Problem 31 - General Chemistry: Principles and Modern Applications

Question

The solubility of 1 -naphthylamine, \(\mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{2}, \mathrm{a}\) substance used in the manufacture of dyes, is given in a handbook as 1 g per \(590 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). What is the approximate \(\mathrm{pH}\) of a saturated aqueous solution of 1-naphthylamine? $$\begin{array}{r} \mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \\ \mathrm{p} K_{\mathrm{b}}=3.92 \end{array}$$

Step-by-Step Solution

Answer

The approximate pH of the saturated solution of 1-naphthylamine is 12.0

Step 1: Understand the problem

The solubility of 1-naphthylamine is given as 1g in 590g of water. The ionization of the compound in water shows that it forms \(OH^-\) ions. The pKb of the compound is also given, which can be used to find the Kb.

Step 2: Find the concentration of the base

First, convert the solubility from g/590g H2O to g/L of H2O by knowing that the density of water is 1g/mL. Then, convert the mass of the compound to moles by dividing by its molar mass. Therefore, \[Concentration = \frac{1g \times 10^3 mL/L}{590g} \times \frac{1 mol}{143g/mol} = 0.012 M\]

Step 3: Find the Kb and [OH-]

Next, find the Kb using the equation \[Kb = 10^{-pKb} = 10^{-3.92} = 1.2 \times 10^{-4}\] Now, because the compound is a weak base and does not ionize completely, we apply the Kb expression: \[Kb = \frac{[OH-][C_{10}H_{7}NH_{3}^{+}]}{[C_{10}H_{7}NH_{2}]}\] As the base is in a 1:1 ratio with the hydroxide and the cation, and is initially present in excess, we get \[Kb = [OH-]^{2}\] From this, we solve for [OH-]: \[[OH-] = \sqrt{Kb} = \sqrt{1.2 \times 10^{-4}} = 1.1 \times 10^{-2} M\]

Step 4: Calculate the pH

The pOH is calculated as the negative logarithm of the OH- concentration: \[pOH = -log[OH-] = -log(1.1 \times 10^{-2}) = 2.0\] To find the pH, use the formula \(pH = 14 - pOH\), which gives \(pH = 14 - 2.0 = 12.0\)

Key Concepts

Solubility

In chemistry, solubility tells us how much of a substance can dissolve in a solvent before reaching saturation. It's like figuring out how much sugar you can stir into a glass of water until no more will dissolve. Solubility is often expressed in terms of grams of solute per volume of solvent, such as grams per liter. When we consider the solubility of 1-naphthylamine, let’s break it down: The given solubility is 1 g per 590 g of water. By understanding the density of water (which is typically 1 g/mL), we can easily convert this to grams per liter. Knowing this value allows us to calculate the concentration in molarity (moles per liter), which is essential for further calculations such as determining the pH of the solution. Understanding solubility is crucial because it sets the foundation for calculating how a substance behaves in a solution, which includes how much it will ionize if it forms ions upon dissolving.

Weak Bases

Weak bases are bases that do not completely dissociate in water. Unlike strong bases, which almost entirely turn into ions in solution, weak bases only partially ionize. This is significant because it influences how we calculate solution properties, such as pH.In the case of 1-naphthylamine, it is a weak base, as indicated by its interaction with water: \[ \mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{2} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{3}^{+} + \mathrm{OH}^{-} \]The pK_b value provided (3.92) is a measure of the base's dissociation tendency. A lower pK_b indicates a stronger base, but in this case, the relatively high pK_b indicates that 1-naphthylamine is a weak base. The dissociation produces hydroxide ions (OH⁻), which are crucial in determining the pH of the solution. Weak bases' partial ionization means we must use equilibrium expressions, like the one involving K_b, to find concentrations of ions like OH^-.

pH Calculation

Calculating the pH of a solution involving a weak base relies on understanding the concentration of hydroxide ions (OH^-) produced. For this, we use the dissociation constant of the base, K_b, and the base's concentration in the solution.Starting from the basic chemical equation:\[ K_b = \frac{[\mathrm{OH}^-][\mathrm{C}_{10}\mathrm{H}_7\mathrm{NH}_3^+]}{[\mathrm{C}_{10}\mathrm{H}_7\mathrm{NH}_2]} \]The concentration of OH⁻ in our case was found to be \(1.1 \times 10^{-2} \; M\). The pOH then is the negative logarithm of this concentration, which calculates as \[ \mathrm{pOH} = -\log(1.1 \times 10^{-2}) = 2.0 \]. With pOH in hand, pH can be directly calculated using the relationship: \[ \mathrm{pH} = 14 - \mathrm{pOH} \]. This implies that the pH for our 1-naphthylamine solution is \( 14 - 2.0 = 12.0 \), indicating a basic (or alkaline) solution.This calculation highlights how the solubility and ionization of weak bases influence pH, allowing chemists to predict the behavior of such solutions in various conditions.