If `K_(sp)` for `HgSO_(4)` is `6.4xx10^(-5)`, then solubility of this substance in mole per `m^(3)` is

Options

A. `8xx10^(-3)`
B. `6.4xx10^(-5)`
C. `8xx10^(-6)`
D. None of these

Answer

To find the solubility of `HgSO4` in moles per cubic meter (m³), we will follow these steps:

### Step 1: Write the dissociation equation
The dissociation of `HgSO4` in water can be represented as:
\[ \text{HgSO}_4 (s) \rightleftharpoons \text{Hg}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

### Step 2: Define solubility
Let the solubility of `HgSO4` be \( S \) moles per cubic meter. When `HgSO4` dissolves, it produces:
- \( S \) moles of \( \text{Hg}^{2+} \)
- \( S \) moles of \( \text{SO}_4^{2-} \)

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for the dissociation of `HgSO4` is given by:
\[ K_{sp} = [\text{Hg}^{2+}][\text{SO}_4^{2-}] \]
Substituting the concentrations in terms of solubility \( S \):
\[ K_{sp} = S \cdot S = S^2 \]

### Step 4: Substitute the value of \( K_{sp} \)
We are given that \( K_{sp} \) for `HgSO4` is \( 6.4 \times 10^{-5} \):
\[ S^2 = 6.4 \times 10^{-5} \]

### Step 5: Solve for \( S \)
To find the solubility \( S \), take the square root of both sides:
\[ S = \sqrt{6.4 \times 10^{-5}} \]

### Step 6: Calculate \( S \)
Calculating the square root:
\[ S = \sqrt{6.4} \times \sqrt{10^{-5}} \]
\[ S \approx 2.5298 \times 10^{-2.5} \]
\[ S \approx 8.0 \times 10^{-3} \, \text{moles/m}^3 \]

Thus, the solubility of `HgSO4` in moles per cubic meter is approximately:
\[ S \approx 8.0 \times 10^{-3} \, \text{moles/m}^3 \]

### Final Answer
The solubility of `HgSO4` is \( 8.0 \times 10^{-3} \, \text{moles/m}^3 \).

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