What is the minimum pH required to prevent the precipitation of `Zn^(+2)` in a solution that is 0.05 M `ZnCl_(2)` and saturated with 0.2M `H_(2)S` (Given : `K_(sp)`( ZnS) = `10^(-20)` and `Ka_(1) xx Ka_(2) ` of `H_(2)S = 10^(-20)` )

Answer

To find the minimum pH required to prevent the precipitation of \( \text{Zn}^{2+} \) in a solution that is 0.05 M \( \text{ZnCl}_2 \) and saturated with 0.2 M \( \text{H}_2\text{S} \), we will follow these steps:

### Step 1: Calculate the sulfide ion concentration \([S^{2-}]\)

The solubility product constant \( K_{sp} \) for zinc sulfide \( \text{ZnS} \) is given as \( 10^{-20} \). The relationship for the solubility product is:

\[ K_{sp} = [Zn^{2+}][S^{2-}] \]

Given that the concentration of \( \text{Zn}^{2+} \) from \( \text{ZnCl}_2 \) is 0.05 M, we can rearrange the equation to find \([S^{2-}]\):

\[ [S^{2-}] = \frac{K_{sp}}{[Zn^{2+}]} = \frac{10^{-20}}{0.05} = 2 \times 10^{-19} \, \text{M} \]

### Step 2: Use the dissociation constants of \( \text{H}_2\text{S} \)

The dissociation of \( \text{H}_2\text{S} \) can be represented as:

1. \( \text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^- \) with \( K_{a1} \)
2. \( \text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-} \) with \( K_{a2} \)

The product of the two dissociation constants is given as:

\[ K_{a1} \cdot K_{a2} = 10^{-20} \]

### Step 3: Relate \( [H^+] \) and \( [S^{2-}] \)

From the equilibrium expressions, we can express the relationship:

\[ K_{a1} \cdot K_{a2 = \frac{[H^+]^2 [S^{2-}]}{[H_2S]}} \]

Rearranging gives:

\[ [H^+]^2 = K_{a1} \cdot K_{a2} \cdot \frac{[H_2S]}{[S^{2-}]} \]

Substituting the values:

\[ [H^+]^2 = 10^{-20} \cdot \frac{0.2}{2 \times 10^{-19}} = 10^{-20} \cdot 10^{1} = 10^{-19} \]

### Step 4: Calculate the concentration of \( [H^+] \)

Taking the square root gives:

\[ [H^+] = \sqrt{10^{-19}} = 10^{-9.5} \, \text{M} \]

### Step 5: Calculate the pH

The pH is calculated using the formula:

\[ \text{pH} = -\log[H^+] \]

Substituting the value of \( [H^+] \):

\[ \text{pH} = -\log(10^{-9.5}) = 9.5 \]

### Conclusion

Thus, the minimum pH required to prevent the precipitation of \( \text{Zn}^{2+} \) is **9.5**.

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