The `K_(SP)` for `Cr(OH)_(3)` is `1.6xx10^(-30)`. The molar solubility of this compound in water is

Options

A. `2sqrt(1.6xx10^(-30))`
B. `4sqrt(1.6xx10^(-30))`
C. `4sqrt((1.6xx10^(-30))/27)`
D. `(1.6xx10^(-30))/27`

Explanation

To find the molar solubility of \( \text{Cr(OH)}_3 \) in water given its solubility product constant \( K_{sp} = 1.6 \times 10^{-30} \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of chromium(III) hydroxide in water can be represented as:
\[ \text{Cr(OH)}_3 (s) \rightleftharpoons \text{Cr}^{3+} (aq) + 3 \text{OH}^- (aq) \]

### Step 2: Define the solubility
Let the molar solubility of \( \text{Cr(OH)}_3 \) be \( s \) mol/L. When \( \text{Cr(OH)}_3 \) dissolves, it produces:
- \( s \) moles of \( \text{Cr}^{3+} \)
- \( 3s \) moles of \( \text{OH}^- \)

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for the dissociation can be expressed as:
\[ K_{sp} = [\text{Cr}^{3+}][\text{OH}^-]^3 \]
Substituting the concentrations in terms of \( s \):
\[ K_{sp} = (s)(3s)^3 = s(27s^3) = 27s^4 \]

### Step 4: Substitute the value of \( K_{sp} \)
Now, substituting the given \( K_{sp} \) value:
\[ 1.6 \times 10^{-30} = 27s^4 \]

### Step 5: Solve for \( s^4 \)
Rearranging the equation gives:
\[ s^4 = \frac{1.6 \times 10^{-30}}{27} \]

### Step 6: Calculate \( s^4 \)
Calculating the right-hand side:
\[ s^4 = \frac{1.6 \times 10^{-30}}{27} \approx 5.9259 \times 10^{-32} \]

### Step 7: Find \( s \)
To find \( s \), we take the fourth root:
\[ s = \left(5.9259 \times 10^{-32}\right)^{1/4} \]

### Step 8: Calculate \( s \)
Calculating the fourth root:
\[ s \approx 1.6 \times 10^{-8} \, \text{mol/L} \]

### Final Answer
The molar solubility of \( \text{Cr(OH)}_3 \) in water is approximately \( 1.6 \times 10^{-8} \, \text{mol/L} \).

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