A weak base BOH of concentration 0.02 mole/litre has a pH value of 10.45. If 100 mL of this base is mixed with 10 mL of 0.1 M HCI, what will be the pH of the mixture?

Answer

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

### Step 1: Calculate the concentration of hydroxide ions \([OH^-]\) from the given pH of the weak base.

Given:
- pH = 10.45

To find pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 10.45 = 3.55 \]

Now, calculate \([OH^-]\):
\[ [OH^-] = 10^{-\text{pOH}} = 10^{-3.55} \approx 2.82 \times 10^{-4} \, \text{M} \]

### Step 2: Calculate the concentration of the weak base BOH.

Given the concentration of BOH is 0.02 M and the volume is 100 mL:
\[ \text{Moles of BOH} = \text{Concentration} \times \text{Volume} = 0.02 \, \text{mol/L} \times 0.1 \, \text{L} = 0.002 \, \text{mol} \]

### Step 3: Calculate the moles of HCl added.

Given:
- Concentration of HCl = 0.1 M
- Volume of HCl = 10 mL = 0.01 L

\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.01 \, \text{L} = 0.001 \, \text{mol} \]

### Step 4: Determine the reaction between BOH and HCl.

The reaction can be represented as:
\[ \text{BOH} + \text{HCl} \rightarrow \text{B}^+ + \text{Cl}^- + \text{H}_2\text{O} \]

Initially, we have:
- Moles of BOH = 0.002 mol
- Moles of HCl = 0.001 mol

After the reaction:
- Moles of BOH remaining = 0.002 - 0.001 = 0.001 mol
- Moles of B\(^+\) produced = 0.001 mol

### Step 5: Calculate the total volume of the mixture.

Total volume = Volume of BOH + Volume of HCl
\[ \text{Total Volume} = 100 \, \text{mL} + 10 \, \text{mL} = 110 \, \text{mL} = 0.110 \, \text{L} \]

### Step 6: Calculate the concentrations of BOH and B\(^+\) in the mixture.

Concentration of BOH:
\[ [\text{BOH}] = \frac{0.001 \, \text{mol}}{0.110 \, \text{L}} \approx 0.00909 \, \text{M} \]

Concentration of B\(^+\):
\[ [\text{B}^+] = \frac{0.001 \, \text{mol}}{0.110 \, \text{L}} \approx 0.00909 \, \text{M} \]

### Step 7: Calculate the pH of the resulting buffer solution.

Since we have equal concentrations of BOH and B\(^+\), we can use the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]

Since \([\text{Base}] = [\text{Acid}]\):
\[ \text{pH} = \text{pKa} + \log(1) = \text{pKa} \]

### Step 8: Calculate pKa from pKb.

We can find pKb using the initial pH:
\[ \text{pKb} = 14 - \text{pH} = 14 - 10.45 = 3.55 \]
Thus,
\[ \text{pKa} = 14 - \text{pKb} = 14 - 3.55 = 10.45 \]

### Final Step: Calculate the final pH.

Since the concentrations of BOH and B\(^+\) are equal, the pH of the mixture will be equal to the pKa:
\[ \text{pH} \approx 10.45 \]

### Conclusion:
The pH of the mixture after mixing 100 mL of 0.02 M BOH with 10 mL of 0.1 M HCl is approximately **10.45**.

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