The pH of a weak mono acidic base, neutralized upto 80% with a strong acid in a dilute solution, is 7.40. The ionization constant of the base is

Options

A. `1.0xx10^(-5)`
B. `1.6xx10^(-7)`
C. `1.0xx10^(-6)`
D. None of these

Explanation

To solve the problem, we need to find the ionization constant (Kb) of a weak monoacidic base that has been neutralized 80% with a strong acid, resulting in a pH of 7.40.

### Step-by-Step Solution:

1. **Understanding the Neutralization**:
- We have a weak base (BOH) and a strong acid (AH). When they react, they form a salt (BA) and water.
- Since the base is neutralized 80%, it means that 80% of the base has reacted with the acid.

2. **Calculating Remaining Base and Formed Salt**:
- If we assume we started with 100 moles of the weak base, then after 80% neutralization:
- Moles of base remaining = 100 - 80 = 20 moles
- Moles of salt formed = 80 moles

3. **Finding pOH**:
- Given the pH = 7.40, we can find the pOH using the relationship:
\[ \text{pOH} = 14 - \text{pH} = 14 - 7.40 = 6.60 \]

4. **Using the Henderson-Hasselbalch Equation**:
- For a basic buffer solution, we can use the equation:
\[ \text{pOH} = \text{pKb} + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
- Here, [Salt] = 80 moles and [Base] = 20 moles.
- Therefore, the ratio of salt to base is:
\[ \frac{[\text{Salt}]}{[\text{Base}]} = \frac{80}{20} = 4 \]

5. **Calculating the Logarithm**:
- Now, we can substitute this into the equation:
\[ \text{pOH} = \text{pKb} + \log(4) \]
- We know that \(\log(4) = 0.6\), so:
\[ 6.60 = \text{pKb} + 0.6 \]

6. **Solving for pKb**:
- Rearranging gives:
\[ \text{pKb} = 6.60 - 0.6 = 6.00 \]

7. **Finding Kb**:
- To find Kb, we use the relationship:
\[ \text{Kb} = 10^{-\text{pKb}} = 10^{-6.00} = 1.00 \times 10^{-6} \]

### Final Answer:
The ionization constant (Kb) of the weak monoacidic base is:
\[ \text{Kb} = 1.00 \times 10^{-6} \]