The Ion Product of Water ($K_w$) and Its Implications
Water, even in its pure form, undergoes a process called self-ionization(or autoionization), where two water molecules interact to form a hydronium ion ($H_3O^+$) and a hydroxide ion ($OH^-$): $$ H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq) $$
For simplicity, we often represent the hydronium ion as $H^+$, so the equation becomes: $$ H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq) $$
The equilibrium constant for this reaction is called the ion product of water, symbolized as $K_w$: $$ K_w = [H^+][OH^-] $$
Here, $[H^+]$ and $[OH^-]$ 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: $$ K_w = 1.0 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6} $$
This means that in pure water at 25°C: $$ [H^+] = [OH^-] = \sqrt{K_w} = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \, \text{mol} \ \text{dm}^{-3} $$
Note
The equal concentrations of $H^+$ and $OH^-$ 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, so both $[H^+]$ and $[OH^-]$ increase equally and the pH of pure water decreases even though it remains neutral.
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 $[OH^-]$ must be balanced to maintain the relationship:
Let’s calculate the pH of a solution with $[H^+] = 2.5 \times 10^{-6} \, \text{mol} \, \text{dm}^{-3}$: $$ pH = -\log[H^+] = -\log(2.5 \times 10^{-6}) \approx 5.60 $$
Since $pH< 7.0$, 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:
Le Châtelier’s Principle
Adding an acid (which increases $[H^+]$) or a base (which increases $[OH^-]$) shifts the equilibrium of water’s self-ionization: $$ H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq) $$
Example
If $[H^+]$ increases, $[OH^-]$ decreases to maintain $K_w$.
This explains why acidic solutions have lower $[OH^-]$ than neutral water.
Example question
Calculate $[OH^-]$ in a 0.010 mol dm$^{-3}$ HCl solution at 298 K.
Solution
Determine $[H^+]$: HCl is a strong acid, so it dissociates completely: $$ [H^+] = 0.010 \, \text{mol} \, \text{dm}^{-3} $$
