The salt `M_x N_y` ionizes as :
`M_x N_y (s) hArr xM^(p+)(aq) + yN^(q-) (xp^(+)=yq^(-))` .
Its solubility may be expressed as :

Options

A. `(K_(sp))^(1//x+y)`
B. `(K_(sp) M^x N_y)^(x+y)`
C. `(K_(sp)//M^x N^y)^(x-y)`
D. `(K_(sp)//M^x N^y)^(1//x+y)`

Answer

To express the solubility of the salt \( M_x N_y \) that ionizes as:

\[ M_x N_y (s) \rightleftharpoons xM^{p+} (aq) + yN^{q-} (aq) \]

we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of the salt \( M_x N_y \) in water can be represented as:

\[ M_x N_y (s) \rightleftharpoons xM^{p+} (aq) + yN^{q-} (aq) \]

### Step 2: Define Solubility
Let the solubility of the salt \( M_x N_y \) be \( S \) mol/L. This means that when \( M_x N_y \) dissolves, it produces \( x \) moles of \( M^{p+} \) and \( y \) moles of \( N^{q-} \) for every mole of \( M_x N_y \) that dissolves.

### Step 3: Calculate the Concentrations
From the dissociation, we can express the concentrations of the ions in terms of \( S \):

- The concentration of \( M^{p+} \) ions will be \( xS \).
- The concentration of \( N^{q-} \) ions will be \( yS \).

### Step 4: Write the Expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for the salt can be expressed as:

\[ K_{sp} = [M^{p+}]^x [N^{q-}]^y \]

Substituting the concentrations from Step 3 into this expression gives:

\[ K_{sp} = (xS)^x (yS)^y \]

### Step 5: Simplify the Expression
This can be simplified to:

\[ K_{sp} = x^x y^y S^{x+y} \]

### Step 6: Solve for Solubility \( S \)
Now, we can solve for \( S \):

\[ S^{x+y} = \frac{K_{sp}}{x^x y^y} \]

Taking the \( (x+y)^{th} \) root of both sides gives:

\[ S = \left( \frac{K_{sp}}{x^x y^y} \right)^{\frac{1}{x+y}} \]

### Final Expression
Thus, the solubility \( S \) of the salt \( M_x N_y \) can be expressed as:

\[ S = \left( \frac{K_{sp}}{x^x y^y} \right)^{\frac{1}{x+y}} \]

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