What is the maximum molarity of `Co^(+2)` ions in `0.1M HC1` saturated with `0.1M H_(2)S. (K_(a) = 4 xx 10^(-21))`. Given: `K_(sp)` of `CoS = 2xx10^(-21)`.

Options

A. `0.10M`
B. `1.00M`
C. `4.48 xx 10^(-11)M`
D. `0.050M`

Explanation

To find the maximum molarity of \( \text{Co}^{2+} \) ions in \( 0.1 \, M \, \text{HCl} \) saturated with \( 0.1 \, M \, \text{H}_2\text{S} \), we can follow these steps:

### Step 1: Determine the concentration of \( \text{S}^{2-} \) ions
The dissociation of \( \text{H}_2\text{S} \) can be represented as:
\[ \text{H}_2\text{S} \rightleftharpoons 2 \text{H}^+ + \text{S}^{2-} \]
The equilibrium expression for this dissociation is given by:
\[ K_a = \frac{[\text{H}^+]^2 [\text{S}^{2-}]}{[\text{H}_2\text{S}]} \]
Given:
- \( K_a = 4 \times 10^{-21} \)
- \( [\text{H}_2\text{S}] = 0.1 \, M \)
- \( [\text{H}^+] = 0.1 \, M \) (from \( 0.1 \, M \, \text{HCl} \))

Substituting the known values into the \( K_a \) expression:
\[ 4 \times 10^{-21} = \frac{(0.1)^2 [\text{S}^{2-}]}{0.1} \]

### Step 2: Solve for \( [\text{S}^{2-}] \)
Rearranging the equation:
\[ [\text{S}^{2-}] = \frac{4 \times 10^{-21} \times 0.1}{(0.1)^2} \]
\[ [\text{S}^{2-}] = \frac{4 \times 10^{-21} \times 0.1}{0.01} \]
\[ [\text{S}^{2-}] = 4 \times 10^{-20} \, M \]

### Step 3: Use \( K_{sp} \) to find \( [\text{Co}^{2+}] \)
The solubility product \( K_{sp} \) for \( \text{CoS} \) is given by:
\[ K_{sp} = [\text{Co}^{2+}][\text{S}^{2-}] \]
Given:
- \( K_{sp} = 2 \times 10^{-21} \)

Substituting \( [\text{S}^{2-}] \) into the \( K_{sp} \) expression:
\[ 2 \times 10^{-21} = [\text{Co}^{2+}] \times (4 \times 10^{-20}) \]

### Step 4: Solve for \( [\text{Co}^{2+}] \)
Rearranging the equation to find \( [\text{Co}^{2+}] \):
\[ [\text{Co}^{2+}] = \frac{2 \times 10^{-21}}{4 \times 10^{-20}} \]
\[ [\text{Co}^{2+}] = \frac{2}{4} \times 10^{-1} \]
\[ [\text{Co}^{2+}] = 0.05 \, M \]

### Final Answer
The maximum molarity of \( \text{Co}^{2+} \) ions in \( 0.1 \, M \, \text{HCl} \) saturated with \( 0.1 \, M \, \text{H}_2\text{S} \) is \( 0.05 \, M \).

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