Solubility if `M_(2)S` type salt is `3.5xx10^(-6)`, then find out its solubility product

Options

A. `1.7xx10^(-6)`
B. `1.7xx10^(-16)`
C. `1.7xx10^(-18)`
D. `1.7xx10^(-12)`

Explanation

To find the solubility product (Ksp) of the salt \( M_2S \) given its solubility, we can follow these steps:

### Step 1: Write the dissociation equation
When the salt \( M_2S \) dissolves in water, it dissociates according to the following equation:
\[ M_2S (s) \rightleftharpoons 2M^+ (aq) + S^{2-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( M_2S \) be \( S \). Given that the solubility of \( M_2S \) is \( 3.5 \times 10^{-6} \, \text{mol/L} \), we can denote this as:
\[ S = 3.5 \times 10^{-6} \, \text{mol/L} \]

### Step 3: Determine the concentrations of the ions
From the dissociation equation, we can see that:
- For every 1 mole of \( M_2S \) that dissolves, it produces 2 moles of \( M^+ \) and 1 mole of \( S^{2-} \).
- Therefore, the concentration of \( M^+ \) ions will be \( 2S \) and the concentration of \( S^{2-} \) ions will be \( S \).

Calculating these concentrations:
\[ [M^+] = 2S = 2 \times (3.5 \times 10^{-6}) = 7.0 \times 10^{-6} \, \text{mol/L} \]
\[ [S^{2-}] = S = 3.5 \times 10^{-6} \, \text{mol/L} \]

### Step 4: Write the expression for the solubility product (Ksp)
The solubility product \( K_{sp} \) is given by the expression:
\[ K_{sp} = [M^+]^2 [S^{2-}] \]

### Step 5: Substitute the concentrations into the Ksp expression
Substituting the values we calculated:
\[ K_{sp} = (7.0 \times 10^{-6})^2 \times (3.5 \times 10^{-6}) \]

### Step 6: Calculate \( Ksp \)
Calculating \( (7.0 \times 10^{-6})^2 \):
\[ (7.0 \times 10^{-6})^2 = 49.0 \times 10^{-12} = 4.9 \times 10^{-11} \]
Now, multiplying by \( (3.5 \times 10^{-6}) \):
\[ K_{sp} = 4.9 \times 10^{-11} \times 3.5 \times 10^{-6} = 17.15 \times 10^{-17} = 1.715 \times 10^{-16} \]

### Final Answer
Thus, the solubility product \( K_{sp} \) of the salt \( M_2S \) is:
\[ K_{sp} = 1.715 \times 10^{-16} \]

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