An aqueous solution of metal chloride `MCI_(2)(0.05M)` is saturated with `H_(2)S (0.1M)`. The minimum `pH` at which metal sulphide will be precipiated is
`[K_(sp)MS = 5 xx 10^(-21),K_(1)(H_(2)S) = 10^(-7),K_(2)(H_(2)S) = 10^(-14)`.

Options

A. `3.25`
B. `2.50`
C. `1.50`
D. `1.25`

Explanation

To solve the problem, we need to determine the minimum pH at which the metal sulfide (MS) will precipitate from a saturated solution of H₂S and MCl₂. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the dissociation reactions
The dissociation of H₂S in water can be represented in two steps:
1. \( H_2S \rightleftharpoons H^+ + HS^- \) with \( K_1 = 10^{-7} \)
2. \( HS^- \rightleftharpoons H^+ + S^{2-} \) with \( K_2 = 10^{-14} \)

### Step 2: Combine the dissociation reactions
By adding the two dissociation reactions, we can derive a combined reaction:
\[ H_2S \rightleftharpoons 2H^+ + S^{2-} \]
The equilibrium constant for this combined reaction, \( K \), is given by:
\[ K = K_1 \times K_2 = 10^{-7} \times 10^{-14} = 10^{-21} \]

### Step 3: Write the expression for K
The equilibrium expression for the combined reaction is:
\[ K = \frac{[H^+]^2 \cdot [S^{2-}]}{[H_2S]} \]
Given that the concentration of H₂S is 0.1 M, we can substitute this into the equation:
\[ 10^{-21} = \frac{[H^+]^2 \cdot [S^{2-}]}{0.1} \]

### Step 4: Determine the concentration of \( S^{2-} \)
For the precipitation of metal sulfide (MS), the ionic product must equal the solubility product (Ksp):
\[ K_{sp} = [M^{2+}][S^{2-}] \]
Given that the concentration of MCl₂ is 0.05 M, the concentration of \( M^{2+} \) is also 0.05 M. Therefore:
\[ 5 \times 10^{-21} = 0.05 \cdot [S^{2-}] \]
Solving for \( [S^{2-}] \):
\[ [S^{2-}] = \frac{5 \times 10^{-21}}{0.05} = 10^{-19} \]

### Step 5: Substitute \( [S^{2-}] \) back into the K expression
Now substituting \( [S^{2-}] \) into the K expression:
\[ 10^{-21} = \frac{[H^+]^2 \cdot 10^{-19}}{0.1} \]
Rearranging gives:
\[ [H^+]^2 = 10^{-21} \cdot 0.1 \cdot 10^{19} = 10^{-21} \cdot 10^{-1} = 10^{-22} \]

### Step 6: Calculate \( [H^+] \)
Taking the square root:
\[ [H^+] = 10^{-11} \]

### Step 7: Calculate pH
The pH is calculated using:
\[ pH = -\log[H^+] = -\log(10^{-11}) = 11 \]

### Conclusion
Thus, the minimum pH at which metal sulfide will precipitate is **11**.

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