Chapter 7: Problem 148
What is the minimum pH needed to prevent the precipitation of \(\mathrm{ZnS}\) in a solution that is \(0.01 \mathrm{M} \mathrm{ZnCl}_{2}\) and saturated with \(0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\) ? \(\left[\mathrm{K}_{\mathrm{SP}}=10^{-21}, \mathrm{~K}_{\mathrm{a}_{1}} \times \mathrm{K}_{\mathrm{a}_{2}}=10^{2}-20\right]\) (a) 0 (b) 1 (c) 3 (d) 5
Short Answer
Step by step solution
Understanding the Reaction
Applying the Solubility Product
Sulfide Ion Equilibrium
Relating the Ion Products and Concentrations
Calculating pH
Choosing the Correct Option
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solubility Product
- \( Zn^{2+} + S^{2-} \rightleftharpoons ZnS \)
- \( [Zn^{2+}] \cdot [S^{2-}] = 10^{-21} \)
Ion Equilibrium
- \( H_2S \rightleftharpoons HS^- + H^+ \)
- \( HS^- \rightleftharpoons S^{2-} + H^+ \)
Weak Acids
- It partially dissociates into \( HS^- \) and \( S^{2-} \).
- Its two-step ionization contributes successive \( H^+ \) ions to the solution.