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 ($H_3O^+$) 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: $$
[H^+] = [OH^-] = \sqrt{K_w} = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \, \text{mol} \ \text{dm}^{-3}
$$
The equal concentrations of $H^+$ and $OH^-$ ions make pure water neutral, with a pH of 7.0.
TipAt 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 $[OH^-]$ must be balanced to maintain the relationship:
$$
K_w = [H^+][OH^-]
$$
This relationship allows us to classify solutions as acidic, neutral, or basic:
Neutral Solutions
- In a neutral solution, $[H^+] = [OH^-]$.
- At 298 K, $[H^+] = [OH^-] = 1.0 \times 10^{-7} \, \text{mol} \, \text{dm}^{-3}$.
- $pH = 7.0$
