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 in a saturated solution of \( \text{Cr(OH)}_3 \), 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 product (Ksp)
The solubility product \( K_{sp} \) for the dissociation reaction is given by:
\[ K_{sp} = [\text{Cr}^{3+}][\text{OH}^-]^3 \]
Where \( [\text{Cr}^{3+}] \) is the concentration of chromium ions and \( [\text{OH}^-] \) is the concentration of hydroxide ions.

### Step 3: Express concentrations in terms of solubility (s)
Let the solubility of \( \text{Cr(OH)}_3 \) be \( s \). Then:
- The concentration of \( \text{Cr}^{3+} \) ions will be \( s \).
- The concentration of \( \text{OH}^- \) ions will be \( 3s \) (since 3 moles of hydroxide ions are produced for each mole of chromium(III) hydroxide that dissolves).

### Step 4: Substitute into the Ksp expression
Substituting these values into the \( K_{sp} \) expression gives:
\[ K_{sp} = [s][3s]^3 = s(27s^3) = 27s^4 \]

### Step 5: Set Ksp equal to the given value
We know from the problem statement that:
\[ K_{sp} = 6.0 \times 10^{-31} \]
Thus, we can set up the equation:
\[ 27s^4 = 6.0 \times 10^{-31} \]

### Step 6: Solve for s
To find \( s \), we rearrange the equation:
\[ s^4 = \frac{6.0 \times 10^{-31}}{27} \]
Calculating the right side:
\[ s^4 = \frac{6.0 \times 10^{-31}}{27} = 2.222 \times 10^{-32} \]
Now, take the fourth root:
\[ s = \left(2.222 \times 10^{-32}\right)^{1/4} \]

### Step 7: Calculate the concentration of hydroxide ions
Since the concentration of hydroxide ions \( [\text{OH}^-] = 3s \):
\[ [\text{OH}^-] = 3 \left(2.222 \times 10^{-32}\right)^{1/4} \]

### Step 8: Final calculation
Calculating \( s \) and then \( [\text{OH}^-] \):
1. Calculate \( s \):
\[ s \approx 1.28 \times 10^{-8} \text{ M} \]
2. Now calculate \( [\text{OH}^-] \):
\[ [\text{OH}^-] = 3 \times 1.28 \times 10^{-8} \approx 3.84 \times 10^{-8} \text{ M} \]

Thus, the concentration of hydroxide ions in a saturated solution of \( \text{Cr(OH)}_3 \) is approximately \( 3.84 \times 10^{-8} \text{ M} \).