The `K_(sp)` of `FeS = 4 xx 10^(-19)` at 298 K. The minimum concentration of `H^(+)` ions required to prevent the precipitation of FeS from a 0.01 M solution `Fe^(2+)` salt by passing `H_(2)S(0.1M)` (Given `H_(2)S k_(a_1) xx k_(b_1) = 10^(-21)`)

Options

A. `1.6 xx 10^(-3)` M
B. `2.5 xx 10^(-4)` M
C. `2.0 xx 10^(-2)` M
D. `1.2 xx 10^(-4)` M

Answer

To solve the problem, we need to find the minimum concentration of \( H^+ \) ions required to prevent the precipitation of \( FeS \) from a 0.01 M solution of \( Fe^{2+} \) when passing \( H_2S \) (0.1 M). Given that \( K_{sp} \) of \( FeS = 4 \times 10^{-19} \) and \( K_{a1} \times K_{b1} = 10^{-21} \) for \( H_2S \), we can follow these steps:

### Step 1: Write down the equilibrium expression for \( K_{sp} \) of \( FeS \)

The solubility product \( K_{sp} \) for the dissociation of \( FeS \) can be represented as:

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

### Step 2: Substitute the known values into the \( K_{sp} \) expression

Given:
- \( K_{sp} = 4 \times 10^{-19} \)
- \( [Fe^{2+}] = 0.01 \, M \)

We can rearrange the equation to find \( [S^{2-}] \):

\[ 4 \times 10^{-19} = (0.01)[S^{2-}] \]

### Step 3: Solve for \( [S^{2-}] \)

\[ [S^{2-}] = \frac{4 \times 10^{-19}}{0.01} = 4 \times 10^{-17} \, M \]

### Step 4: Write the equilibrium expression for the dissociation of \( H_2S \)

The dissociation of \( H_2S \) can be represented as:

1. \( H_2S \rightleftharpoons H^+ + HS^- \) (with \( K_{a1} \))
2. \( HS^- \rightleftharpoons H^+ + S^{2-} \) (with \( K_{a2} \))

The product of these two reactions gives us:

\[ K_{a1} \times K_{b1} = [H^+]^2 [S^{2-}] / [H_2S] \]

### Step 5: Substitute known values into the \( K_{a1} \times K_{b1} \) expression

Given \( K_{a1} \times K_{b1} = 10^{-21} \) and \( [H_2S] = 0.1 \, M \):

\[ 10^{-21} = \frac{[H^+]^2 [S^{2-}]}{[H_2S]} \]

Substituting \( [S^{2-}] \):

\[ 10^{-21} = \frac{[H^+]^2 (4 \times 10^{-17})}{0.1} \]

### Step 6: Rearrange to find \( [H^+] \)

\[ 10^{-21} = \frac{[H^+]^2 \times 4 \times 10^{-17}}{0.1} \]

Multiplying both sides by \( 0.1 \):

\[ 10^{-22} = [H^+]^2 \times 4 \times 10^{-17} \]

Now, divide both sides by \( 4 \times 10^{-17} \):

\[ [H^+]^2 = \frac{10^{-22}}{4 \times 10^{-17}} = \frac{10^{-22 + 17}}{4} = \frac{10^{-5}}{4} \]

### Step 7: Take the square root to find \( [H^+] \)

\[ [H^+] = \sqrt{\frac{10^{-5}}{4}} = \frac{10^{-2.5}}{2} = \frac{1}{2} \times 10^{-2.5} \approx 1.58 \times 10^{-3} \, M \]

### Final Answer

Thus, the minimum concentration of \( H^+ \) ions required to prevent the precipitation of \( FeS \) is approximately:

\[ [H^+] \approx 1.6 \times 10^{-3} \, M \]

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