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. `(18 xx 10^(-31))^(1/4) M`
B. `(18 xx 10^(-31))^(1/2) M`
C. `(6 xx 10^(-31))^(1/4) M`
D. `(27 xx 10^(-31))^(1/4) M`

Answer

To determine the concentration of hydroxide ions \([OH^-]\) from the solubility product constant \(K_{sp}\) of 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:
\[ Cr(OH)_3 (s) \rightleftharpoons Cr^{3+} (aq) + 3 OH^- (aq) \]

### Step 2: Define solubility
Let the solubility of \(Cr(OH)_3\) be \(S\). From the dissociation equation, we can see that for every mole of \(Cr(OH)_3\) that dissolves, it produces:
- 1 mole of \(Cr^{3+}\)
- 3 moles of \(OH^-\)

Thus, at equilibrium:
\[ [Cr^{3+}] = S \quad \text{and} \quad [OH^-] = 3S \]

### Step 3: Write the expression for \(K_{sp}\)
The solubility product constant \(K_{sp}\) for the dissociation of \(Cr(OH)_3\) is given by:
\[ K_{sp} = [Cr^{3+}][OH^-]^3 \]
Substituting the expressions for the concentrations:
\[ K_{sp} = S \cdot (3S)^3 = S \cdot 27S^3 = 27S^4 \]

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

### Step 5: Solve for \(S\)
Rearranging gives:
\[ S^4 = \frac{6 \times 10^{-31}}{27} \]
Calculating the right-hand side:
\[ S^4 = \frac{6}{27} \times 10^{-31} = \frac{2}{9} \times 10^{-31} \]

Now, taking the fourth root:
\[ S = \left(\frac{2}{9} \times 10^{-31}\right)^{1/4} \]

### Step 6: Calculate \([OH^-]\)
Since \([OH^-] = 3S\), we can substitute \(S\) into this equation:
\[ [OH^-] = 3 \left(\frac{2}{9} \times 10^{-31}\right)^{1/4} \]

### Step 7: Simplify the expression
Calculating the value:
\[ [OH^-] = 3 \cdot \left(\frac{2^{1/4}}{3^{1/2}}\right) \cdot 10^{-31/4} \]
This can be simplified to:
\[ [OH^-] = 3 \cdot \frac{2^{1/4}}{3^{1/2}} \cdot 10^{-31/4} \]

### Final Result
The concentration of hydroxide ions \([OH^-]\) is:
\[ [OH^-] = 3 \cdot \frac{2^{1/4}}{3^{1/2}} \cdot 10^{-31/4} \text{ mol/L} \]