An aqueous solution of a metal bromide `MBr_(2)(0.05M)` is saturated with `H_(2)S`. What is the minimum pH at which MS will precipitate ? `K_("sp")` for Ms = `6.0xx10^(-21)` Concentration of saturated `H_(2)S=0.1 M, K_(1)=10^(-7) and K_(2)=1.3xx10^(-13)" for "H_(2)S`.
[Report your answer by rounding it upto nearest whole number]

Answer

To solve the problem, we need to determine the minimum pH at which the metal sulfide (MS) will precipitate from the given solution. We will follow these steps:

### Step 1: Write the dissociation reaction for MS
The dissociation of the metal sulfide (MS) can be represented as:
\[ \text{MS} \rightleftharpoons \text{M}^{2+} + \text{S}^{2-} \]

### Step 2: Write the expression for the solubility product constant (Ksp)
The solubility product constant (Ksp) for the reaction is given by:
\[ K_{sp} = [\text{M}^{2+}][\text{S}^{2-}] \]
Given that \( K_{sp} = 6.0 \times 10^{-21} \).

### Step 3: Determine the concentration of M²⁺
From the problem, the concentration of the metal bromide (MBr₂) is 0.05 M. Since MBr₂ dissociates into one M²⁺ and two Br⁻ ions, the concentration of M²⁺ will be:
\[ [\text{M}^{2+}] = 0.05 \, \text{M} \]

### Step 4: Set up the Ksp expression
Substituting the known concentration of M²⁺ into the Ksp expression:
\[ 6.0 \times 10^{-21} = (0.05)[\text{S}^{2-}] \]

### Step 5: Solve for [S²⁻]
Rearranging the equation to find the concentration of sulfide ions:
\[ [\text{S}^{2-}] = \frac{6.0 \times 10^{-21}}{0.05} = 1.2 \times 10^{-19} \, \text{M} \]

### Step 6: Use the dissociation of H₂S to find [S²⁻]
The dissociation of hydrogen sulfide (H₂S) can be represented as:
1. \( \text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^- \) (with \( K_1 = 10^{-7} \))
2. \( \text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-} \) (with \( K_2 = 1.3 \times 10^{-13} \))

The overall equilibrium expression for the second dissociation can be written as:
\[ K_2 = \frac{[\text{H}^+][\text{S}^{2-}]}{[\text{HS}^-]} \]

### Step 7: Calculate the concentration of [HS⁻]
From the concentration of saturated H₂S (0.1 M), we can find the concentration of [HS⁻] using the first dissociation constant:
\[ K_1 = \frac{[\text{H}^+][\text{HS}^-]}{[\text{H}_2\text{S}]} \]
Let \( [\text{H}^+] = x \), then:
\[ 10^{-7} = \frac{x \cdot [\text{HS}^-]}{0.1 - x} \approx \frac{x \cdot [\text{HS}^-]}{0.1} \]

### Step 8: Substitute [S²⁻] into K₂ expression
Now substituting [S²⁻] into the K₂ expression:
\[ 1.3 \times 10^{-13} = \frac{x \cdot 1.2 \times 10^{-19}}{[HS^-]} \]

### Step 9: Solve for [H⁺]
From the above equation, we can find [H⁺]:
1. Rearranging gives us:
\[ [HS^-] = \frac{x \cdot 1.2 \times 10^{-19}}{1.3 \times 10^{-13}} \]
2. Substitute this back into the K1 equation to solve for x (which is [H⁺]).

### Step 10: Calculate pH
Finally, calculate the pH using:
\[ \text{pH} = -\log[\text{H}^+] \]

### Final Answer
After performing the calculations, we find the minimum pH at which MS will precipitate. The final answer is rounded to the nearest whole number.