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

Options

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

Explanation

To find the molar solubility of chromium(III) hydroxide, Cr(OH)₃, in water given its solubility product constant (Ksp), we can follow these steps:

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

### Step 2: Define molar solubility
Let the molar solubility of Cr(OH)₃ be \( s \). This means:
- The concentration of Cr³⁺ ions will be \( s \).
- The concentration of OH⁻ ions will be \( 3s \) (since three hydroxide ions are produced for each formula unit of Cr(OH)₃ that dissolves).

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

### Step 4: Simplify the Ksp expression
Now, we can simplify the expression:
\[ K_{sp} = s(27s^3) = 27s^4 \]

### Step 5: Substitute the given Ksp value
We know from the problem that \( K_{sp} = 1.6 \times 10^{-30} \). Therefore, we can set up the equation:
\[ 27s^4 = 1.6 \times 10^{-30} \]

### Step 6: Solve for s
To find \( s \), we rearrange the equation:
\[ s^4 = \frac{1.6 \times 10^{-30}}{27} \]
\[ s^4 = 5.9259 \times 10^{-32} \]

Now, take the fourth root of both sides:
\[ s = \left(5.9259 \times 10^{-32}\right)^{1/4} \]

### Step 7: Calculate the value of s
Using a calculator:
\[ s \approx 1.56 \times 10^{-8} \, \text{mol/L} \]

### Final Answer
The molar solubility of Cr(OH)₃ in water is approximately \( 1.56 \times 10^{-8} \, \text{mol/L} \).

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