Problem 42 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An aqueous solution of a metal bromide \(M \mathrm{Br}_{2}(0.05 \mathrm{M})\) is saturated with \(\mathrm{H}_{2} \mathrm{~S}\). What is the minimum \(\mathrm{pH}\) at which \(M \mathrm{~S}\) will precipitate? [1993 - 3 Marks] \(K_{s p}\) for \(M \mathrm{~S}=6.0 \times 10^{-21} ;\) concentration of saturated \(\mathrm{H}_{2} \mathrm{~S}=0.1 \mathrm{M}\) \(K_{1}=10^{-7}\) and \(K_{2}=1.3 \times 10^{-13}\), for \(\mathrm{H}_{2} \mathrm{~S}\).

Step-by-Step Solution

Answer

The minimum pH for precipitation is approximately 10.38.

Step 1: Understanding the Problem

The problem asks us to determine the minimum pH at which the metal sulfide, \( M \mathrm{~S} \), will begin to precipitate from an aqueous solution of \( M \mathrm{Br}_2 \). Given the solubility product \( K_{sp} \) for \( M \mathrm{~S} \) and the dissociation constants \( K_1 \) and \( K_2 \) for \( \mathrm{H}_{2} \mathrm{~S} \), we need to find the concentration of sulfide ions \( \text{[S}^{2-}] \) needed to initiate precipitation, and then relate this to pH.

Step 2: Calculate Sulfide Ion Concentration Required for Precipitation

The precipitation of \( M \mathrm{~S} \) occurs when the ionic product \( [M^{2+}][S^{2-}] \) exceeds the solubility product \( K_{sp} = 6.0 \times 10^{-21} \).\[ [S^{2-}] = \frac{K_{sp}}{[M^{2+}]} = \frac{6.0 \times 10^{-21}}{0.05} = 1.2 \times 10^{-19} \]

Step 3: Calculate Sulfide Ion from Dissociation of \( \mathrm{H}_{2} \mathrm{S} \)

The dissociation of \( \mathrm{H}_{2} \mathrm{~S} \) is in two steps:\(\mathrm{H}_2\mathrm{~S} \rightleftharpoons \mathrm{HS}^{-} + \mathrm{H}^{+},\) with \( K_1 = 10^{-7} \) \(\mathrm{HS}^{-} \rightleftharpoons \mathrm{S}^{2-} + \mathrm{H}^{+},\) with \( K_2 = 1.3 \times 10^{-13} \)The overall equilibrium for \( \mathrm{S}^{2-} \) is given by:\[ \mathrm{H}_2\mathrm{~S} \rightleftharpoons \mathrm{S}^{2-} + 2\mathrm{H}^{+} \text{ with } K = K_1 \times K_2 = 1.3 \times 10^{-20} \]

Step 4: Relate Sulfide Concentration to pH

Combining the equilibrium constant for \( \mathrm{S}^{2-} \) with the required \( [\mathrm{S}^{2-}] \),\[ [\mathrm{S}^{2-}] = K \frac{1}{[\mathrm{H}^{+}]^2} = 1.2 \times 10^{-19} \]This implies,\[ K_{1} \times K_{2} \times [\mathrm{H}_{2}\mathrm{~S}] = 1.2 \times 10^{-19}[\mathrm{H}^{+}]^2 \]\[ 1.3 \times 10^{-20} \times 0.1 = 1.2 \times 10^{-19}[\mathrm{H}^{+}]^2 \]Solving for \([\mathrm{H}^{+}]\) yields:\[ [\mathrm{H}^{+}] = \sqrt{\frac{1.3 \times 10^{-21}}{1.2 \times 10^{-19}}} \approx 0.104 \times 10^{-1} \approx 4.17 \times 10^{-11} \]Then,\[ \text{pH} = -\log(4.17 \times 10^{-11}) \approx 10.38 \]

Step 5: Final Step: Calculate the Minimum pH

The minimum pH required for \( M \mathrm{~S} \) to precipitate is approximately \( 10.38 \). Ensure that calculations are correct and double-check assumptions such as concentrations and reactions involved.

Key Concepts

Solubility Product (Ksp)

The solubility product, represented as \(K_{sp}\), is a constant for a given sparingly soluble compound at a particular temperature. It indicates the maximum amount that can dissolve in solution. When dealing with precipitation, understanding \(K_{sp}\) is essential because it helps predict when a compound will start to form a solid, that is, precipitate out of solution. For a compound \(M \mathrm{~S}\), the solubility product expression is defined as:\[K_{sp} = [M^{2+}][S^{2-}]\]The square brackets denote concentrations of metal ions \(M^{2+}\) and sulfide ions \(S^{2-}\). When the product of these ion concentrations exceeds \(K_{sp}\), the compound begins to precipitate. It's essentially a balancing act—maintaining product concentrations below this threshold prevents precipitation.Naively increasing the concentration of either ion will incline the balance towards precipitation, understanding \(K_{sp}\) allows chemists to control this with precision.

Dissociation Constants

Dissociation constants are vital in understanding how molecules dissociate in solution, particularly weak acids and bases. For hydrogen sulfide \(\mathrm{H}_{2} \mathrm{~S}\), its dissociation occurs in two steps, each with its own dissociation constant, specifically \(K_1\) and \(K_2\). In the first step, \(\mathrm{H}_2\mathrm{~S}\) dissociates into hydrosulfide ions (\(\mathrm{HS}^-\)) and hydrogen ions (\(\mathrm{H}^+\)) with:\[K_1 = [\mathrm{HS}^-][\mathrm{H}^+]/[\mathrm{H}_2\mathrm{~S}]\]The second step involves \(\mathrm{HS}^-\) further dissociating into \(\mathrm{S}^{2-}\) and \(\mathrm{H}^+\):\[K_2 = [\mathrm{S}^{2-}][\mathrm{H}^+]/[\mathrm{HS}^-]\]These constants help us derive the overall equilibrium constant \(K\) for the full dissociation to \(\mathrm{S}^{2-}\). Mathematically, this is the product of \(K_1\) and \(K_2\), giving a direct insight into the equilibrium concentration of \(\mathrm{S}^{2-}\) ions, pivotal in precipitation calculations.This two-step dissociation model embodies the complexity of weak acid behavior and allows prediction of specific ionic concentrations under various conditions.

Hydrogen Sulfide Equilibrium

Hydrogen sulfide equilibrium plays a significant role in environments where \(\mathrm{H}_2 \mathrm{~S}\) gas is used. It involves the balance between the molecular form \(\mathrm{H}_2\mathrm{~S}\), the hydrosulfide ion \(\mathrm{HS}^-\), and the sulfide ion \(\mathrm{S}^{2-}\).The equilibrium for this full dissociation is depicted by:\[\mathrm{H}_2\mathrm{~S} \rightleftharpoons \mathrm{S}^{2-} + 2\mathrm{H}^{+}\]The equilibrium constant \(K\) for this specific reaction combines both dissociation constants \(K_1\) and \(K_2\), giving:\[K = K_1 \times K_2\]This equilibrium helps determine the concentration of \(\mathrm{S}^{2-}\) in solution, especially as it relates to pH interactions.Since the extent of sulfide ion formation is dependent on \(\mathrm{H}^{+}\) concentration, changes in pH notably shift this equilibrium, dictating whether \(\mathrm{S}^{2-}\) concentration becomes sufficient to trigger precipitation.