Problem 252 - A Complete Resource Book in Chemistry for JEE Main

Question

If the solubility of \(\mathrm{RNH}_{2}(\mathrm{~g})\) in water at \(1 \mathrm{~atm}\) and \(273 \mathrm{~K}\) is \(22.4 \mathrm{~L}\) volumes of \(\mathrm{RNH}_{2}\) per unit volume of water. If \(\mathrm{pK}_{\mathrm{b}}\) of \(\mathrm{RNH}_{2}\) is 4 . Find the maximum \(\mathrm{pOH}\) that can be attained by dissolving \(\mathrm{RNH}_{2}\) in water.

Step-by-Step Solution

Answer

The maximum pOH is 2.

Step 1: Determine the Molarity of Gas in Water

Given the solubility is 22.4 L of \( \text{RNH}_2 \) gas per unit volume of water at STP, convert this volume to moles using the molar volume of a gas at STP, which is 22.4 L/mol. Therefore, the molarity is \( \frac{22.4 \text{ L of gas}}{22.4 \text{ L/mol}} = 1 \text{ mol/L} \). Thus, the initial concentration of \( \text{RNH}_2 \) in the solution is 1 M.

Step 2: Write the Base Dissociation Reaction

Record the reaction: \[ \text{RNH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{RNH}_3^+ + \text{OH}^- \] This equation represents the base \( \text{RNH}_2 \) dissociating in water to form \( \text{RNH}_3^+ \) and hydroxide ions \( \text{OH}^- \).

Step 3: Use \( pK_b \) to Find \( K_b \)

The \( pK_b \) is given as 4, so the base dissociation constant \( K_b \) is calculated using: \[ K_b = 10^{-pK_b} = 10^{-4} \] So, \( K_b = 10^{-4} \).

Step 4: Set Up the ICE Table for the Reaction

Initially, the concentration of \( \text{RNH}_2 \) is 1 M, and \( \text{RNH}_3^+ \) and \( \text{OH}^- \) are 0. As the base dissolves, let \( x \) be the change in concentration.At equilibrium:- \( [\text{RNH}_2] = 1 - x \)- \( [\text{RNH}_3^+] = x \)- \( [\text{OH}^-] = x \)

Step 5: Apply Equilibrium Expression

Using the equilibrium expression for the dissociation reaction: \[ K_b = \frac{[\text{RNH}_3^+][\text{OH}^-]}{[\text{RNH}_2]} = \frac{x^2}{1 - x} \] Since \( K_b \) is small, assume \( 1 - x \approx 1 \). Thus, simplify to: \[ x^2 = K_b \] Substituting \( K_b = 10^{-4} \), we get: \[ x = \sqrt{10^{-4}} = 10^{-2} \]. So, \( [\text{OH}^-] = 0.01 \text{ M} \).

Step 6: Calculate \( pOH \) Using \( [OH]^- \)

Finally, find \( pOH \) using: \[ \text{pOH} = -\log[\text{OH}^-] = -\log(0.01) = 2 \].

Key Concepts

Solubility in Water

Solubility is the ability of a substance to dissolve in a solvent, forming a solution. In this exercise, we deal with the solubility of the gaseous base \( \text{RNH}_2 \) in water. We are given that the solubility is \( 22.4 \text{ L} \) of \( \text{RNH}_2 \) gas per unit volume of water at standard temperature and pressure (STP). At STP, one mole of an ideal gas occupies \( 22.4 \text{ L} \). Thus, when \( \text{RNH}_2 \) dissolves in water, its molarity— a measure of the concentration of solute in a solution—can be calculated as \( 1 \text{ M}\). This indicates that every liter of water can dissolve enough \( \text{RNH}_2 \) to reach a concentration of \( 1 \text{ mol/L}\). This initial concentration is crucial for analyzing how the base will behave once it's dissolved.

Base Dissociation Reaction

In an aqueous solution, the dissolved \( \text{RNH}_2 \) undergoes a base dissociation reaction. This reaction is the process in which a base accepts a proton from water, forming hydroxide ions \( \text{OH}^- \) and conjugate acids. The reaction for \( \text{RNH}_2 \) in water is:\[ \text{RNH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{RNH}_3^+ + \text{OH}^- \]Here, \( \text{RNH}_2 \) accepts a proton from water, creating \( \text{RNH}_3^+ \), its conjugate acid, and hydroxide ions \( \text{OH}^- \). The production of \( \text{OH}^- \) increases the solution's basicity, which in turn decreases its acidity (since pH increases with decreasing hydrogen ion concentration). Understanding dissociation reactions helps us comprehend how acids and bases interact and change the composition of a solution.

Equilibrium Constant

The equilibrium constant \( K_b \) for a reaction describes the ratio of the concentration of products to reactants at equilibrium. In this exercise, given that the \( pK_b \) of \( \text{RNH}_2 \) is 4, we find \( K_b \) using the formula:\[ K_b = 10^{-pK_b} = 10^{-4} \]This constant is small, indicating that at equilibrium, the concentration of products \( (\text{RNH}_3^+ \) and \( \text{OH}^-) \) is much smaller compared to the reactant \( \text{RNH}_2 \). Small \( K_b \) values suggest a weak base, which means it does not dissociate extensively in water. Equilibrium constants are essential in predicting the direction in which a reaction will proceed and how far the reaction will go to reach equilibrium.

ICE Table

The ICE table is a valuable tool used to organize and calculate concentrations of reactants and products for reactions reaching equilibrium. ICE stands for Initial, Change, and Equilibrium. For the dissociation of \( \text{RNH}_2 \):- **Initial:** \( [\text{RNH}_2] = 1 \text{ M}\), while \( [\text{RNH}_3^+] \) and \( [\text{OH}^-] \) are 0 M.- **Change:** As the reaction proceeds, \( \text{RNH}_2 \) decreases by \( x \), and \( \text{OH}^- \) and \( \text{RNH}_3^+ \) each increase by \( x \).- **Equilibrium:** We define the equilibrium concentrations as, \( [\text{RNH}_2] = 1 - x \), \( [\text{RNH}_3^+] = x \), and \( [\text{OH}^-] = x \).By using an ICE table, we can better manage complex chemical reactions and make accurate calculations regarding the amounts of substances at equilibrium.

Base pK_b

The \( pK_b \) value of a base is a logarithmic measure of its strength as a base: \[ pK_b = -\log(K_b) \]For \( \text{RNH}_2 \), \( pK_b \) is given as 4, meaning \( K_b = 10^{-4} \). This value illustrates the weak nature of the base, since lower \( K_b \) values generally correlate with weaker bases. The significance of \( pK_b \) is that it provides a straightforward method to evaluate and compare the strength of different bases. Smaller \( pK_b \) values suggest a stronger base, which generally dissociates more readily to form \( \text{OH}^- \) ions in water. Understanding \( pK_b \) is crucial for estimating the pOH and the degree to which a base can increase the basicity of a solution.