When a 20 mL of 0.08 M weak base BOH is titrated with 0.08 M HCl, the pH of the solution at the end point is 5. What will be the pOH if 10 mL of 0.04 M NaOH is added to the resulting solution?
`[Given : log 2=0.30 and log 3=0.48]`

Options

A. `5.40`
B. `5.88`
C. `4.92`
D. None of these

Answer

To solve the problem step-by-step, we will follow these steps:

### Step 1: Calculate the moles of weak base BOH
Given:
- Volume of BOH = 20 mL = 0.020 L
- Concentration of BOH = 0.08 M

Using the formula for moles:
\[ \text{Moles of BOH} = \text{Volume (L)} \times \text{Concentration (M)} = 0.020 \, \text{L} \times 0.08 \, \text{mol/L} = 0.0016 \, \text{mol} \]

### Step 2: Calculate the moles of HCl
Given:
- Concentration of HCl = 0.08 M
- We need to find the volume of HCl at the endpoint.

Using the equivalence point condition (N1V1 = N2V2):
\[ \text{Moles of HCl} = \text{Moles of BOH} = 0.0016 \, \text{mol} \]
\[ \text{Volume of HCl} = \frac{\text{Moles}}{\text{Concentration}} = \frac{0.0016 \, \text{mol}}{0.08 \, \text{mol/L}} = 0.020 \, \text{L} = 20 \, \text{mL} \]

### Step 3: Determine the total volume after titration
The total volume after mixing BOH and HCl:
\[ \text{Total Volume} = 20 \, \text{mL (BOH)} + 20 \, \text{mL (HCl)} = 40 \, \text{mL} \]

### Step 4: Calculate the concentration of BCl formed
At the endpoint, all BOH reacts with HCl to form BCl:
\[ \text{Concentration of BCl} = \frac{\text{Moles of BCl}}{\text{Total Volume}} = \frac{0.0016 \, \text{mol}}{0.040 \, \text{L}} = 0.04 \, \text{M} \]

### Step 5: Use the given pH to find pKb
Given that the pH at the endpoint is 5:
\[ \text{pH} = 5 \]
Using the relationship for weak base and strong acid titration:
\[ \text{pH} = \frac{1}{2} (pK_w - pK_b) - \log[C] \]
Where \( C = 0.04 \, \text{M} \):
\[ \text{pH} = 5 = \frac{1}{2} (14 - pK_b) - \log(0.04) \]
Calculating \( \log(0.04) \):
\[ \log(0.04) = \log(4 \times 10^{-2}) = \log(4) - 2 = 0.60 - 2 = -1.40 \]
Substituting this back:
\[ 5 = \frac{1}{2} (14 - pK_b) + 1.40 \]
\[ 5 - 1.40 = \frac{1}{2} (14 - pK_b) \]
\[ 3.60 = \frac{1}{2} (14 - pK_b) \]
Multiplying both sides by 2:
\[ 7.20 = 14 - pK_b \]
\[ pK_b = 14 - 7.20 = 6.80 \]

### Step 6: Adding NaOH to the resulting solution
Now, we add 10 mL of 0.04 M NaOH:
\[ \text{Moles of NaOH} = 0.010 \, \text{L} \times 0.04 \, \text{mol/L} = 0.0004 \, \text{mol} \]

### Step 7: Calculate the new concentrations after adding NaOH
The total volume after adding NaOH:
\[ \text{Total Volume} = 40 \, \text{mL} + 10 \, \text{mL} = 50 \, \text{mL} \]

Calculating moles of BCl remaining:
\[ \text{Moles of BCl} = 0.0016 \, \text{mol} - 0.0004 \, \text{mol} = 0.0012 \, \text{mol} \]

Calculating concentrations:
\[ \text{Concentration of BCl} = \frac{0.0012 \, \text{mol}}{0.050 \, \text{L}} = 0.024 \, \text{M} \]
\[ \text{Concentration of BOH} = \frac{0.0004 \, \text{mol}}{0.050 \, \text{L}} = 0.008 \, \text{M} \]

### Step 8: Calculate pOH using Henderson-Hasselbalch equation
Using the Henderson-Hasselbalch equation:
\[ \text{pOH} = pK_b + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
Substituting the values:
\[ \text{pOH} = 6.80 + \log \left( \frac{0.024}{0.008} \right) = 6.80 + \log(3) = 6.80 + 0.48 = 7.28 \]

### Final Answer
\[ \text{pOH} = 7.28 \]