Home
Class 12
CHEMISTRY
An acidified of 0.05 M Zn^(2+) is satura...

An acidified of `0.05 M Zn^(2+)` is saturated with 0.1 M `H_2S`. What is the minimum molar concentration (M) or `H^+` required to prevent the precipitation of ZnS ? Use `K_(sp)(ZnS) = 1.25 xx10^(-22)` and overall dissociation constant of `H_2S, K_("NET") =K_1K_2=1xx10^(-21)` .

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine the minimum molar concentration of \( H^+ \) required to prevent the precipitation of \( ZnS \) from a solution that is saturated with \( H_2S \) and contains \( 0.05 M \) \( Zn^{2+} \). ### Step-by-Step Solution: 1. **Identify the Equilibrium Expression for \( ZnS \)**: The solubility product constant \( K_{sp} \) for \( ZnS \) is given by: \[ K_{sp} = [Zn^{2+}][S^{2-}] \] Given \( K_{sp}(ZnS) = 1.25 \times 10^{-22} \). 2. **Determine the Concentration of \( S^{2-} \)**: Since the solution is saturated with \( H_2S \), we need to find the concentration of \( S^{2-} \) in equilibrium. We can use the \( K_{sp} \) expression: \[ K_{sp} = [Zn^{2+}][S^{2-}] \] Rearranging gives: \[ [S^{2-}] = \frac{K_{sp}}{[Zn^{2+}]} \] Substituting the known values: \[ [S^{2-}] = \frac{1.25 \times 10^{-22}}{0.05} = 2.5 \times 10^{-21} \, M \] 3. **Use the Dissociation Constant of \( H_2S \)**: The overall dissociation constant \( K_{NET} \) for \( H_2S \) is given as: \[ K_{NET} = K_1 K_2 = 1 \times 10^{-21} \] The dissociation of \( H_2S \) can be represented as: \[ H_2S \rightleftharpoons H^+ + HS^- \] \[ HS^- \rightleftharpoons H^+ + S^{2-} \] The equilibrium expression for the dissociation of \( H_2S \) is: \[ K_{NET} = \frac{[H^+][S^{2-}]}{[H_2S]} \] 4. **Substituting Known Values**: We know \( [S^{2-}] = 2.5 \times 10^{-21} \, M \) and \( [H_2S] = 0.1 \, M \). We can substitute these into the \( K_{NET} \) expression: \[ 1 \times 10^{-21} = \frac{[H^+](2.5 \times 10^{-21})}{0.1} \] 5. **Solving for \( [H^+] \)**: Rearranging gives: \[ [H^+] = \frac{1 \times 10^{-21} \times 0.1}{2.5 \times 10^{-21}} = \frac{1 \times 10^{-22}}{2.5 \times 10^{-21}} = 0.04 \, M \] 6. **Conclusion**: The minimum molar concentration of \( H^+ \) required to prevent the precipitation of \( ZnS \) is: \[ [H^+] = 0.04 \, M \]

To solve the problem, we need to determine the minimum molar concentration of \( H^+ \) required to prevent the precipitation of \( ZnS \) from a solution that is saturated with \( H_2S \) and contains \( 0.05 M \) \( Zn^{2+} \). ### Step-by-Step Solution: 1. **Identify the Equilibrium Expression for \( ZnS \)**: The solubility product constant \( K_{sp} \) for \( ZnS \) is given by: \[ K_{sp} = [Zn^{2+}][S^{2-}] ...
Promotional Banner

Topper's Solved these Questions

  • JEE ADVANCED 2020

    JEE ADVANCED PREVIOUS YEAR|Exercise SECTION-2|6 Videos
  • JEE ADVANCED

    JEE ADVANCED PREVIOUS YEAR|Exercise CHEMISTRY SECTION - IV : Matric Match Type|2 Videos
  • JEE ADVANCED 2021

    JEE ADVANCED PREVIOUS YEAR|Exercise QUESTION|38 Videos

Similar Questions

Explore conceptually related problems

What is the minimum pH required to prevent the precipitation of ZnS in a solution which is 0.01 M ZnCl_2 and saturated with 0.1 M H_2S ? K_(sp)" of " (ZnS)=10^(-21),K_(a_1)xxK_(a_2)(H_2S)=10^(-20)

A solution containing 0.1 M Zn^(2+) and 0.01 M Cu^(2+) is saturated with H_2S . The S^(2-) concentration is 8.1 xx 10^(-31) M. Will ZnS or CuS precipitate ? K_(sp)(ZnS)=3.0xx10^(-23) and K_(sp)(CuS)=8.0xx10^(-34) .