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Step-by-Step Solution
Step 1: Identify the Relevant Equilibria
We have two equilibria involving the dissociation of hydrogen sulfide:
1) $H_{2}S \rightleftharpoons H^{+} + HS^{-}$ with equilibrium constant $K_{1} = 1.0 \times 10^{-7}.$
2) $HS^{-} \rightleftharpoons H^{+} + S^{2-}$ with equilibrium constant $K_{2} = 1.2 \times 10^{-13}.$
Together, these can be combined into the overall reaction:
$H_{2}S \rightleftharpoons 2H^{+} + S^{2-}.$
Step 2: Calculate the Overall Equilibrium Constant
When we combine the two equilibria, the overall equilibrium constant $K$ is given by the product:
$K = K_{1} \times K_{2}.$
Substituting the given values:
$K = (1.0 \times 10^{-7}) \times (1.2 \times 10^{-13}) = 1.2 \times 10^{-20}.$
Step 3: Note the Initial Concentrations
• The solution contains $0.10\text{ M }H_{2}S.$
• The solution also contains $0.20\text{ M }HCl,$ which dissociates completely to give $[H^{+}] = 0.20\text{ M}.$
Since $K_{1}$ and $K_{2}$ are very small, the dissociation of $H_{2}S$ contributes a negligible amount of additional $H^{+}$ ions. Therefore, we assume $[H^{+}] \approx 0.20\text{ M}.$
Step 4: Apply the Overall Equilibrium Expression
For the overall reaction
$H_{2}S \rightleftharpoons 2H^{+} + S^{2-},$
the equilibrium constant expression is:
$K = \dfrac{[H^{+}]^{2}\,[S^{2-}]}{[H_{2}S]}.$
Rearrange to find $[S^{2-}]$:
$[S^{2-}] = \dfrac{K \times [H_{2}S]}{[H^{+}]^{2}}.$
Step 5: Substitute the Values
Substitute $K = 1.2 \times 10^{-20},\, [H_{2}S] = 0.10\text{ M},$ and $[H^{+}] = 0.20\text{ M}:$
$[S^{2-}] = \dfrac{(1.2 \times 10^{-20}) \times (0.10)}{(0.20)^{2}}.$
Calculate step by step:
1) $(0.20)^{2} = 0.04.$
2) Numerator $= 1.2 \times 10^{-20} \times 0.10 = 1.2 \times 10^{-21}.$
3) Divide by $0.04:$
$\dfrac{1.2 \times 10^{-21}}{0.04} = 3.0 \times 10^{-20}.$
Hence,
$[S^{2-}] = 3 \times 10^{-20}\,\text{M}.$
Final Answer
The concentration of $S^{2-}$ ions in the aqueous solution is $3 \times 10^{-20}\,\text{M}.$
