R3.1.5 The ion product of water (Kw)

The Ion Product of Water ( $K_{w}$ ) and Its Implications

The Ion Product of Water ( $K_{w}$ )

Water, even in its pure form, undergoes a process called self-ionization (or autoionization), where two water molecules interact to form a hydronium ion ( $H_{3} O^{+}$ ) and a hydroxide ion ( $O H^{-}$ ):

$H_{2} O (l) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + O H^{-} (a q)$

For simplicity, we often represent the hydronium ion as $H^{+}$ , so the equation becomes:

$H_{2} O (l) ⇌ H^{+} (a q) + O H^{-} (a q)$

Tip

This reaction is an equilibrium process, meaning the concentrations of $H^{+}$ and $O H^{-}$ ions remain constant in pure water at any given temperature.

The equilibrium constant for this reaction is called the ion product of water, symbolized as $K_{w}$ :

$K_{w} = [H^{+}] [O H^{-}]$

Here, $[H^{+}]$ and $[O H^{-}]$ are the molar concentrations of hydrogen ions and hydroxide ions, respectively.

$K_{w}$ at 298 K (25°C)

At 298 K (room temperature), the value of $K_{w}$ is:

$K_{w} = 1.0 \times 10^{- 14} {mol}^{2} {dm}^{- 6}$

This means that in pure water at 25°C:

$[H^{+}] = [O H^{-}] = \sqrt{K_{w}} = \sqrt{1.0 \times 10^{- 14}} = 1.0 \times 10^{- 7} mol {dm}^{- 3}$

Note

The equal concentrations of $H^{+}$ and $O H^{-}$ ions make pure water neutral, with a pH of 7.0.

Tip

At temperatures other than 298 K, the value of $K_{w}$ changes because the self-ionization of water is endothermic. For example, as temperature increases, $K_{w}$ increases, making water slightly more acidic.

Interpreting $K_{w}$ : Acidic, Neutral, and Basic Solutions

The value of $K_{w}$ is constant for a given temperature, so any change in $[H^{+}]$ or $[O H^{-}]$ must be balanced to maintain the relationship:

$K_{w} = [H^{+}] [O H^{-}]$

This relationship allows us to classify solutions as acidic, neutral, or basic:

1. Neutral Solutions

In a neutral solution, $[H^{+}] = [O H^{-}]$ .
At 298 K, $[H^{+}] = [O H^{-}] = 1.0 \times 10^{- 7} mol {dm}^{- 3}$ .
$p H = 7.0$

2. Acidic Solutions

In an acidic solution, $[H^{+}] > [O H^{-}]$ .
For example, if $[H^{+}] = 1.0 \times 10^{- 4} mol {dm}^{- 3}$ , then:
$[O H^{-}] = \frac{K_{w}}{[H^{+}]} = \frac{1.0 \times 10^{- 14}}{1.0 \times 10^{- 4}} = 1.0 \times 10^{- 10} mol {dm}^{- 3}$
$p H < 7.0$

3. Basic Solutions

In a basic solution, $[H^{+}] < [O H^{-}]$ .
For example, if $[O H^{-}] = 1.0 \times 10^{- 3} mol {dm}^{- 3}$ , then:
$[H^{+}] = \frac{K_{w}}{[O H^{-}]} = \frac{1.0 \times 10^{- 14}}{1.0 \times 10^{- 3}} = 1.0 \times 10^{- 11} mol {dm}^{- 3}$
$p H > 7.0$

Example

Let’s calculate the pH of a solution with $[H^{+}] = 2.5 \times 10^{- 6} mol {dm}^{- 3}$ :
$p H = - \log [H^{+}] = - \log (2.5 \times 10^{- 6}) \approx 5.60$
Since $p H < 7.0$ , the solution is acidic.

Why Does $K_{w}$ Matter?

Understanding $K_{w}$ allows us to predict the behavior of acids and bases in aqueous solutions. Here are some practical implications:

1. pH and pOH Scales

The pH and pOH scales are logarithmic representations of $[H^{+}]$ and $[O H^{-}]$ , respectively:

$p H = - \log [H^{+}] and p O H = - \log [O H^{-}]$

Since $K_{w} = [H^{+}] [O H^{-}]$ , we can derive the relationship:

$p H + p O H = 14 (at 298 K)$

Common Mistake

Many students forget that $p H + p O H = 14$ is valid only at 298 K. At other temperatures, the sum will differ because $K_{w}$ changes.

2. Le Châtelier’s Principle

Adding an acid (which increases $[H^{+}]$ ) or a base (which increases $[O H^{-}]$ ) shifts the equilibrium of water’s self-ionization:

$H_{2} O (l) ⇌ H^{+} (a q) + O H^{-} (a q)$

Example

If $[H^{+}]$ increases, $[O H^{-}]$ decreases to maintain $K_{w}$ . This explains why acidic solutions have lower $[O H^{-}]$ than neutral water

Example question

Calculating $[O H^{-}]$ in an Acidic Solution

Calculate $[O H^{-}]$ in a 0.010 mol dm $^{- 3}$ HCl solution at 298 K.

Solution

Determine $[H^{+}]$ : HCl is a strong acid, so it dissociates completely:
$[H^{+}] = 0.010 mol {dm}^{- 3}$
Use $K_{w}$ to find $[O H^{-}]$ : $[O H^{-}] = \frac{K_{w}}{[H^{+}]} = \frac{1.0 \times 10^{- 14}}{0.010} = 1.0 \times 10^{- 12} mol {dm}^{- 3}$
Interpret the result: Since $[H^{+}] > [O H^{-}]$ , the solution is acidic.

Reflection:

Self review

What happens to $[H^{+}]$ and $[O H^{-}]$ if the temperature of water increases? How does this affect pH?

Theory of Knowledge

How does the concept of equilibrium in chemistry reflect broader ideas of balance in nature?
Consider how this principle applies in biology, ecosystems, or even economics.

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R3.1.5 The ion product of water (Kw)

The Ion Product of Water (Kw) and Its Implications

The Ion Product of Water (Kw)

Kw at 298 K (25°C)

Interpreting Kw: Acidic, Neutral, and Basic Solutions