Determine the solubility of `Cr(OH)_(3)` in `mol L^(-1)`, If its `K_(sp)` is `2.7 xx 10^(-31) M^(4)`.

Options

A. `1 xx 10^(-8)`
B. `8 xx 10^(-8)`
C. `1.1 xx 10^(-8)`
D. `0.18 xx 10^(-8)`

Answer

To determine the solubility of `Cr(OH)_(3)` in `mol L^(-1)`, given that its `K_(sp)` is `2.7 × 10^(-31) M^(4)`, we can follow these steps:

### Step 1: Write the dissociation equation
`Cr(OH)_(3)` is a salt that dissociates in water as follows:
\[ Cr(OH)_{3 (s)} \rightleftharpoons Cr^{3+} (aq) + 3 OH^{-} (aq) \]

### Step 2: Define solubility
Let the solubility of `Cr(OH)_(3)` be \( S \) mol/L. When it dissolves, it produces:
- 1 mole of \( Cr^{3+} \)
- 3 moles of \( OH^{-} \)

Thus, at equilibrium:
- Concentration of \( Cr^{3+} \) = \( S \)
- Concentration of \( OH^{-} \) = \( 3S \)

### Step 3: Write the expression for Ksp
The solubility product \( K_{sp} \) is given by the expression:
\[ K_{sp} = [Cr^{3+}][OH^{-}]^3 \]
Substituting the concentrations from the solubility:
\[ K_{sp} = S \cdot (3S)^3 \]

### Step 4: Simplify the expression
Expanding the expression:
\[ K_{sp} = S \cdot 27S^3 = 27S^4 \]

### Step 5: Substitute the value of Ksp
We know that:
\[ K_{sp} = 2.7 × 10^{-31} \]
So we can set up the equation:
\[ 27S^4 = 2.7 × 10^{-31} \]

### Step 6: Solve for S
Now, we can solve for \( S \):
\[ S^4 = \frac{2.7 × 10^{-31}}{27} \]
Calculating the right side:
\[ S^4 = 1 × 10^{-32} \]

Now, taking the fourth root:
\[ S = (1 × 10^{-32})^{1/4} \]
\[ S = 10^{-8} \]

### Final Answer
The solubility of `Cr(OH)_(3)` is:
\[ S = 10^{-8} \, \text{mol L}^{-1} \]