Problem 86 - A Complete Resource Book in Chemistry for JEE Main

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) \(1 \times 10^{-12}\)

Step-by-Step Solution

Answer

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

Step 1: Understanding the given information

We are told that at 80°C, the concentration of \( \text{H}_3\text{O}^+ \) is \( 1 \times 10^{-6} \text{ mol/L} \). This refers to the concentration of hydronium ions in distilled water.

Step 2: Relation between Hydronium and Hydroxide ions

In pure water, the concentration of hydronium ions \( [\text{H}_3\text{O}^+] \) is equal to the concentration of hydroxide ions \( [\text{OH}^-] \). Thus, \( [\text{OH}^-] = 1 \times 10^{-6} \text{ mol/L} \) as well.

Step 3: Recalling the formula for \( \text{K}_w \)

The ion product of water, \( K_w \), is given by the product of the concentrations of hydronium and hydroxide ions: \[ K_w = [\text{H}_3\text{O}^+] \times [\text{OH}^-] \]

Step 4: Calculating \( \text{K}_w \) at 80°C

Substitute the given values into the formula: \[ K_w = (1 \times 10^{-6}) \times (1 \times 10^{-6}) = 1 \times 10^{-12} \] mol²/L².

Step 5: Matching to the provided options

Comparing the calculated value of \( K_w \) (\( 1 \times 10^{-12} \)) with the answer choices, it matches option (d).

Key Concepts

Concentration of Hydronium Ions

In any aqueous solution, the concentration of hydronium ions (represented as \([\text{H}_3\text{O}^+]\)) is crucial to understanding the acid-base nature of the solution. Hydronium ions are formed when hydrogen ions (protons, \(\text{H}^+\)) associate with water molecules. This occurs naturally in water, leading to the equation: \[\text{H}_2\text{O} + \text{H}^+ \rightarrow \text{H}_3\text{O}^+\] At different temperatures, the concentration of these ions changes, affecting various properties of water.For example, at \(80^{\circ} \text{C}\), the concentration of hydronium ions in distilled water might be \(1 \times 10^{-6} \text{ mol/L}\). This is a higher concentration than at \(25^{\circ} \text{C}\), where it is typically \(1 \times 10^{-7} \text{ mol/L}\). These changes indicate that temperature can influence the degree of ionization in water.

Concentration of Hydroxide Ions

Hydroxide ions (\([\text{OH}^-]\)) are also a key component of water's ion equilibrium. In pure water, the concentration of hydronium ions is always equal to the concentration of hydroxide ions due to the balanced process of water self-ionization: \[\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-\] As a result, if the concentration of \([\text{H}_3\text{O}^+]\) at \(80^{\circ} \text{C}\) is \(1 \times 10^{-6} \text{ mol/L}\), then the \([\text{OH}^-]\) concentration is also \(1 \times 10^{-6} \text{ mol/L}\).This relationship helps us establish the concept of neutral water, where \([\text{H}_3\text{O}^+] = [\text{OH}^-]\). It's necessary to remember that both ion concentrations are dependent on specific conditions, such as temperature. The balance between hydronium and hydroxide ions is crucial for maintaining the neutrality of water.

Temperature Dependence of Kw

The ion product of water, \(K_w\), is calculated as the product of the hydronium and hydroxide ion concentrations: \[K_w = [\text{H}_3\text{O}^+] \times [\text{OH}^-]\] At \(25^{\circ} \text{C}\), \(K_w\) is typically \(1 \times 10^{-14} \mathrm{~mol}^2/\text{L}^2\). However, this value changes with temperature.At higher temperatures, like \(80^{\circ} \text{C}\), the \(K_w\) increases because thermal energy facilitates the increased ionization of water molecules. In our scenario, the \(K_w\) at \(80^{\circ} \text{C}\) is \(1 \times 10^{-12} \mathrm{~mol}^2/\text{L}^2\). Understanding this temperature dependence is critical for interpreting how water behaves under different thermal conditions.

pH and pOH at Different Temperatures

The concepts of pH and pOH are derived from the concentrations of hydronium and hydroxide ions, respectively. They provide a way to express the acidity or basicity of a solution as logarithmic scales:

• \(\text{pH} = -\log([\text{H}_3\text{O}^+])\)
• \(\text{pOH} = -\log([\text{OH}^-])\)

At standard room temperature (\(25^{\circ} \text{C}\)), the neutral pH is 7 because both \([\text{H}_3\text{O}^+]\) and \([\text{OH}^-]\) concentrations are \(1 \times 10^{-7} \text{ mol/L}\).As temperature increases, such as to \(80^{\circ} \text{C}\), the concentrations of both ions increase to \(1 \times 10^{-6} \text{ mol/L}\). This results in a lower neutral pH value of 6, which still signifies neutrality due to equal ion concentrations.Therefore, while the actual pH value changes with temperature, the state of neutrality depends on the equal balance of \([\text{H}_3\text{O}^+]\) and \([\text{OH}^-]\), not on the pH value itself. This distinction is key in accurately assessing the acid-base status of solutions at various temperatures.