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 its `Ksp` value of `2.7 x 10^(-31) M^(4)`, we can follow these steps:

### Step-by-Step Solution:

1. **Dissociation of Chromium Hydroxide**:
The dissociation of chromium hydroxide in water can be represented as:
\[ Cr(OH)_3 (s) \rightleftharpoons Cr^{3+} (aq) + 3OH^{-} (aq) \]
From this equation, we see that one mole of `Cr(OH)_(3)` produces one mole of `Cr^{3+}` and three moles of `OH^{-}`.

2. **Define Molar Solubility**:
Let the molar solubility of `Cr(OH)_(3)` be `S` (in mol/L). Therefore:
- The concentration of `Cr^{3+}` ions will be `S`.
- The concentration of `OH^{-}` ions will be `3S` (since three hydroxide ions are produced).

3. **Write the Expression for Ksp**:
The solubility product constant (`Ksp`) for `Cr(OH)_(3)` can be expressed as:
\[ Ksp = [Cr^{3+}][OH^{-}]^3 \]
Substituting the concentrations in terms of `S`, we get:
\[ Ksp = S \cdot (3S)^3 \]
Simplifying this gives:
\[ Ksp = S \cdot 27S^3 = 27S^4 \]

4. **Substitute the Given Ksp Value**:
We know that `Ksp = 2.7 x 10^{-31}`. Therefore, we can set up the equation:
\[ 27S^4 = 2.7 \times 10^{-31} \]

5. **Solve for S**:
Rearranging the equation to solve for `S`:
\[ S^4 = \frac{2.7 \times 10^{-31}}{27} \]
Calculating the right side:
\[ S^4 = \frac{2.7}{27} \times 10^{-31} = 0.1 \times 10^{-31} = 1.0 \times 10^{-32} \]

6. **Taking the Fourth Root**:
Now, we need to find `S` by taking the fourth root:
\[ S = (1.0 \times 10^{-32})^{1/4} \]
This simplifies to:
\[ S = 10^{-32/4} = 10^{-8} \]

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

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