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 pOH that can be attained by dissolving \(\mathrm{RNH}_{2}\) in water.

Step-by-Step Solution

Answer

The maximum pOH that can be attained is approximately 1.33.

Step 1: Understanding the Problem

We need to find the maximum pOH that can be obtained when a given amount of gaseous amine (RNH_2) is dissolved in water. Given that the solubility of 10 (RNH_2) in water at 1 atm and 273 K is 22.4 L per unit volume of water, and the pK_b of 10 (10) is 4.

Step 2: Analyze Solubility Data

The solubility of 10 (RNH_2) is given as 22.4 L per unit volume of water, which implies that at maximum solubility, 22.4 moles of 10 (RNH_2) are dissolved in 1 mole of water for ideal gas behavior. This means the concentration of 10 (RNH_2) in solution is 22.4 M.

Step 3: Relate pK_b to K_b

The relationship between pK_b and K_b is given by the formula: pK_b = -log(K_b). Since pK_b for 10 (10) is 4, we can calculate K_b as: \[ K_b = 10^{-4} \]

Step 4: Set up the Equilibrium Expression

The equilibrium reaction for 10 (RNH_2) in water can be written as: \[ \mathrm{RNH}_2 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{RNH}_3^+ + \mathrm{OH}^- \]At equilibrium, assuming an initial concentration of 10 (RNH_2) as 22.4 M, and a very small x dissociates, the expression for K_b is:\[ K_b = \frac{[\mathrm{RNH}_3^+][\mathrm{OH}^-]}{[\mathrm{RNH}_2]} \approx \frac{x^2}{22.4} \]

Step 5: Solve for x, the Concentration of OH-

Setting the equation for K_b:\[ 10^{-4} = \frac{x^2}{22.4} \]Solving for x gives:\[ x = \sqrt{10^{-4} \times 22.4} \]\[ x \approx 0.047 \] M. Therefore, the concentration of OH^- at maximum solubility is 0.047 M.

Step 6: Calculate the pOH

The formula for pOH is \[ \text{pOH} = -\log[\mathrm{OH}^-] \].Substitute the value:\[ \text{pOH} = -\log(0.047) \]\[ \text{pOH} \approx 1.33 \]

Key Concepts

Solubility

Solubility refers to the ability of a substance to dissolve in a solvent to form a homogenous solution. In the context of our exercise, we are considering the solubility of a gaseous amine, \( \mathrm{RNH}_2(\mathrm{~g}) \), in water. The solubility is provided as 22.4 L per unit volume of water under specific conditions (1 atm and 273 K). This value suggests that at full solubility, 22.4 volumes of gaseous amine dissolve in a volume of water. Solubility is crucial in determining how much of the amine can interact with water, which directly influences chemical reactions, such as amine's dissociation into its ionic forms. A high solubility means more \( \mathrm{RNH}_2 \) is available to participate in equilibrium reactions, which relates to the calculation of maximum pOH.

pKb

The \( \text{pK}_\mathrm{b} \) represents the basic dissociation constant's negative logarithm. It's a critical parameter in understanding the strength of a base in its ability to generate hydroxide ions (\(\mathrm{OH}^-\)) in solution. For gaseous amine \( \mathrm{RNH}_2 \), a \( \text{pK}_\mathrm{b} = 4 \) indicates a certain level of intrinsic basicity. A lower \( \text{pK}_\mathrm{b} \) means a stronger base because it implies a higher \( K_\mathrm{b} \). Use the formula \( \text{pK}_\mathrm{b} = -\log(K_\mathrm{b}) \) to find \( K_\mathrm{b} \):- Calculate \( K_\mathrm{b} \) as: \[ K_\mathrm{b} = 10^{-4} \]This constant helps us set up the equilibria involving the amine and predict concentrations of produced ions at equilibrium which we use in calculating pOH.

Gaseous Amine

A gaseous amine like \( \mathrm{RNH}_2(\mathrm{~g}) \) is an amine in the gaseous state dissolved in a solvent like water. These compounds can act as weak bases, meaning they accept protons and increase the concentration of hydroxide ions \(\mathrm{OH}^-\) in the solution.When gauging the effects of a gaseous amine, focus on:- Its ability to dissolve (solubility)- Its basic strength (via \( \text{pK}_\mathrm{b} \))Dissolving gaseous amines affects the chemical equilibrium in solution, as they can dissociate and influence the resulting pOH, which measures a solution's alkalinity.

Equilibrium Expression

An equilibrium expression helps us understand how species in a chemical reaction dynamically balance. For \( \mathrm{RNH}_2 \) dissolving in water, the equilibrium can be written as:\[ \mathrm{RNH}_2 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{RNH}_3^+ + \mathrm{OH}^- \]This expression reflects how \( \mathrm{RNH}_2 \), once dissolved, can form \( \mathrm{RNH}_3^+ \) and \(\mathrm{OH}^-.\) Setting up the equilibrium as \( K_\mathrm{b} = \frac{[\mathrm{RNH}_3^+][\mathrm{OH}^-]}{[\mathrm{RNH}_2]} \) emphasizes the concentrations' relationship at equilibrium.At maximum solubility (22.4 M concentration of \( \mathrm{RNH}_2 \)), we assume minimal change in its concentration due to dissociation (small \( x \)). Solving gives us \( x \approx 0.047 \) M for \( \mathrm{OH}^- \). Calculating pOH from this illustrates how equilibrium constants like \( K_\mathrm{b} \) dictate the resulting alkali level in a solution.