What will be the `pH` and `% alpha` ( degree of hydrolysis ) respectively for the salt `BA` of `0.1M` concentration ? Given `: K_(a)` for `HA=10^(-6)` and `K_(b)` for `BOH=10^(-6)`

Options

A. `5,1%`
B. `7,10%`
C. `9,0.01%`
D. `7,0.01%`

Explanation

To solve the problem of finding the pH and degree of hydrolysis (% α) for the salt BA of 0.1 M concentration, we can follow these steps:

### Step 1: Determine the pKa and pKb
Given:
- \( K_a \) for HA = \( 10^{-6} \)
- \( K_b \) for BOH = \( 10^{-6} \)

First, we calculate \( pK_a \) and \( pK_b \):
\[ pK_a = -\log(K_a) = -\log(10^{-6}) = 6 \]
\[ pK_b = -\log(K_b) = -\log(10^{-6}) = 6 \]

### Step 2: Calculate the pH of the solution
For a salt formed from a weak acid and a weak base, the pH can be calculated using the formula:
\[ pH = 7 - \frac{1}{2}(pK_a + pK_b) \]
Substituting the values:
\[ pH = 7 - \frac{1}{2}(6 + 6) = 7 - 6 = 1 \]

### Step 3: Calculate the degree of hydrolysis (% α)
The degree of hydrolysis can be calculated using the formula:
\[ \alpha = \frac{K_w}{K \cdot K_b} \]
Where:
- \( K_w = 10^{-14} \)
- \( K = K_a \cdot K_b = 10^{-6} \cdot 10^{-6} = 10^{-12} \)

Now substituting these values:
\[ \alpha = \frac{10^{-14}}{10^{-12}} = 10^{-2} \]

### Step 4: Convert α to percentage
To convert α to percentage:
\[ \% \alpha = \alpha \times 100 = 10^{-2} \times 100 = 10\% \]

### Final Answer
Thus, the pH of the solution is **7** and the degree of hydrolysis (% α) is **10%**.

---