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.
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 $[OH^-]$ 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, $[H^+] = [OH^-] = 1.0 \times 10^{-7} \, \text{mol} \, \text{dm}^{-3}$.
- $pH = 7.0$
Acidic Solutions
- In an acidic solution, $[H^+] > [OH^-]$.
- For example, if $[H^+] = 1.0 \times 10^{-4} \, \text{mol} \, \text{dm}^{-3}$, then:
$$
[OH^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-4}} = 1.0 \times 10^{-10} \, \text{mol} \, \text{dm}^{-3}
$$ - $pH< 7.0$
Basic Solutions
- In a basic solution, $[H^+] < [OH^-]$.
- For example, if $[OH^-] = 1.0 \times 10^{-3} \, \text{mol} \, \text{dm}^{-3}$, then:
$$
[H^+] = \frac{K_w}{[OH^-]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-3}} = 1.0 \times 10^{-11} \, \text{mol} \, \text{dm}^{-3}
$$ - $pH >7.0$
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:
pH and pOH Scales
- The pH and pOH scales are logarithmic representations of $[H^+]$ and $[OH^-]$, respectively: $$
pH = -\log[H^+] \quad \text{and} \quad pOH = -\log[OH^-]
$$ - Since $K_w = [H^+][OH^-]$, we can derive the relationship: $$
pH + pOH = 14 \, \text{(at 298 K)}
$$
- Many students forget that $pH + pOH = 14$ is valid only at 298 K.
- At other temperatures, the sum will differ because $K_w$ changes.
Le Châtelier’s Principle
Adding an acid (which increases $[H^+]$) or a base (which increases $[OH^-]$) shifts the equilibrium of water’s self-ionization: $$
H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)
$$
- If $[H^+]$ increases, $[OH^-]$ decreases to maintain $K_w$.
- This explains why acidic solutions have lower $[OH^-]$ than neutral water.
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:
$$
[H^+] = 0.010 \, \text{mol} \, \text{dm}^{-3}
$$ - Use $K_w$ to find $[OH^-]$:$$
[OH^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{0.010} = 1.0 \times 10^{-12} \, \text{mol} \, \text{dm}^{-3}
$$ - Interpret the result: Since $[H^+] > [OH^-]$, the solution is acidic.
What happens to $[H^+]$ and $[OH^-]$ if the temperature of water increases? How does this affect pH?
