Problem 71 - General Chemistry: Principles and Modern Applications

Question

Determine the molar solubility of lead(II) azide, \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2},\) in a buffer solution with \(\mathrm{pH}=3.00,\) given that \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{N}_{3}^{-}(\mathrm{aq})\) \(K_{\mathrm{sp}}=2.5 \times 10^{-9}\) \(\mathrm{HN}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{N}_{3}^{-}(\mathrm{aq})\) \(K_{\mathrm{a}}=1.9 \times 10^{-5}\)

Step-by-Step Solution

Answer

The molar solubility is calculated by using the formula for the solubility product while considering the azide ion concentration from the buffer. The answer depends on the initial concentration of the hydrazoic acid, which can be found using the Henderson – Hasselbalch equation. Solving for 's' in \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{N}_3^-]^{2}\) will give the molar solubility of Lead(II) azide.

Step 1: Equilibrium Expression for Lead(II) Azide

First, set up the equilibrium expression for the lead(II) azide. Lead(II) azide dissociates according to the equation: \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons\mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{N}_{3}^{-}(\mathrm{aq})\)where the solubility product constant is:\[K_{\mathrm{sp}} = [\mathrm{Pb}^{2+}][\mathrm{N}_3^-]^{2}\],which equals to \(2.5 \times 10^{-9}\). The molar solubility of lead(II) azide can be represented by 's', then at equilibrium, we can get:\(\mathrm{Pb}^{2+} = s\) and \(\mathrm{N}_3^- = 2s\)

Step 2: Set Up the Equilibrium Expressions for the Buffer Solution

With the buffer solution of pH = 3.00, we know that \([\mathrm{H}_{3}\mathrm{O}^{+}]=10^{-3.00}\). The equilibrium equation for the hydrazoic acid is:\[\mathrm{HN}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(\mathrm{aq})+\mathrm{N}_{3}^{-}(\mathrm{aq}),\]with the acid dissociation constant \(K_{a}\) is given as \(1.9 \times 10^{-5}\).We can set up the Henderson – Hasselbalch equation (since this solution acts as buffer solution):\(\mathrm{pH}=\mathrm{pKa} + \log\left(\frac{[\mathrm{A^{-}]}{[\mathrm{HA}]}\right)\)which can be re-arranged to find the concentration of \(\mathrm{N}_{3}^{-}\):\[ [\mathrm{N}_3^-] = [\mathrm{HA}]\times 10^{(\mathrm{pH}-\mathrm{pKa})}\]

Step 3: Calculate the Molar Solubility of Lead(II) Azide

We know from Step 1 that \(\mathrm{N}_3^- = 2s\) and from Step 2 that \(\mathrm{N}_3^-\) has a fixed concentration due to the buffer. So, we have \(\mathrm{Pb}^{2+} = s\) and \(\mathrm{N}_3^- = 2s + [\mathrm{N}_3^-]_{\mathrm{buffer}}\).Substituting these back into \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{N}_3^-]^{2}\) and solving for 's' (molar solubility) gives the final answer.

Key Concepts

Solubility Product Constant

The solubility product constant (\( K_{sp} \) ), is an essential concept in the study of solubility equilibria. It is a special case of the equilibrium constant and applies to sparingly soluble salts, like lead(II) azide in our exercise. The constant quantifies the degree to which a compound can dissolve in water by relating the molar concentrations of the ionic components at saturation point.

In the formula \( K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{N}_3^-]^2 \) for lead(II) azide, \( K_{sp} \) signifies the product of the molar concentrations of lead (\( \mathrm{Pb}^{2+} \) ) ions and azide (\( \mathrm{N}_3^- \) ) ions raised to the power of their stoichiometric coefficients when the salt is in a dynamic equilibrium with its dissolved ions. If you increase the concentration of one ion, the other must decrease to maintain the value of \( K_{sp} \). This balance is vital when calculating molar solubility, especially when external factors, such as buffer solutions, come into play.

Buffer Solution pH

A buffer solution is a system that resists changes in pH when small amounts of an acid or a base are added. It is typically composed of a weak acid and its conjugate base or vice versa. In this scenario, the buffer helps maintain the pH at 3.00, which is crucial for predicting the behavior of ions in the solution.

The Henderson-Hasselbalch equation is a formula derived from the acid dissociation constant that allows us to calculate the pH of a buffer solution: \( \mathrm{pH} = \mathrm{pKa} + \log\left(\frac{[\mathrm{A^{-}}]}{[\mathrm{HA}]}\right) \). This equation demonstrates the relationship between the pH of the solution, the concentration of the acid and its conjugate base, and the acid's dissociation constant (\( pKa \)). The buffer's pH can influence the solubility of substances in solution. Knowing the pH allows us to compute the concentration of the azide ion (\( \mathrm{N}_3^- \)) in the buffer, which is an essential step towards finding the molar solubility in a buffered environment.

Acid Dissociation Constant

The acid dissociation constant (\( K_a \)) is a quantitative measure of the strength of an acid in solution. It refers to the equilibrium constant for the dissociation of a weak acid into its conjugate base and a hydrogen ion in an aqueous solution. The larger the value of \( K_a \), the stronger the acid because it indicates a greater tendency for the acid to lose its proton.

In the given exercise, hydrazoic acid (\(\mathrm{HN}_3\)) dissociates into azide ions (\(\mathrm{N}_3^-\)) and hydronium ions (\(\mathrm{H}_3\mathrm{O}^+\)), with the degree of dissociation being dictated by \(K_a\). This dissociation impacts the concentration of the azide ions in the buffer solution, which in turn affects the molar solubility of lead(II) azide. By understanding the concept of acid dissociation and \(K_a\), we can better grasp how chemical equilibria shift in response to conditions like pH, ultimately influencing the solubility of ionic compounds in solution.