The number of `S^(2-)` ions present in `1L` of `0.1 MH_(2)S [K_(a(H_(2)S)) = 10^(-21)]` solution having `[H^(o+)] = 0.1M` is:

Options

A. (a) `6.023 xx 10^(3)`
B. (b) `6.023 xx 10^(4)`
C. (c) `6.023 xx 10^(5)`
D. (d) `6.023 xx 10^(6)`

Explanation

To solve the problem of finding the number of \( S^{2-} \) ions present in 1 L of 0.1 M \( H_2S \) solution, we can follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of \( H_2S \) can be represented as:
\[ H_2S \rightleftharpoons 2H^+ + S^{2-} \]

### Step 2: Write the expression for the equilibrium constant \( K_a \)
The expression for the equilibrium constant \( K_a \) for this reaction is given by:
\[ K_a = \frac{[H^+]^2 \cdot [S^{2-}]}{[H_2S]} \]

### Step 3: Substitute known values into the \( K_a \) expression
We know:
- \( K_a = 10^{-21} \)
- \( [H^+] = 0.1 \, M \)
- \( [H_2S] \) is approximately \( 0.1 \, M \) because the dissociation is very small.

Substituting these values into the \( K_a \) expression:
\[ 10^{-21} = \frac{(0.1)^2 \cdot [S^{2-}]}{0.1} \]

### Step 4: Simplify the equation
This simplifies to:
\[ 10^{-21} = \frac{0.01 \cdot [S^{2-}]}{0.1} \]
\[ 10^{-21} = 0.1 \cdot [S^{2-}] \]

### Step 5: Solve for \( [S^{2-}] \)
Now, we can solve for \( [S^{2-}] \):
\[ [S^{2-}] = \frac{10^{-21}}{0.1} = 10^{-20} \, M \]

### Step 6: Calculate the number of moles of \( S^{2-} \)
Since the volume of the solution is 1 L, the number of moles of \( S^{2-} \) is equal to its concentration:
\[ \text{Moles of } S^{2-} = [S^{2-}] \times \text{Volume} = 10^{-20} \, \text{mol/L} \times 1 \, \text{L} = 10^{-20} \, \text{mol} \]

### Step 7: Convert moles to number of ions
To find the number of \( S^{2-} \) ions, we multiply the number of moles by Avogadro's number (\( N_A = 6.022 \times 10^{23} \)):
\[ \text{Number of } S^{2-} \text{ ions} = 10^{-20} \, \text{mol} \times 6.022 \times 10^{23} \, \text{ions/mol} \]
\[ = 6.022 \times 10^{3} \text{ ions} \]

### Final Answer
The number of \( S^{2-} \) ions present in 1 L of 0.1 M \( H_2S \) solution is approximately \( 6022 \) ions.

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