The molar solubility (in mol `L^(-1) ) ` of a sparingly soluble salt `MX_4` is 's' The corresponding solubility product is `K_(sp)` .s is given in terms of `K_(sp)` by the relation.

Options

A. ` s= ((K_(sp))/(128))^(1//4) `
B. ` s= (128K_(sp) )^(1//4)`
C. ` s= ( 256 K_(sp)) ^(1//5)`
D. ` s= ((K_(sp))/(256))^(1//5) `

Explanation

To find the molar solubility (s) of the sparingly soluble salt \( MX_4 \) in terms of its solubility product \( K_{sp} \), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of the salt \( MX_4 \) in water can be represented as:
\[ MX_4 (s) \rightleftharpoons M^{4+} (aq) + 4X^{-} (aq) \]

### Step 2: Define Molar Solubility
Let the molar solubility of \( MX_4 \) be \( s \) mol/L. This means:
- The concentration of \( M^{4+} \) ions will be \( s \) mol/L.
- The concentration of \( X^{-} \) ions will be \( 4s \) mol/L (since there are 4 moles of \( X^{-} \) for every mole of \( MX_4 \)).

### Step 3: Write the Expression for Solubility Product
The solubility product \( K_{sp} \) is given by the expression:
\[ K_{sp} = [M^{4+}][X^{-}]^4 \]
Substituting the concentrations in terms of \( s \):
\[ K_{sp} = (s)(4s)^4 \]

### Step 4: Simplify the Expression
Now, simplify the expression:
\[ K_{sp} = s \cdot (4^4 \cdot s^4) = s \cdot 256s^4 = 256s^5 \]

### Step 5: Solve for Molar Solubility (s)
Now, we can express \( s \) in terms of \( K_{sp} \):
\[ s^5 = \frac{K_{sp}}{256} \]
Taking the fifth root gives:
\[ s = \left(\frac{K_{sp}}{256}\right)^{\frac{1}{5}} \]

### Final Result
Thus, the molar solubility \( s \) in terms of the solubility product \( K_{sp} \) is:
\[ s = \frac{K_{sp}^{1/5}}{256^{1/5}} \]