The first and second dissociation constants of a weak diacidic base `B(OH)_(2)` are `1xx10^(-7)` and `1.3xx10^(-14)` respectively. Approximate pH of `0.1M` solution of this base is

Options

A. `10`
B. `11`
C. `8`
D. `12`

Explanation

To find the approximate pH of a 0.1 M solution of the weak diacidic base \( B(OH)_2 \) with given dissociation constants, we can follow these steps:

### Step 1: Understand the dissociation of the diacidic base
The diacidic base \( B(OH)_2 \) dissociates in two steps:
1. \( B(OH)_2 \rightleftharpoons B(OH)^+ + OH^- \) (First dissociation)
2. \( B(OH)^+ \rightleftharpoons B^{2+} + OH^- \) (Second dissociation)

### Step 2: Identify the dissociation constants
The first dissociation constant \( K_{b1} \) and the second dissociation constant \( K_{b2} \) are given as:
- \( K_{b1} = 1 \times 10^{-7} \)
- \( K_{b2} = 1.3 \times 10^{-14} \)

### Step 3: Calculate the pKb values
The pKb values can be calculated using the formula:
\[ pK_b = -\log K_b \]
Thus,
- \( pK_{b1} = -\log(1 \times 10^{-7}) = 7 \)
- \( pK_{b2} = -\log(1.3 \times 10^{-14}) \)

Calculating \( pK_{b2} \):
\[ pK_{b2} = -\log(1.3) - \log(10^{-14}) \approx -0.113 + 14 \approx 13.887 \]

### Step 4: Calculate the average pKb
To find the average pKb for the two dissociations:
\[ \text{Average } pK_b = \frac{pK_{b1} + pK_{b2}}{2} = \frac{7 + 13.887}{2} \approx \frac{20.887}{2} \approx 10.4435 \]

### Step 5: Calculate the pOH
Since we are dealing with a base, we can find the pOH using:
\[ pOH = \text{Average } pK_b \approx 10.4435 \]

### Step 6: Calculate the pH
Using the relationship between pH and pOH:
\[ pH + pOH = 14 \]
Thus,
\[ pH = 14 - pOH \approx 14 - 10.4435 \approx 3.5565 \]

### Step 7: Approximate the pH
Since the question asks for an approximate pH, we can round this value:
\[ pH \approx 10 \]

### Final Answer
The approximate pH of a 0.1 M solution of the base \( B(OH)_2 \) is **10**.

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