The solubility product of `Cr(OH)_(3)` at 298 K is `6.0xx10^(-31)`. The concentration of hydroxide ions in a saturated solution of `Cr(OH)_(3)` will be :

Options

A. `(18xx10^(-31))^(1//4)`
B. `(4.86xx10^(-29))^(1//4)`
C. `(2.22xx10^(-31))^(1//4)`
D. `(18xx10^(-31))^(1//2)`

Explanation

To find the concentration of hydroxide ions \([OH^-]\) in a saturated solution of chromium hydroxide \((Cr(OH)_3)\) given its solubility product \((K_{sp})\) at 298 K, we can follow these steps:

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

### Step 2: Define the solubility
Let the solubility of \(Cr(OH)_3\) be \(S\) mol/L. From the dissociation equation, we can see that for every 1 mole of \(Cr(OH)_3\) that dissolves, it produces 1 mole of \(Cr^{3+}\) ions and 3 moles of \(OH^-\) ions. Therefore:
- The concentration of \(Cr^{3+}\) ions in solution will be \(S\).
- The concentration of \(OH^-\) ions in solution will be \(3S\).

### Step 3: Write the expression for \(K_{sp}\)
The solubility product \(K_{sp}\) for the dissociation of \(Cr(OH)_3\) can be expressed as:
\[ K_{sp} = [Cr^{3+}][OH^-]^3 \]
Substituting the concentrations from Step 2:
\[ K_{sp} = S \cdot (3S)^3 \]
This simplifies to:
\[ K_{sp} = S \cdot 27S^3 = 27S^4 \]

### Step 4: Substitute the value of \(K_{sp}\)
Given that \(K_{sp} = 6.0 \times 10^{-31}\), we can set up the equation:
\[ 27S^4 = 6.0 \times 10^{-31} \]

### Step 5: Solve for \(S\)
To find \(S\), we rearrange the equation:
\[ S^4 = \frac{6.0 \times 10^{-31}}{27} \]
Calculating the right side:
\[ S^4 = \frac{6.0}{27} \times 10^{-31} = 0.2222 \times 10^{-31} \]
Now, we take the fourth root:
\[ S = \left(0.2222 \times 10^{-31}\right)^{1/4} \]

### Step 6: Calculate \(S\)
Calculating \(S\):
\[ S \approx 0.22 \times 10^{-7.75} \approx 0.22 \times 10^{-8} \approx 2.2 \times 10^{-8} \]

### Step 7: Find the concentration of \(OH^-\)
Now, since the concentration of hydroxide ions \([OH^-]\) is \(3S\):
\[ [OH^-] = 3S = 3 \times 2.2 \times 10^{-8} \approx 6.6 \times 10^{-8} \text{ mol/L} \]

### Final Answer
The concentration of hydroxide ions in a saturated solution of \(Cr(OH)_3\) is approximately:
\[ [OH^-] \approx 18 \times 10^{-6} \text{ mol/L} \]