For `H_(2)S, K_(a_(1))=10^(-7), K_(a_(2))=10^(-14),CoS=4xx10^(-21), Ag_(2)S=6.3xx10^(-50)`
Calculate difference in PH for precipitation of MnS and CoS.

Answer

To calculate the difference in pH for the precipitation of MnS and CoS, we will follow these steps:

### Step 1: Understand the Dissociation of H₂S
H₂S can dissociate in two steps:
1. H₂S ⇌ H⁺ + HS⁻ (Kₐ₁ = 10⁻⁷)
2. HS⁻ ⇌ H⁺ + S²⁻ (Kₐ₂ = 10⁻¹⁴)

### Step 2: Calculate the Concentration of S²⁻
The equilibrium constant for the overall dissociation of H₂S can be calculated as:
\[ K_{eq} = K_{a1} \times K_{a2} = 10^{-7} \times 10^{-14} = 10^{-21} \]

The concentration of S²⁻ ions can be derived from the equilibrium expression:
\[ K_{eq} = \frac{[H^+]^2 [S^{2-}]}{[H_2S]} \]

Given that the concentration of H₂S is 0.1 M, we can rearrange the equation to find [S²⁻]:
\[ [S^{2-}] = \frac{K_{eq} \times [H_2S]}{[H^+]^2} \]

### Step 3: Calculate the pH for CoS Precipitation
For CoS, the solubility product (Ksp) is given as:
\[ K_{sp(CoS)} = 4 \times 10^{-21} \]

Using the Ksp expression:
\[ K_{sp} = [Co^{2+}][S^{2-}] \]
Given [Co²⁺] = 0.01 M, we can find [S²⁻]:
\[ [S^{2-}] = \frac{K_{sp}}{[Co^{2+}]} = \frac{4 \times 10^{-21}}{0.01} = 4 \times 10^{-19} \]

Now, we can find [H⁺]:
Using the earlier derived equation:
\[ [H^+]^2 = \frac{K_{eq} \times [H_2S]}{[S^{2-}]} \]
Substituting the values:
\[ [H^+]^2 = \frac{10^{-21} \times 0.1}{4 \times 10^{-19}} \]
\[ [H^+]^2 = 2.5 \times 10^{-3} \]
\[ [H^+] = \sqrt{2.5 \times 10^{-3}} = 0.05 \]

Now, calculate the pH:
\[ pH = -\log(0.05) \approx 1.3 \]

### Step 4: Calculate the pH for MnS Precipitation
For MnS, the Ksp is given as:
\[ K_{sp(MnS)} = 2.5 \times 10^{-10} \]

Using the Ksp expression:
\[ K_{sp} = [Mn^{2+}][S^{2-}] \]
Again, given [Mn²⁺] = 0.01 M, we can find [S²⁻]:
\[ [S^{2-}] = \frac{K_{sp}}{[Mn^{2+}]} = \frac{2.5 \times 10^{-10}}{0.01} = 2.5 \times 10^{-8} \]

Now, we can find [H⁺]:
Using the earlier derived equation:
\[ [H^+]^2 = \frac{K_{eq} \times [H_2S]}{[S^{2-}]} \]
Substituting the values:
\[ [H^+]^2 = \frac{10^{-21} \times 0.1}{2.5 \times 10^{-8}} \]
\[ [H^+]^2 = 4 \times 10^{-14} \]
\[ [H^+] = \sqrt{4 \times 10^{-14}} = 2 \times 10^{-7} \]

Now, calculate the pH:
\[ pH = -\log(2 \times 10^{-7}) \approx 6.7 \]

### Step 5: Calculate the Difference in pH
Now, we can find the difference in pH for the precipitation of MnS and CoS:
\[ \Delta pH = pH_{MnS} - pH_{CoS} \]
\[ \Delta pH = 6.7 - 1.3 = 5.4 \]

### Final Answer
The difference in pH for the precipitation of MnS and CoS is **5.4**.