Problem 115 - A Complete Resource Book in Chemistry for JEE Main

Question

If equilibrium constant for the reaction, \(\mathrm{XO}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) \rightleftharpoons \mathrm{HXO}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\) is \(0.36 \times 10^{-6}\) then find the value of dissociation constant \(\left(\mathrm{K}_{\alpha}\right)\) for \(\mathrm{HXO} ?\) (a) \(0.36 \times 10^{-8}\) (b) \(2.8 \times 10^{-8}\) (c) \(2.8 \times 10^{-10}\) (d) \(0.36 \times 10^{-6}\)

Step-by-Step Solution

Answer

The value of the dissociation constant \( K_a \) for \( \mathrm{HXO} \) is approximately \( 2.8 \times 10^{-8} \), option (b).

Step 1: Understand the Reaction and Equilibrium Constant

The given reaction is:\[ \mathrm{XO}^{-}(\mathrm{aq}) + \mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{HXO}(\mathrm{aq}) + \mathrm{OH}^{-}(\mathrm{aq}) \]The equilibrium constant \( K_c \) for this reaction is defined as:\[ K_c = \frac{[\mathrm{HXO}][\mathrm{OH}^{-}]}{[\mathrm{XO}^{-}]} \]Given, \( K_c = 0.36 \times 10^{-6} \). This equilibrium constant is for the ionization of \( \mathrm{XO}^{-} \).

Step 2: Relate to the Dissociation Constant

The dissociation constant, \( K_a \), for the reverse reaction, where \( \mathrm{HXO} \) dissociates into \( \mathrm{XO}^{-} \) and \( \mathrm{H}^+ \), is related to the equilibrium constant \( K_c \) of the given reaction by the relationship:\[ K_w = K_a \times K_b \]where, \( K_w \) is the ion product of water (\( 1.0 \times 10^{-14} \) at 25°C). The equilibrium constant of the given reaction is effectively \( K_b \).

Step 3: Calculate \( K_a \) for \( \mathrm{HXO} \)

Use the relationship \( K_w = K_a \times K_b \) to find \( K_a \):\[ K_a = \frac{K_w}{K_b} \]Given, \( K_b = 0.36 \times 10^{-6} \) and \( K_w = 1.0 \times 10^{-14} \).Substitute the values:\[ K_a = \frac{1.0 \times 10^{-14}}{0.36 \times 10^{-6}} \]\[ K_a = 2.78 \times 10^{-8} \]

Step 4: Match with Options

The calculated value for the dissociation constant \( K_a \) is \( 2.78 \times 10^{-8} \). Comparing with the provided options, the closest match is option (b): \( 2.8 \times 10^{-8} \).

Key Concepts

Dissociation Constant

Understanding the dissociation constant is crucial in studying acid-base reactions. It involves the equilibrium condition of acid dissociation. Let's break it down. When an acid, like \( \mathrm{HXO} \), dissociates in water, it breaks down into its ions: \( \mathrm{XO}^{-} \) and \( \mathrm{H}^{+} \). The dissociation constant \( K_a \), is a measure of the strength of an acid in solution. In simpler words, \( K_a \) tells us how easily an acid gives up its proton to form \( \mathrm{H}^{+} \).

• A large \( K_a \) indicates a strong acid, which dissociates completely.
• A small \( K_a \) suggests a weak acid, with only partial dissociation.

In the given problem, we need to convert the equilibrium constant \( K_c \) to \( K_a \) using the relationship with \( K_w \).

Ion Product of Water

The ion product of water, denoted as \( K_w \), is an important constant in chemistry. At 25°C, \( K_w \) is always \( 1.0 \times 10^{-14} \). This constant represents the equilibrium condition of water's autoionization: \[ \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}^+ + \mathrm{OH}^- \]
Water naturally dissociates into \( \mathrm{H}^+ \) and \( \mathrm{OH}^- \) ions, although to a very small extent. Therefore, \( K_w \) is extremely important when calculating the concentrations of hydrogen and hydroxide ions in a solution. When dealing with other reactions, \( K_w \) acts as a linking point for finding relationships between \( K_a \) and \( K_b \).

• \( K_w = K_a \times K_b \)

This relationship helps in calculating unknown equilibrium constants such as in our exercise where we use \( K_w \) to find \( K_a \).

Chemical Equilibrium

Chemical equilibrium in a reaction means that the rate of the forward reaction equals the rate of the backward reaction. During equilibrium, the concentration of reactants and products remains constant, though not necessarily equal. In our problem, the equilibrium constant \( K_c \) indicates the position of equilibrium: \[ K_c = \frac{[\mathrm{HXO}][\mathrm{OH}^{-}]}{[\mathrm{XO}^{-}]} \]

• A high \( K_c \) means more products are formed, favoring forward reaction.
• A low \( K_c \) suggests the reaction favors the reactants.

The given exercise focuses on converting \( K_c \) into \( K_a \), which involves knowing how these equilibria relate in terms of weak bases and weak acids.

Acid-Base Reactions

In chemistry, acid-base reactions involve protons (\( \mathrm{H}^{+} \)) and the formation of water and its ions. These reactions are essential for many biological and chemical processes. In our context, we see \( \mathrm{HXO} \) reacting in water, highlighting a classical acid-base interaction where:

• An acid donates a proton (\( \mathrm{HXO} \) to \( \mathrm{XO}^- \))
• The base accepts a proton (\( \mathrm{H}_2\mathrm{O} \) to form \( \mathrm{OH}^- \))

The ability to understand and calculate the dissociation constant using the equilibrium constant and the ion product of water is critical for analyzing such reactions. This demonstrates how acids and bases interact reversibly at equilibrium.