Problem 56 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An acidified solution of \(0.05 \mathrm{M} \mathrm{Zn}^{2+}\) is saturated with \(0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\). What is the minimum molar concentration (M) of \(\mathrm{H}^{+}\)required to prevent the precipitation of \(\mathrm{ZnS}\) ? Use \(K_{\mathrm{sp}}(\mathrm{ZnS})=1.25 \times 10^{-22}\) and overall dissociation constant of \(\mathrm{H}_{2} \mathrm{~S}, K_{\mathrm{NET}}\) \(=K_{1} K_{2}=1 \times 10^{-21} . \quad\)

Step-by-Step Solution

Answer

The minimum [H⁺] required is 0.2 M.

Step 1: Write the Relevant Chemical Reaction

To prevent the precipitation of ZnS, we need to analyze the equilibrium of the reaction \(\mathrm{Zn}^{2+} + \mathrm{S}^{2-} \rightleftharpoons \mathrm{ZnS}\). ZnS will precipitate if the product of the ion concentrations exceeds the solubility product \(K_{\mathrm{sp}}\). Thus, for no precipitation, \([\mathrm{Zn}^{2+}][\mathrm{S}^{2-}] < K_{\mathrm{sp}}(\mathrm{ZnS})\).

Step 2: Use the Solubility Product Constant

Given \(K_{\mathrm{sp}}(\mathrm{ZnS})=1.25 \times 10^{-22}\), the condition \([\mathrm{Zn}^{2+}][\mathrm{S}^{2-}] < 1.25 \times 10^{-22}\) must be satisfied. The concentration of \(\mathrm{Zn}^{2+}\) is \(0.05 \mathrm{M}\). Therefore, \([\mathrm{S}^{2-}] < \frac{1.25 \times 10^{-22}}{0.05}\).

Step 3: Calculate the Maximum [S²⁻] Allowed

Calculate \([\mathrm{S}^{2-}] < \frac{1.25 \times 10^{-22}}{0.05} = 2.5 \times 10^{-21}\). This is the maximum concentration of \(\mathrm{S}^{2-}\) allowed to prevent precipitation.

Step 4: Understand Dissociation of H2S

The ionization of \(\mathrm{H_2S}\) in water is described by\(\mathrm{H_2S}\rightleftharpoons \mathrm{H}^+ + \mathrm{HS}^-\) and \(\mathrm{HS}^-\rightleftharpoons \mathrm{H}^+ + \mathrm{S}^{2-}\). Using the given \(K_{\mathrm{NET}} \) for \(\mathrm{H_2S}\), calculate the concentration of \([\mathrm{S}^{2-}]\).

Step 5: Use the Overall Dissociation Constant

Given \(K_{\mathrm{NET}}=K_1 \times K_2=1 \times 10^{-21}\), the relation is \([\mathrm{S}^{2-}] = \frac{K_{\mathrm{NET}}}{([\mathrm{H}]^+)^2}[\mathrm{H}_2S]\). With \([\mathrm{S}^{2-}] < 2.5 \times 10^{-21}\) and \([\mathrm{H}_2S] = 0.1 \mathrm{M}\), rearrange the expression to find \([\mathrm{H}^+]\).

Step 6: Solve for Minimum [H⁺] Concentration

Substituting into the relation gives \(2.5 \times 10^{-21} = \frac{1 \times 10^{-21}}{([\mathrm{H}^+]^2)} \times 0.1\). Solving this equation: \([\mathrm{H}^+]^2 = \frac{1 \times 10^{-21} \times 0.1}{2.5 \times 10^{-21}} = 0.04\), so \([\mathrm{H}^+] = \sqrt{0.04} = 0.2 \mathrm{M}\).

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, or \( K_{sp} \), is an important concept in chemistry that describes how soluble a compound can be in solution before it begins to precipitate. Essentially, it is the product of the concentrations of the ions in a saturated solution. For a compound like zinc sulfide (ZnS), the reaction can be depicted as \( \text{Zn}^{2+} + \text{S}^{2-} \rightleftharpoons \text{ZnS} \).
\( K_{sp} \) determines the maximum product of the concentrations of the ions that the solution can sustain before the solid starts forming, i.e., precipitating.

• This means that for \( \text{ZnS} \), the equation is \([\text{Zn}^{2+}][\text{S}^{2-}] = K_{sp} \).
• If \([\text{Zn}^{2+}][\text{S}^{2-}] < K_{sp}(\text{ZnS})\), no precipitation occurs.
• If \([\text{Zn}^{2+}][\text{S}^{2-}] > K_{sp}(\text{ZnS})\), precipitation will happen.
  

Understanding \( K_{sp} \) is crucial as it helps predict whether a mix of ions in solution will form a precipitate based on their concentrations.

Dissociation Constants

Dissociation constants are values that indicate the degree to which a compound dissociates into its components in solution. For \( \text{H}_2\text{S} \), the dissociation into \( \text{H}^+ \), \( \text{HS}^- \), and \( \text{S}^{2-} \) occurs in two steps.
The first step is \( \text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^- \), while the second step is \( \text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-} \).

• Each step has a specific dissociation constant, termed \( K_1 \) and \( K_2 \) respectively.
• The overall dissociation constant \( K_{\text{NET}} \) for \( \text{H}_2\text{S} \) is the product of these, or \( K_1 \times K_2 \).
• \( K_{\text{NET}} \) is important for calculating the concentration of sulfide ions \([\text{S}^{2-}] \) in solution, especially when other ions affect the balance.
  

Knowing the dissociation constants helps to understand how much \( \text{H}_2\text{S} \) will remain intact or dissociate into ions in different conditions.

Precipitation Reaction

Precipitation reactions involve the formation of a solid, called a precipitate, from ions present in solution. For the reaction \( \text{Zn}^{2+} + \text{S}^{2-} \rightleftharpoons \text{ZnS} \), precipitation occurs if the ion concentration products exceed \( K_{sp} \).
The solid form of \( \text{ZnS} \) will appear when the concentration of zinc ions and sulfide ions are too high for the solution to hold.

• A useful way to predict if a precipitate will form is by applying the \( K_{sp} \) value in calculations.
• If the solution's \([\text{Zn}^{2+}][\text{S}^{2-}] \) product exceeds \( 1.25 \times 10^{-22} \), \( \text{ZnS} \) begins to precipitate. Otherwise, the solution remains clear and unsaturated with zinc sulfide.
  

Understanding precipitation reactions is key as they are critical in areas ranging from water treatment to mineral formation in nature.

Equilibrium Concentration

Equilibrium concentration refers to the steady state reached by the system, where the rates of the forward and reverse reactions are equal. In the context of our exercise, this involves the concentrations of \( \text{Zn}^{2+} \) and \( \text{S}^{2-} \) reaching a balance with \( \text{ZnS} \) in solution.

• For no precipitation to occur, the equilibrium concentration of \( \text{S}^{2-} \) must be less than \( 2.5 \times 10^{-21} \) M.
• This requires using \( K_{sp} \) and the concentrations of other present ions to ensure equilibrium in reactions.
• The calculations lead to understanding how much of each species, ion or undissociated acid, is present once equilibrium is established.
  

Grasping equilibrium concentration is essential, as it allows predictions on system behavior under various conditions and how it approaches balance.

Acidic Solution Chemistry

Acidic solution chemistry is vital in understanding how an increase in hydrogen ion \( \text{H}^+ \) concentration can affect other ions' solubility and reaction propensity. In an acidic solution, \( \text{H}_2\text{S} \) dissociates less, reducing \( \text{S}^{2-} \) availability, therefore preventing precipitation.

• An increase in \( \text{H}^+ \) concentration shifts the equilibrium of \( \text{H}_2\text{S} \) dissociation backward, reducing sulfide ion amounts.
• For \( \text{ZnS} \), maintaining \([\text{H}^+] \approx 0.2 \text{ M} \) is needed to keep \( [\text{S}^{2-}] \) lower than necessary for precipitation.
• This practice is common in keeping sulfide and other potential precipitants soluble in industrial and scientific processes.
  

Understanding acidic solution chemistry helps manage reactions in different pH environments and is essential for reactions requiring precise ion concentrations.