For the electrolyte of type, `A_(2)B, K_(sp)` is given then its solubility is calculated by

Options

A. `K_(sp)//4`
B. `3sqrt((K_(sp))/(4))`
C. `3sqrt(K_(sp))`
D. `sqrt((K_(sp))/(4))`

Explanation

To calculate the solubility of the electrolyte \( A_2B \) in terms of its solubility product \( K_{sp} \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of the electrolyte \( A_2B \) in water can be represented as:
\[ A_2B \rightleftharpoons 2A^+ + B^{2-} \]

### Step 2: Define solubility
Let the solubility of \( A_2B \) be \( S \) mol/L. This means that when \( A_2B \) dissolves, it produces:
- \( 2S \) moles of \( A^+ \) ions (since there are 2 moles of \( A^+ \) for every mole of \( A_2B \))
- \( S \) moles of \( B^{2-} \) ions (since there is 1 mole of \( B^{2-} \) for every mole of \( A_2B \))

### Step 3: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is defined as the product of the concentrations of the ions, each raised to the power of their coefficients in the balanced equation:
\[ K_{sp} = [A^+]^2[B^{2-}] \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (2S)^2 \cdot (S) = 4S^2 \cdot S = 4S^3 \]

### Step 4: Solve for \( S \)
To find the solubility \( S \) in terms of \( K_{sp} \), we rearrange the equation:
\[ 4S^3 = K_{sp} \]
Dividing both sides by 4 gives:
\[ S^3 = \frac{K_{sp}}{4} \]
Taking the cube root of both sides, we find:
\[ S = \sqrt[3]{\frac{K_{sp}}{4}} \]

### Final Answer
Thus, the solubility \( S \) of the electrolyte \( A_2B \) in terms of its solubility product \( K_{sp} \) is:
\[ S = \sqrt[3]{\frac{K_{sp}}{4}} \]