At `25^@C, K_b` for BOH=`1.0xx10^(-12)`.0.01 M solution of BOH has `[OH^-]` :

Options

A. `1.0xx10^(-6) M`
B. `1.0xx10^(-7) M`
C. `1.0xx10^(-5) M`
D. `2.0xx10^(-6) M`

Answer

To find the concentration of hydroxide ions \([OH^-]\) in a 0.01 M solution of BOH at 25°C, given that \(K_b\) for BOH is \(1.0 \times 10^{-12}\), we can follow these steps:

### Step 1: Write the dissociation equation
BOH dissociates in water as follows:
\[ \text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^- \]

### Step 2: Set up the initial and equilibrium concentrations
Let the initial concentration of BOH be \(C = 0.01 \, \text{M}\). At equilibrium, if \( \alpha \) is the degree of dissociation, the concentrations will be:
- \([BOH] = 0.01(1 - \alpha)\)
- \([B^+] = 0.01\alpha\)
- \([OH^-] = 0.01\alpha\)

### Step 3: Write the expression for \(K_b\)
The expression for \(K_b\) is given by:
\[ K_b = \frac{[B^+][OH^-]}{[BOH]} \]
Substituting the equilibrium concentrations into the \(K_b\) expression:
\[ K_b = \frac{(0.01\alpha)(0.01\alpha)}{0.01(1 - \alpha)} = \frac{0.01\alpha^2}{0.01(1 - \alpha)} = \frac{\alpha^2}{1 - \alpha} \]

### Step 4: Substitute the known value of \(K_b\)
Given \(K_b = 1.0 \times 10^{-12}\), we can set up the equation:
\[ 1.0 \times 10^{-12} = \frac{\alpha^2}{1 - \alpha} \]

### Step 5: Assume \(\alpha\) is small
Since \(K_b\) is very small, we can assume that \(\alpha\) is small compared to 1, thus \(1 - \alpha \approx 1\):
\[ 1.0 \times 10^{-12} \approx \alpha^2 \]

### Step 6: Solve for \(\alpha\)
Taking the square root of both sides:
\[ \alpha = \sqrt{1.0 \times 10^{-12}} = 1.0 \times 10^{-6} \]

### Step 7: Calculate \([OH^-]\)
Now, we can find the concentration of hydroxide ions:
\[ [OH^-] = 0.01 \times \alpha = 0.01 \times 1.0 \times 10^{-6} = 1.0 \times 10^{-8} \, \text{M} \]

### Final Answer
Thus, the concentration of hydroxide ions \([OH^-]\) in the solution is:
\[ [OH^-] = 1.0 \times 10^{-8} \, \text{M} \]

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