Chapter 8: Problem 224
\(\mathrm{K}_{\mathrm{a}} \times \mathrm{K}_{\mathrm{a}}\) for \(\mathrm{H}_{2} \mathrm{~S}=1.0 \times 10^{-21} \mathrm{M}^{2} .\) The concentration of \(\left[\mathrm{S}^{2-}\right]\) ion present in \(1 \mathrm{~L}\) of \(0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\) having \(\left[\mathrm{H}^{+}\right]\) equal to \(0.1 \mathrm{M}\) is \(\mathrm{x} \times 10^{-20}\). The value of \(\mathrm{x}\) is
Short Answer
Step by step solution
Understand the dissociation of H2S
Express in terms of equilibrium concentrations
Use ion product to find [S2-]
Calculate [S2-] concentration
Determine the value of x in terms of exponent
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Dissociation Constants
- First, H_2 S dissociates into hydrogen ions ( H^+ ) and hydrosulfide ions ( HS^- ).
- Then, the hydrosulfide ions further dissociate into hydrogen ions and sulfide ions ( S^{2-} ).
Equilibrium Concentrations
Hydrogen Sulfide Equilibrium
- The solution's equilibrium is significantly influenced by the H^+ concentration, provided as 0.1 M.
- Attaining equilibrium means the formation and reformation of molecules occurs at the same rate.