A cetrain ion `B^(-)` has an Arrhenius constant for basic character (eq. constnat `2.8xx10^(-7)`). The equilibrium constant for Lowry-Bronsed basic character is:

Options

A. `2.8 xx10^(-7)`
B. `3.57 xx10^(-8)`
C. `3.57 xx10^(8)`
D. `2.8 xx10^(7)`

Explanation

To find the equilibrium constant for the Lowry-Brønsted basic character of the ion \( B^{-} \), we can follow these steps:

### Step 1: Understand the Given Information
We are provided with the Arrhenius constant for the basic character of the ion \( B^{-} \), which is given as:
\[ K_b = 2.8 \times 10^{-7} \]

### Step 2: Recall the Relationship Between \( K_b \) and \( K_{eq} \)
For a base \( B^{-} \) reacting with water, the equilibrium reaction can be represented as:
\[ B^{-} + H_2O \rightleftharpoons HB + OH^{-} \]
The equilibrium constant \( K_{eq} \) for this reaction can be expressed in terms of \( K_b \) and \( K_w \) (the ion product of water):
\[ K_{eq} = \frac{K_b}{K_w} \]

### Step 3: Identify the Value of \( K_w \)
The value of \( K_w \) at 25°C is known to be:
\[ K_w = 1.0 \times 10^{-14} \]

### Step 4: Substitute the Values into the Equation
Now, we can substitute the values of \( K_b \) and \( K_w \) into the equation for \( K_{eq} \):
\[ K_{eq} = \frac{2.8 \times 10^{-7}}{1.0 \times 10^{-14}} \]

### Step 5: Perform the Calculation
Calculating the above expression:
\[ K_{eq} = 2.8 \times 10^{7} \]

### Step 6: Conclusion
Thus, the equilibrium constant for the Lowry-Brønsted basic character of the ion \( B^{-} \) is:
\[ K_{eq} = 2.8 \times 10^{7} \]

### Final Answer
The equilibrium constant for the Lowry-Brønsted basic character is \( 2.8 \times 10^{7} \).

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