pH of Ba`(OH)_(2)` Ssolution is 12. Its solubility product is :

Options

A. `10^(-6) M^(3)`
B. `4 xx 10^(-6) M^(3)`
C. `0.5 xx 10^(-7) M^(3)`
D. `5 xx 10^(-7) M^(3)`

Answer

To find the solubility product (Ksp) of barium hydroxide (Ba(OH)₂) given that the pH of the solution is 12, we can follow these steps:

### Step 1: Calculate the pOH
We know that:
\[ \text{pH} + \text{pOH} = 14 \]
Given that the pH is 12, we can calculate the pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 12 = 2 \]

### Step 2: Calculate the concentration of OH⁻ ions
The concentration of hydroxide ions (OH⁻) can be calculated using the pOH:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Thus,
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2} \, \text{M} \]

### Step 3: Write the dissociation equation for Ba(OH)₂
Barium hydroxide dissociates in water as follows:
\[ \text{Ba(OH)}_2 \rightleftharpoons \text{Ba}^{2+} + 2\text{OH}^- \]
From this equation, we can see that for every 1 mole of Ba(OH)₂ that dissolves, it produces 2 moles of OH⁻ ions.

### Step 4: Relate the concentration of Ba²⁺ to the concentration of OH⁻
Let the solubility of Ba(OH)₂ be \( s \) M. Then, the concentration of Ba²⁺ ions will be:
\[ [\text{Ba}^{2+}] = s \]
And since 2 moles of OH⁻ are produced for every mole of Ba(OH)₂:
\[ [\text{OH}^-] = 2s \]

From our previous calculation, we found that:
\[ [\text{OH}^-] = 10^{-2} \, \text{M} \]
Thus,
\[ 2s = 10^{-2} \]
So,
\[ s = \frac{10^{-2}}{2} = 5 \times 10^{-3} \, \text{M} \]

### Step 5: Write the expression for Ksp
The solubility product (Ksp) is given by:
\[ Ksp = [\text{Ba}^{2+}][\text{OH}^-]^2 \]
Substituting the values we found:
\[ Ksp = (5 \times 10^{-3}) \times (10^{-2})^2 \]
\[ Ksp = (5 \times 10^{-3}) \times (10^{-4}) \]
\[ Ksp = 5 \times 10^{-7} \]

### Final Answer
The solubility product (Ksp) of Ba(OH)₂ is:
\[ Ksp = 5 \times 10^{-7} \]

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