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: Understand the dissociation of the salt
The salt \( M_2S \) dissociates in water as follows:
\[ M_2S \rightarrow 2M^+ + S^{2-} \]

### Step 2: Define the solubility
Let the solubility of \( M_2S \) be \( S \). According to the dissociation, for every mole of \( M_2S \) that dissolves, it produces:
- 2 moles of \( M^+ \) ions
- 1 mole of \( S^{2-} \) ions

### Step 3: Express the concentrations at equilibrium
At equilibrium:
- The concentration of \( M^+ \) ions will be \( 2S \)
- The concentration of \( S^{2-} \) ions will be \( S \)

### Step 4: Write the expression for Ksp
The solubility product \( K_{sp} \) is given by the formula:
\[ K_{sp} = [M^+]^2 [S^{2-}] \]
Substituting the equilibrium concentrations:
\[ K_{sp} = (2S)^2 \cdot (S) \]
\[ K_{sp} = 4S^2 \cdot S = 4S^3 \]

### Step 5: Substitute the given solubility
We are given that the solubility \( S = 3.5 \times 10^{-6} \). Now, we can substitute this value into the Ksp expression:
\[ K_{sp} = 4(3.5 \times 10^{-6})^3 \]

### Step 6: Calculate \( Ksp \)
Now we calculate \( (3.5 \times 10^{-6})^3 \):
\[ (3.5)^3 = 42.875 \]
\[ (10^{-6})^3 = 10^{-18} \]
Thus,
\[ (3.5 \times 10^{-6})^3 = 42.875 \times 10^{-18} \]

Now, we can calculate \( K_{sp} \):
\[ K_{sp} = 4 \times 42.875 \times 10^{-18} = 171.5 \times 10^{-18} = 1.715 \times 10^{-16} \]

### Step 7: Final answer
Therefore, the solubility product \( K_{sp} \) of the salt \( M_2S \) is:
\[ K_{sp} \approx 1.7 \times 10^{-16} \]