If `K_sp` of `HgSO_4, is 6.4 x 10^-5 M^2`, then solubility of the `HgSO_4`, in water will be

Options

A. `5.4*10^-5 M`
B. `8*10^-3 M`
C. `8* 10^-4 M`
D. `6.4* 10^-3 M`

Answer

To find the solubility of \( \text{HgSO}_4 \) in water given its \( K_{sp} \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \( \text{HgSO}_4 \) 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 \( \text{HgSO}_4 \) be \( s \) M. This means that at equilibrium:
- The concentration of \( \text{Hg}^{2+} \) ions will be \( s \) M.
- The concentration of \( \text{SO}_4^{2-} \) ions will also be \( s \) M.

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for the dissociation of \( \text{HgSO}_4 \) can be expressed as:
\[ K_{sp} = [\text{Hg}^{2+}][\text{SO}_4^{2-}] = s \cdot s = s^2 \]

### Step 4: Substitute the given \( K_{sp} \)
We are given:
\[ K_{sp} = 6.4 \times 10^{-5} \, M^2 \]
Thus, we can set up the equation:
\[ s^2 = 6.4 \times 10^{-5} \]

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

### Step 6: Calculate the value
Calculating the square root:
\[ s = \sqrt{6.4} \times \sqrt{10^{-5}} = 2.52982 \times 10^{-2.5} = 2.52982 \times 10^{-3} \approx 8 \times 10^{-3} \, M \]

### Final Answer
The solubility of \( \text{HgSO}_4 \) in water is approximately:
\[ s \approx 8 \times 10^{-3} \, M \]

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