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 \(\mathrm{pK}_{b}\) 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 \(\mathrm{pOH}\) is 2.

Step 1: Understand Solubility at STP

The solubility of a gas in water is given as \(22.4\) liters (under standard temperature and pressure conditions), meaning \(22.4\) liters of \(\mathrm{RNH}_2\) gas can dissolve in one liter of water. This signifies that at these conditions, the concentration of \(\mathrm{RNH}_2\) in water is \(1\, \mathrm{mol/L}\) due to the molar volume at STP being \(22.4\, \mathrm{L/mol}\).

Step 2: Write the Base Ionization Equation

The dissolution in water can be described by the following equilibrium reaction: \[\mathrm{RNH}_2(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{RNH}_3^+(aq) + \mathrm{OH}^-(aq)\]Here, \(\mathrm{RNH}_2\) acts as a base accepting a proton from water.

Step 3: Use \(\mathrm{pK}_b\) to Find \(K_b\)

Since \(\mathrm{pK}_b = 4\), we find the base dissociation constant, \(K_b\), using the formula:\[K_b = 10^{-\mathrm{pK}_b}\]Plugging in the value: \[K_b = 10^{-4}\]

Step 4: Set Up the Equilibrium Expression

At equilibrium, define \(x\) as the concentration of \(\mathrm{OH}^-\) formed:\[\frac{x^2}{1-x} = 10^{-4}\]Approximately, since \(x\) will be small compared to \(1\): \[x^2 \approx 10^{-4}\]

Step 5: Solve for \(x\)

We solve \(x^2 = 10^{-4}\), giving:\[x = 10^{-2}\]Thus, the \([\mathrm{OH}^-]\) concentration is \(0.01 \mathrm{~mol/L}\).

Step 6: Calculate \(\mathrm{pOH}\)

Using the \(\mathrm{pOH}\) definition,\[\mathrm{pOH} = -\log[\mathrm{OH}^-]\]we plug in the concentration:\[\mathrm{pOH} = -\log(0.01) = 2\]

Key Concepts

Base Ionization Equation

When a weak base like \( \mathrm{RNH}_2 \) is dissolved in water, it undergoes an ionization process. This reaction occurs because the base accepts protons from the water molecules, forming ions. The reaction can be simply represented as:\[ \mathrm{RNH}_2(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{RNH}_3^+(aq) + \mathrm{OH}^-(aq) \]This equation represents an equilibrium state. In this state, the forward and backward reactions occur at the same rate. Therefore, there is a constant concentration of the species involved. Since \( \mathrm{RNH}_2 \) is a base, it increases the concentration of hydroxide ions, \( \mathrm{OH}^- \), which is crucial for calculating the \( \mathrm{pOH} \). Remember, the higher the concentration of \( \mathrm{OH}^- \), the lower the \( \mathrm{pOH} \), indicating a strongly basic solution.
Understanding this equation helps us grasp why dissolving bases in water affects pH levels.

Solubility of Gases

The solubility of gases, like \( \mathrm{RNH}_2 \), in water is affected by temperature and pressure. At standard temperature and pressure (STP), gases exhibit predictable behavior. According to the problem, \( \mathrm{RNH}_2 \) has a solubility of 22.4 L per unit volume of water at these conditions.This means that during the gas dissolution process, 1 mol/L of \( \mathrm{RNH}_2 \) is present, thanks to the molar volume of 22.4 L/mol at STP.
This knowledge is pivotal because it allows us to calculate the concentration of dissolved gas in terms of moles per liter (Molarity), which directly impacts the reaction's equilibrium and the resulting hydroxide ion concentration.

Equilibrium Expression

The concept of equilibrium involves understanding how the concentrations of reactants and products stabilize over time during a reaction. For our base ionization reaction, we use an equilibrium expression related to the base dissociation constant, \( K_b \).Given that \( \mathrm{pK}_b = 4 \), we find \( K_b \) through:\[ K_b = 10^{-\mathrm{pK}_b} = 10^{-4} \]Now, set up the expression for equilibrium. Assume \( x \) is the concentration of \( \mathrm{OH}^- \) at equilibrium: \[ \frac{x^2}{1-x} = 10^{-4} \]Since \( x \) is small compared to 1, simplify to:\[ x^2 \approx 10^{-4} \]Solving this gives \( x = 10^{-2} \). Thus, the hydroxide ion concentration is 0.01 mol/L.This minor adjustment simplifies calculations without losing accuracy, allowing us to easily determine the \( \mathrm{pOH} \), calculated as\[ \mathrm{pOH} = -\log(x) = 2 \].
This process illustrates how base ionization and equilibrium principles help describe and predict properties of dissolving bases in water.