Question

At \(80^{\circ} \mathrm{C}\), distilled water \(\left(\mathrm{H}_{3} \mathrm{O}^{+}\right)\) concentration is equal to \(1 \times 10^{-6} \mathrm{~mol} /\) litre. At the same temperature the value of \(\mathrm{Kw}\) is: (a) \(1 \times 10^{-3}\) (b) \(1 \times 10^{-6}\) (c) \(1 \times 10^{-9}\) (d) \(1 \times 10^{-12}\)

Short answer

The value of \(\text{K}_w\) at \(80^\circ \text{C}\) is \(1 \times 10^{-12}\). Option (d) is correct.

Step-by-step solution

Understanding the equation for water dissociation

The self-ionization of water can be represented as: \[ \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^- \]This equilibrium process is characterized by the ion product of water, \(\text{K}_w\). At any temperature, \(\text{K}_w\) is defined as:\[ \text{K}_w = [\text{H}_3\text{O}^+][\text{OH}^-] \]

Substituting known values

We know from the problem that at \(80^\circ \text{C}\), the concentration of \(\text{H}_3\text{O}^+\) is given to be \(1 \times 10^{-6}\text{ mol/L}\). Since pure water is neutral, the concentration of \(\text{OH}^-\) will also be the same as \(\text{H}_3\text{O}^+\):\[ [\text{OH}^-] = 1 \times 10^{-6}\text{ mol/L} \]

Calculating \(\text{K}_w\) at \(80^\circ \text{C}\)

Now, we use the formula for \(\text{K}_w\):\[ \text{K}_w = [\text{H}_3\text{O}^+][\text{OH}^-] \]Substitute the known concentrations:\[ \text{K}_w = (1 \times 10^{-6})(1 \times 10^{-6}) = 1 \times 10^{-12} \]

Selecting the correct choice from options

From the options, the value closest to our calculated \(\text{K}_w\) is:(d) \(1 \times 10^{-12}\)

Key concepts

Self-ionization of water

Water has a remarkable ability to undergo a reaction with itself, a process called self-ionization or autoionization. This means that water molecules can split into ions without any external influence. The reaction can be represented by the equation:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^- \]This reversible reaction shows that a small fraction of water molecules dissociate to form hydronium ions \(\text{H}_3\text{O}^+\) and hydroxide ions \(\text{OH}^-\). Even in pure water, this dynamic process occurs continuously. However, only a very small number of water molecules actually ionize.
This balance between the forward and reverse reactions maintains the system at equilibrium at a given temperature. As temperature changes, so does the equilibrium position, affecting how much water dissociates into ions.
The constant that characterizes this is the ion product of water, \(K_w\), and it varies with temperature.

Hydronium ion concentration

The hydronium ion concentration, represented as \([\text{H}_3\text{O}^+]\), provides a measure of how acidic or basic a solution is. In the case of pure water, the concentration of hydronium ions is a central factor in determining its pH.
At 25°C, in pure water, the concentration of \(\text{H}_3\text{O}^+\) is typically \(1 \times 10^{-7} \text{ mol/L}\). But as temperature increases, the amount of ionization generally increases, causing this concentration to change. For instance, at 80°C, it's given as \(1 \times 10^{-6} \text{ mol/L}\) in the problem.
Understanding the concentration of hydronium ions is important because:

• It directly influences the solution's pH. pH is calculated using the formula: \( \text{pH} = -\log[\text{H}_3\text{O}^+] \).
• The higher the concentration of hydronium ions, the more acidic the solution is.
• Knowing \([\text{H}_3\text{O}^+]\) allows us to calculate \(K_w\), the ion product of water at various temperatures.

The challenge with temperatures beyond the standard conditions is that these concentrations change, making it essential to consider the impact of temperature on the equilibrium.

Neutrality of pure water

Under standard conditions, pure water is considered neutral. This means that its pH is 7 at 25°C because the concentration of hydronium ions \([\text{H}_3\text{O}^+]\) equals the concentration of hydroxide ions \([\text{OH}^-]\). At this state, water does not favor an acidic or basic nature.
The concept of neutrality adjusts with varying temperatures. For example, at 80°C, despite the increased concentration of \([\text{H}_3\text{O}^+]\) to \(1 \times 10^{-6} \text{ mol/L}\), water remains neutral. This neutrality is ensured by the simultaneous increase in \([\text{OH}^-]\) to match \([\text{H}_3\text{O}^+]\). The equilibrium ensures that any change in \([\text{H}_3\text{O}^+]\) is balanced by changes in \([\text{OH}^-]\), keeping the product \([\text{H}_3\text{O}^+][\text{OH}^-]\) constant and equal to \(K_w\). Key points about neutrality in pure water include:

• It is defined by the equality \([\text{H}_3\text{O}^+] = [\text{OH}^-]\).
• The pH of neutral water varies with temperature but always indicates equal concentrations of the ions.
• Any disruption to this balance (like adding acids or bases) shifts the pH away from 7, defining the solution's acidity or alkalinity.