Given `K_(sp)` for `Cr(OH)_3` is `6 xx 10^(-31)` then determine `[OH^- ]`. (Neglect the contribution of `OH^-` ions from `H_2O`)

Options

A. `(2.22 xx 10^(-31))^(1//4)`
B. `(4.86 xx 10^(-29))^(1//4)`
C. `(18xx10^(-31))^(1//4)`
D. `(18xx10^(-31))^(1//2)`

Answer

To solve the problem of determining the concentration of hydroxide ions \([OH^-]\) from the given solubility product constant \(K_{sp}\) for chromium(III) hydroxide \((Cr(OH)_3)\), we can follow these steps:

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

### Step 2: Define the solubility
Let the solubility of \(Cr(OH)_3\) be \(S\). According to the dissociation equation:
- The concentration of \(Cr^{3+}\) ions will be \(S\).
- The concentration of \(OH^-\) ions will be \(3S\) (since 3 moles of \(OH^-\) are produced for every mole of \(Cr(OH)_3\)).

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

### Step 4: Substitute the given \(K_{sp}\) value
Given that \(K_{sp} = 6 \times 10^{-31}\), we can set up the equation:
\[ 27S^4 = 6 \times 10^{-31} \]

### Step 5: Solve for \(S\)
Rearranging the equation gives:
\[ S^4 = \frac{6 \times 10^{-31}}{27} \]
Calculating the right side:
\[ S^4 = \frac{6}{27} \times 10^{-31} = \frac{2}{9} \times 10^{-31} \]
Calculating \(S\):
\[ S = \left(\frac{2}{9} \times 10^{-31}\right)^{1/4} \]
Calculating the numerical value:
\[ S \approx 1.2209 \times 10^{-8} \text{ M} \]

### Step 6: Calculate \([OH^-]\)
Now, we can find the concentration of hydroxide ions:
\[ [OH^-] = 3S = 3 \times (1.2209 \times 10^{-8}) \approx 3.6627 \times 10^{-8} \text{ M} \]

### Final Answer
Thus, the concentration of hydroxide ions \([OH^-]\) is approximately:
\[ [OH^-] \approx 3.66 \times 10^{-8} \text{ M} \]

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