Problem 31 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(90^{\circ} \mathrm{C}\), pure water has \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] 10^{-6}\) mole litre \(^{-1}\). What is the value of \(\mathrm{K}_{\mathrm{w}}\) at \(90^{\circ} \mathrm{C}\) ? (a) \(10^{-8}\) (b) \(10^{-6}\) (c) \(10^{-12}\) (d) \(10^{-14}\)

Step-by-Step Solution

Answer

The value of \( \mathrm{K}_{\mathrm{w}} \) at \(90^\circ \mathrm{C}\) is \(10^{-12}\). Option (c).

Step 1: Understanding Ion Product of Water

The ion product of water (\( \mathrm{K}_{\mathrm{w}} \)) is the equilibrium constant for the self-ionization of water, and is given by the equation \( \mathrm{K}_{\mathrm{w}} = [\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{OH}^{-}] \).

Step 2: Given Hydronium Ion Concentration

According to the exercise, at \(90^\circ \mathrm{C}\), the concentration of \([\mathrm{H}_{3}\mathrm{O}^{+}]\) is \(10^{-6} \text{ mol L}^{-1}\).

Step 3: Applying Neutrality in Pure Water

In pure water, \([\mathrm{H}_{3}\mathrm{O}^{+}]\) equals \([\mathrm{OH}^{-}]\) due to the neutrality condition. Therefore, \([\mathrm{OH}^{-}]\) is also \(10^{-6} \text{ mol L}^{-1}\) at \(90^\circ \mathrm{C}\).

Step 4: Calculating K_{w} at 90°C

Apply the formula for \( \mathrm{K}_{\mathrm{w}} \): \( \mathrm{K}_{\mathrm{w}} = [\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{OH}^{-}] = (10^{-6})(10^{-6}) = 10^{-12} \).

Step 5: Conclusion and Selection of the Correct Answer

From the calculation, \( \mathrm{K}_{\mathrm{w}} = 10^{-12} \) at \(90^\circ \mathrm{C}\), which matches option (c).

Key Concepts

Equilibrium Constant

In chemistry, an equilibrium constant is fundamental for understanding reactions that can reach an equilibrium state. An equilibrium constant, noted as \( K \), quantifies the balance between reactants and products in a chemical reaction at equilibrium. It indicates the relative concentrations of substances at equilibrium.
For any given reversible chemical reaction: \[ aA + bB \rightleftharpoons cC + dD \]

1. \( A \) and \( B \) are reactants.
2. \( C \) and \( D \) are products.
3. \( a \), \( b \), \( c \), and \( d \) are coefficients indicating the number of moles.

The equilibrium constant is defined as: \[ K = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}} \] This formula implies that the ratio of product concentrations to reactant concentrations is constant at a given temperature.
In the case of water's self-ionization, we call this equilibrium constant the ion product of water \( (\mathrm{K}_{\mathrm{w}}) \), which specifically involves hydronium and hydroxide ions.

Self-ionization of Water

The self-ionization of water is a fascinating process. It describes how water, despite being a neutral molecule, can split into ions. In pure water, there is a small, spontaneous amount of water molecules dissociating into hydronium \( ([\mathrm{H}_3\mathrm{O}^+]) \) and hydroxide ions \( ([\mathrm{OH}^-]) \). This reaction is represented as: \[ 2\mathrm{H}_2\mathrm{O} \rightleftharpoons [\mathrm{H}_3\mathrm{O}^+] + [\mathrm{OH}^-] \] This self-ionization reaches an equilibrium state where the ion concentrations balance each other, and is described by the ion product constant for water, \( \mathrm{K}_{\mathrm{w}} = [\mathrm{H}_3\mathrm{O}^+][\mathrm{OH}^-] \).
The interesting part is that \( \mathrm{K}_{\mathrm{w}} \) varies with temperature. At room temperature (25°C), \( \mathrm{K}_{\mathrm{w}} \) is traditionally \( 1.0 \times 10^{-14} \). However, as discovered in the exercise, at 90°C, \( \mathrm{K}_{\mathrm{w}} \) becomes \( 10^{-12} \). This increase reflects that water becomes slightly more dissociated at higher temperatures.

Hydronium Ion Concentration

Understanding hydronium ion concentration is crucial in chemistry, especially when discussing acids and bases. The hydronium ion \((\mathrm{H}_3\mathrm{O}^+)\) is essentially a water molecule with an additional proton. It's a common way to express the acidic nature of a solution.
In water, the concentration of hydronium ions is pivotal in determining the acidity or basicity of a solution. When the concentration is higher than \( 1.0 \times 10^{-7} \text{ mol L}^{-1} \) (which is the neutral concentration at 25°C under standard conditions), the solution is acidic. Conversely, if it's lower, the solution is basic.
In the given exercise, at 90°C, the hydronium ion concentration in pure water was \( 10^{-6} \text{ mol L}^{-1} \). This concentration indicates more ions than at 25°C, reflecting the effect of temperature on self-ionization. This reinforces the idea that, even in pure water, the number of hydronium and hydroxide ions can vary significantly with temperature changes, hence altering the \( \mathrm{K}_{\mathrm{w}} \).