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 \( \mathrm{Kw} \) at \( 80^{\circ} \mathrm{C} \) is \( 1 \times 10^{-12} \), option (d).

Step 1: Understand the Concept of Ion Product of Water

The ion product of water, denoted by \( K_w \), is the constant that represents the product of the concentrations of hydronium ions \( \left( [\mathrm{H}_3\mathrm{O}^+] \right) \) and hydroxide ions \( \left( [\mathrm{OH}^-] \right) \) in water. Mathematically, it is expressed as \[ K_w = [\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^-]. \]

Step 2: Identify the Given Values

We are given that at \( 80^{\circ} \mathrm{C} \), the concentration of \( \mathrm{H}_3\mathrm{O}^+ \) is \( 1 \times 10^{-6} \) mol/L.

Step 3: Apply the Concept to Find the Hydroxide Concentration

For neutral water, the concentration of \( \mathrm{H}_3\mathrm{O}^+ \) is equal to that of \( \mathrm{OH}^- \). Thus, \( [\mathrm{OH}^-] = 1 \times 10^{-6} \) mol/L.

Step 4: Calculate the Ion Product \( K_w \)

Substitute the values into the equation for \( K_w \): \[ K_w = [\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^-] = (1 \times 10^{-6})(1 \times 10^{-6}) = 1 \times 10^{-12}. \]

Step 5: Select the Correct Answer

By comparing with the options, the value of \( K_w = 1 \times 10^{-12} \) corresponds to option (d).

Key Concepts

Hydronium Ion Concentration

In distilled water, the concentration of hydronium ions, \(\mathrm{H}_3\mathrm{O}^+\), plays a crucial role in understanding pH and related chemical properties. Hydronium ions are formed when water molecules dissociate, a process described by the reaction: \(\mathrm{H_2O} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O}^+ + \mathrm{OH}^-\).

At a specific temperature, such as \(80^{\circ} \mathrm{C}\), the balance of this dissociation reaches a state where the concentration of hydronium ions in pure water can be defined. In the given exercise, this concentration is provided as \(1 \times 10^{-6}\) mol/L. This value becomes an anchor point for calculating other important properties related to the water's ionization.

• The hydronium ion concentration is often directly related to pH values, where \(\text{pH} = -\log[\mathrm{H}_3\mathrm{O}^+]\).
• In pure water, an equal concentration of hydroXholide ions \(\mathrm{OH}^-\) accompanies the hydronium ions.

In essence, understanding the concentration of hydronium ions is foundational to exploring the chemical nature of aqueous environments.

Hydroxide Ion Concentration

Hydroxide ions, represented as \(\mathrm{OH}^-\), are as vital as hydronium ions for determining the nature of water's ionization. In neutral water, the concentration of hydronium ions \(\mathrm{H}_3\mathrm{O}^+\) matches that of hydroxide ions.

This intrinsic balance implies that if \(\mathrm{H}_3\mathrm{O}^+\) concentration is given, the \(\mathrm{OH}^-\) concentration will be the same. For instance, in the provided problem, with a hydronium concentration of \(1 \times 10^{-6}\) mol/L at \(80^{\circ} \mathrm{C}\), the hydroxide concentration is also \(1 \times 10^{-6}\) mol/L.

• This equivalence is due to the self-ionization property of water, ensuring neutrality in pure water.
• Should the concentration of hydronium ions alter, the hydroxide concentration follows to maintain the balance \(\text{pH + pOH = 14}\) under standard conditions.

Understanding hydroxide ion concentration in relation to hydronium is essential for exploring the water's ion product, \(K_w\), as well as for addressing the acidic or basic nature of solutions.

Kw Calculation

The ion product of water, known as \(K_w\), is an equilibrium constant that is central to acid-base chemistry. At any temperature, it reflects the product of the concentrations of hydronium and hydroxide ions in water.

Mathematically, \(K_w\) is expressed as:\[K_w = [\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^-]\]Given the task setup, where each ion concentration is \(1 \times 10^{-6}\) mol/L at \(80^{\circ} \mathrm{C}\), the calculation for \(K_w\) becomes straightforward:

Substitute the given values:
\[K_w = (1 \times 10^{-6}) \times (1 \times 10^{-6}) = 1 \times 10^{-12}\]

This result shows us that at \(80^{\circ} \mathrm{C}\), the ion product, \(K_w\), is \(1 \times 10^{-12}\). The ability to calculate \(K_w\) accurately provides insight into the reaction environment and the water's neutral, acidic, or basic state at a given temperature, deviating from standard conditions (where \(K_w\) is \(1 \times 10^{-14}\) at \(25^{\circ} \mathrm{C}\)).

• Adjustments in temperature influence the degree of water's ionization, thereby altering \(K_w\).
• Knowing \(K_w\) allows for a deeper exploration into solution chemistry and its implications on biological and environmental systems.