The `pK_a` of a weak acid HA and `pk_b` of a weak base BOH are 4.60 and 4.80 respectively. The pH of 0.1 M solution of the salt, BA, formed from the acid HA and base BOH is

Options

A. 7.1
B. 9.4
C. 6.9
D. 0.2

Answer

To find the pH of a 0.1 M solution of the salt BA formed from the weak acid HA and weak base BOH, we will follow these steps:

### Step 1: Understand the relationship between pKa, pKb, and pKw
The relationship between the pKa of the weak acid and the pKb of the weak base is given by:
\[ pK_w = pK_a + pK_b \]
where \( pK_w \) is the ion product constant of water, which is 14 at 25°C.

### Step 2: Calculate pKw
Given:
- \( pK_a = 4.60 \)
- \( pK_b = 4.80 \)

We can calculate \( pK_w \):
\[ pK_w = pK_a + pK_b \]
\[ pK_w = 4.60 + 4.80 = 9.40 \]

### Step 3: Calculate the pH of the salt solution
For a salt formed from a weak acid and a weak base, the pH can be calculated using the formula:
\[ pH = \frac{1}{2} (pK_w - (pK_a + pK_b)) \]

Substituting the values we have:
\[ pH = \frac{1}{2} (14 - (4.60 + 4.80)) \]
\[ pH = \frac{1}{2} (14 - 9.40) \]
\[ pH = \frac{1}{2} (4.60) \]
\[ pH = 2.30 \]

### Step 4: Final Calculation
Since we need to find the pH of the salt solution, we can also use the formula:
\[ pH = \frac{1}{2} (pK_a + pK_b) \]
So,
\[ pH = \frac{1}{2} (4.60 + 4.80) \]
\[ pH = \frac{1}{2} (9.40) \]
\[ pH = 4.70 \]

### Conclusion
Thus, the pH of the 0.1 M solution of the salt BA is **4.70**.

---