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Solubility is all about how much of a substance can dissolve in a solvent at a given temperature and pressure. In the context of the problem, we consider the solubility of a gas, \( \mathrm{RNH}_2 \), in water. The solubility provided is a ratio—22.41 volumes of \( \mathrm{RNH}_2 \) gas per volume of water. This means that under the specified conditions, you can dissolve a large amount of gas in a small amount of water.

• Temperature and Pressure: For gases, solubility usually decreases with increasing temperature and increases with pressure, following Henry's Law.
• Measurement Unit: Volumes per volume is a common way to express gas solubility, but it can also be converted into moles per liter using the Ideal Gas Law.

Understanding solubility helps us predict what happens when a gas meets a solution. Knowing solubility is crucial when calculating concentrations, especially for gases like \( \mathrm{RNH}_2 \) that react further in solution.

The Ideal Gas Law is a fundamental equation in chemistry used to relate the pressure, volume, temperature, and number of moles of a gas. The equation is given by:\[ \text{PV} = nRT \]where:

• \( P \) is the pressure of the gas
• \( V \) is the volume of the gas
• \( n \) is the number of moles of the gas
• \( R \) is the ideal gas constant (0.0821 L·atm/mol·K)
• \( T \) is the temperature in Kelvin

In our solution, we're converting volumes of \( \mathrm{RNH}_2 \) gas into moles. At standard conditions, 1 mole of any gas occupies 22.41 L. This means that when 22.41 L of \( \mathrm{RNH}_2 \) is dissolved, it equates to 1 mole per liter of solution.This conversion is crucial because it allows chemical reactions in solutions to be quantitatively analyzed.

The base dissociation constant, denoted as \( K_b \), expresses the ability of a base to dissociate into its ions in solution. It provides insight into the strength of a base. A higher \( K_b \) value indicates a stronger base.For the reaction involving \( \mathrm{RNH}_2 \):\[ \mathrm{RNH}_2 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{RNH}_3^+ + \mathrm{OH}^- \]\( K_b \) is calculated using the formula:\[ K_b = \frac{K_w}{K_a} \]where \( K_w \) is the ionization constant of water \( (10^{-14}) \) and \( K_a \) is the acid dissociation constant for the conjugate acid \( \mathrm{RNH}_3^+ \).In this problem, \( K_a \) is given by \( \mathrm{pK}_a = 4 \), so \( K_a = 10^{-4} \), making:\[ K_b = \frac{10^{-14}}{10^{-4}} = 10^{-10} \]This value reflects the weak basic strength of \( \mathrm{RNH}_2 \) and helps calculate the equilibrium concentrations needed for finding the \( \text{pOH} \).

The \( \text{pOH} \) is a measure of the hydroxide ion concentration in a solution, similar to \( \text{pH} \), but for basic solutions. It provides insight into the solution's basicity.To calculate \( \text{pOH} \) for \( \mathrm{RNH}_2 \)'s dissolution, we determine the amount of \( \mathrm{OH}^- \) in solution. From the equilibrium expression:\[ K_b = \frac{[\mathrm{RNH}_3^+][\mathrm{OH}^-]}{[\mathrm{RNH}_2]} = \frac{x^2}{1-x} \]We simplify for small \( x \) values as \( x^2 = 10^{-10} \) and solve to find:\[ x = \sqrt{10^{-10}} = 10^{-5} \]The \( \text{pOH} \) is then calculated using:\[ \text{pOH} = -\log([\mathrm{OH}^-]) \]\[ \text{pOH} = -\log(10^{-5}) = 5 \]Understanding \( \text{pOH} \) complements \( \text{pH} \), as their sum equals \( 14 \) for aqueous solutions at 25°C. This calculation shows how solute concentration influences a solution’s acidity or basicity.