mathrm{x}{10}^{-14}}\phantom{\rule{0ex}{0ex}}\left[{\mathrm{H}}^{+}\right]=1.45\mathrm{x}{10}^{-7}$

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

$\mathrm{pH}=-\mathrm{log}\left(\right[{\mathrm{H}}^{+}\left]\right)\phantom{\rule{0ex}{0ex}}\mathrm{pH}=-\mathrm{log}(1.45\mathrm{x}{10}^{-7})=6.84$

Remember that this solution is still neutral. It contains equal concentrations of hydrogen and hydroxide ions.

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 ${\mathrm{pK}}_{\mathrm{w}}$.

Calculate the pH of a $0.05\mathrm{mol}{\mathrm{dm}}^{-3}$ solution of potassium hydroxide, $\mathrm{KOH}$, at 25℃.

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:

$\left[{\mathrm{H}}^{+}\right]=\frac{1.00\mathrm{x}{10}^{-14}}{0.05}=2\mathrm{x}{10}^{-13}\phantom{\rule{0ex}{0ex}}\mathrm{pH}=-\mathrm{log}\left(\right[{\mathrm{H}}^{+}\left]\right)\phantom{\rule{0ex}{0ex}}\mathrm{pH}=-\mathrm{log}(2\mathrm{x}{10}^{-13})=12.70$

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, **p****O****H** 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:

$\mathrm{pOH}=-\mathrm{log}\left(\right[\mathrm{OH}\left]\right)\phantom{\rule{0ex}{0ex}}\mathrm{pH}=-\mathrm{log}(0.05)=1.30$

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

${\mathrm{pK}}_{\mathrm{w}}=-\mathrm{log}\left({\mathrm{K}}_{\mathrm{w}}\right)\phantom{\rule{0ex}{0ex}}{\mathrm{pK}}_{\mathrm{w}}=-\mathrm{log}(1.00\mathrm{x}{10}^{-14})=14\phantom{\rule{0ex}{0ex}}\mathrm{pH}={\mathrm{pK}}_{\mathrm{w}}-\mathrm{pOH}\phantom{\rule{0ex}{0ex}}\mathrm{pH}=14-1.30=12.70$

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

Both methods used to find pH work equally well. Find out which one your exam board wants you to know and practice using that one.

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

- Water is an amphoteric substance, meaning it acts as both an acid and a base.
- Water dissociates to form hydronium ions and hydroxide ions.
- The ionic product of water, ${\mathrm{K}}_{\mathrm{w}}$, is a modified equilibrium constant for the dissociation of water. It is affected by temperature. ${\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$ and has the units ${\mathrm{mol}}^{2}{\mathrm{dm}}^{-6}$
- ${\mathrm{pK}}_{\mathrm{w}}=-\mathrm{log}\left({\mathrm{K}}_{\mathrm{w}}\right)$
- We can use ${\mathrm{K}}_{\mathrm{w}}$ to find the pH of water and solutions containing strong bases.

What is an amphoteric substance?

A substance that can behave as both an acid and a base.

The forward reaction for the dissociation of water is _________.

Endothermic

At temperature A, \(K_w = 6.81 \cdot 10^{-15}\). At temperature B, \(K_w = 1.471 \cdot 10^{-14}\). Which of the following statements is true?

Temperature B is higher and has a lower pH.

What is a neutral solution?

A neutral solution has equal concentrations of hydrogen and hydroxide ions.

What is the pH of a neutral solution?

It depends

How many decimal places do we give pH to?

2

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