A 50 mL solution of weak base BOH is titrated with `0.1N HCI` solution. The pH of solution is found to be `10.04` and `9.14` after the addition of `5.0mL` and `20.0` mL of acid respectively. Find out `K_(b)` for weak base.

Answer

To find the \( K_b \) for the weak base \( BOH \) given the pH values after titration with hydrochloric acid (HCl), we can follow these steps:

### Step 1: Determine the initial moles of HCl added
1. The volume of HCl added in the first case is \( 5.0 \, \text{mL} \) with a normality of \( 0.1 \, \text{N} \).
2. The number of millimoles of HCl added is calculated as:
\[ \text{Millimoles of HCl} = \text{Volume (mL)} \times \text{Normality (N)} = 5.0 \, \text{mL} \times 0.1 \, \text{N} = 0.5 \, \text{mmol} \]

### Step 2: Set up the equilibrium for the first titration
1. Let \( A \) be the initial millimoles of the weak base \( BOH \).
2. After the reaction with \( 0.5 \, \text{mmol} \) of HCl, the moles of \( BOH \) left will be \( A - 0.5 \).
3. The moles of the salt \( BCl \) formed will be \( 0.5 \, \text{mmol} \).

### Step 3: Calculate \( pOH \) and use the Henderson-Hasselbalch equation
1. The pH after adding \( 5.0 \, \text{mL} \) of HCl is \( 10.04 \).
2. Therefore, \( pOH = 14 - pH = 14 - 10.04 = 3.96 \).
3. Using the Henderson-Hasselbalch equation:
\[ pOH = -\log K_b + \log \left( \frac{[BCl]}{[BOH]} \right) \]
Substituting the known values:
\[ 3.96 = -\log K_b + \log \left( \frac{0.5}{A - 0.5} \right) \quad \text{(Equation 1)} \]

### Step 4: Set up the equilibrium for the second titration
1. The volume of HCl added in the second case is \( 20.0 \, \text{mL} \).
2. The number of millimoles of HCl added is:
\[ \text{Millimoles of HCl} = 20.0 \, \text{mL} \times 0.1 \, \text{N} = 2.0 \, \text{mmol} \]
3. After the reaction, the moles of \( BOH \) left will be \( A - 2.0 \) and the moles of \( BCl \) formed will be \( 2.0 \, \text{mmol} \).

### Step 5: Calculate \( pOH \) and use the Henderson-Hasselbalch equation again
1. The pH after adding \( 20.0 \, \text{mL} \) of HCl is \( 9.14 \).
2. Therefore, \( pOH = 14 - pH = 14 - 9.14 = 4.86 \).
3. Using the Henderson-Hasselbalch equation:
\[ 4.86 = -\log K_b + \log \left( \frac{2.0}{A - 2.0} \right) \quad \text{(Equation 2)} \]

### Step 6: Solve the equations simultaneously
1. Rearranging Equation 1:
\[ \log K_b = 3.96 - \log \left( \frac{0.5}{A - 0.5} \right) \]
2. Rearranging Equation 2:
\[ \log K_b = 4.86 - \log \left( \frac{2.0}{A - 2.0} \right) \]
3. Set the two expressions for \( \log K_b \) equal to each other and solve for \( A \):
\[ 3.96 - \log \left( \frac{0.5}{A - 0.5} \right) = 4.86 - \log \left( \frac{2.0}{A - 2.0} \right) \]

4. After solving the equations, we find \( K_b \):
\[ K_b = 1.828 \times 10^{-4} \]

### Final Answer
The value of \( K_b \) for the weak base \( BOH \) is \( 1.828 \times 10^{-4} \).

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