The molar solubility of `M(OH)_(3)` in `0.4 M M(NO_(3))_(3)` solution inters of solubility product of `M(OH)_(3)`

Options

A. `(K_(sp)//10.8)^(1//3)`
B. `(K_(sp)//3.6)^(1//3)`
C. `(K_(sp)//10.8)^(1//4)`
D. `(K_(sp)//0.4)^(1//3)`

Answer

To find the molar solubility of \( M(OH)_3 \) in a \( 0.4 \, M \, M(NO_3)_3 \) solution in terms of its solubility product \( K_{sp} \), we can follow these steps:

### Step 1: Write the dissociation reaction of \( M(OH)_3 \)
The dissociation of \( M(OH)_3 \) can be represented as:
\[ M(OH)_3 (s) \rightleftharpoons M^{3+} (aq) + 3 OH^{-} (aq) \]

### Step 2: Define the solubility product \( K_{sp} \)
The solubility product \( K_{sp} \) for \( M(OH)_3 \) is given by:
\[ K_{sp} = [M^{3+}][OH^{-}]^3 \]

### Step 3: Determine the concentration of \( M^{3+} \) from \( M(NO_3)_3 \)
In a \( 0.4 \, M \, M(NO_3)_3 \) solution, \( M(NO_3)_3 \) dissociates completely to give:
\[ M(NO_3)_3 \rightarrow M^{3+} + 3 NO_3^{-} \]
Thus, the concentration of \( M^{3+} \) ions in the solution is \( 0.4 \, M \).

### Step 4: Let the molar solubility of \( M(OH)_3 \) be \( S \)
When \( M(OH)_3 \) dissolves, it contributes \( S \) moles of \( M^{3+} \) and \( 3S \) moles of \( OH^{-} \):
- The total concentration of \( M^{3+} \) will be \( 0.4 + S \).
- The concentration of \( OH^{-} \) will be \( 3S \).

### Step 5: Substitute into the \( K_{sp} \) expression
Substituting these concentrations into the \( K_{sp} \) expression gives:
\[ K_{sp} = (0.4 + S)(3S)^3 \]
Expanding this:
\[ K_{sp} = (0.4 + S)(27S^3) \]

### Step 6: Assuming \( S \) is small compared to \( 0.4 \)
If we assume \( S \) is much smaller than \( 0.4 \), we can approximate:
\[ K_{sp} \approx 0.4 \cdot 27S^3 \]
This simplifies to:
\[ K_{sp} \approx 10.8S^3 \]

### Step 7: Solve for \( S \)
Rearranging gives:
\[ S^3 = \frac{K_{sp}}{10.8} \]
Taking the cube root:
\[ S = \left(\frac{K_{sp}}{10.8}\right)^{\frac{1}{3}} \]

### Final Answer
The molar solubility \( S \) of \( M(OH)_3 \) in terms of its solubility product \( K_{sp} \) is:
\[ S = \left(\frac{K_{sp}}{10.8}\right)^{\frac{1}{3}} \]

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