`K_(SP)` of `MX_(4)` and solubility of `MX_(4)` is `S mol//L` is related by:

Options

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

Explanation

To find the relationship between the solubility product \( K_{sp} \) of \( MX_4 \) and its solubility \( S \) (in mol/L), we can follow these steps:

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

### Step 2: Define Solubility
Let the solubility of \( MX_4 \) be \( S \) mol/L. This means that when \( MX_4 \) dissolves, it produces:
- \( M^{4+} \) ions: \( S \) mol/L
- \( X^{-} \) ions: \( 4S \) mol/L (since there are 4 moles of \( X^{-} \) produced for every mole of \( MX_4 \))

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

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

### Step 5: Rearranging the Equation
To express \( S \) in terms of \( K_{sp} \):
\[ S^5 = \frac{K_{sp}}{256} \]
Taking the fifth root:
\[ S = \left(\frac{K_{sp}}{256}\right)^{\frac{1}{5}} \]

### Final Result
Thus, the relationship between the solubility \( S \) and the solubility product \( K_{sp} \) is:
\[ S = \left(\frac{K_{sp}}{256}\right)^{\frac{1}{5}} \]