If pH of a saturated solution of `Ba(OH)_(2)` is 12, the value of its `K_((SP))` is

Options

A. `5.00 xx 10^(-7) M^(-3)`
B. `4.00 xx 10^(-6) M^(3)`
C. `4.00 xx 10^(-7) M^(3)`
D. `5.00 xx 10^(-6) M^(3)`

Answer

To find the value of \( K_{sp} \) for a saturated solution of \( Ba(OH)_2 \) with a given pH of 12, we can follow these steps:

### Step 1: Determine the concentration of hydroxide ions \([OH^-]\)
The pH of the solution is given as 12. We can find the pOH using the relationship:
\[ pOH = 14 - pH \]
Substituting the given pH:
\[ pOH = 14 - 12 = 2 \]
Now, we can find the concentration of hydroxide ions:
\[ [OH^-] = 10^{-pOH} = 10^{-2} \, \text{M} \]

### Step 2: Relate the concentration of hydroxide ions to solubility
The dissociation of \( Ba(OH)_2 \) in water can be represented as:
\[ Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^- \]
If we let the solubility of \( Ba(OH)_2 \) be \( S \), then:
- The concentration of \( Ba^{2+} \) will be \( S \)
- The concentration of \( OH^- \) will be \( 2S \)

From Step 1, we know:
\[ 2S = [OH^-] = 10^{-2} \, \text{M} \]
Thus, we can solve for \( S \):
\[ S = \frac{10^{-2}}{2} = 5 \times 10^{-3} \, \text{M} \]

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for \( Ba(OH)_2 \) is given by:
\[ K_{sp} = [Ba^{2+}][OH^-]^2 \]
Substituting the values we found:
\[ K_{sp} = S \cdot (2S)^2 \]
Substituting \( S \):
\[ K_{sp} = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the value of \( S \) into the \( K_{sp} \) expression
Now substituting \( S = 5 \times 10^{-3} \):
\[ K_{sp} = 4(5 \times 10^{-3})^3 \]
Calculating \( (5 \times 10^{-3})^3 \):
\[ (5 \times 10^{-3})^3 = 125 \times 10^{-9} = 1.25 \times 10^{-7} \]
Thus,
\[ K_{sp} = 4 \times 1.25 \times 10^{-7} = 5 \times 10^{-7} \]

### Final Answer
The value of \( K_{sp} \) for \( Ba(OH)_2 \) is:
\[ K_{sp} = 5 \times 10^{-7} \]