Question

100 mL of 0.1 M HA (weak acid) and 100 mL of 0.2 M NaA are mixed. What is the pH of the resultant solution? \((K_a \text{ of HA is } 10^{-5}, \log 2 = 0.3)\)

• A. \( 4.7 \)
• B. \( 5.0 \)
• C. \( 5.3 \)
• D. \( 4.0 \)

Hint:

For buffer solutions, use the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \] If the salt concentration is twice that of the acid, pH increases by \( \log 2 \approx 0.3 \).

Answer 1

The Correct Option is C

Step 1: Identify the Buffer System 
- The given solution consists of a weak acid (\( HA \)) and its salt (\( NaA \)), forming a buffer solution. 
- The pH of a buffer solution is given by the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \] 

Step 2: Calculate pKa 
- Given \( K_a = 10^{-5} \), we calculate: \[ \text{pKa} = -\log (10^{-5}) = 5 \] 

Step 3: Calculate the Concentrations 
- Moles of HA (Acid): \[ \text{Moles} = M \times V = (0.1 \times 0.1) = 0.01 \] - Moles of NaA (Salt): \[ \text{Moles} = M \times V = (0.2 \times 0.1) = 0.02 \] - Total Volume after Mixing: \[ V_{\text{total}} = 100 + 100 = 200 \text{ mL} = 0.2 \text{ L} \] - Final Concentrations: \[ [\text{HA}] = \frac{0.01}{0.2} = 0.05 \quad , \quad [\text{NaA}] = \frac{0.02}{0.2} = 0.1 \] 

Step 4: Apply the Henderson-Hasselbalch Equation \[ \text{pH} = 5 + \log \left( \frac{0.1}{0.05} \right) \] \[ = 5 + \log 2 \] \[ = 5 + 0.3 \] \[ = 5.3 \]

Final Answer: The pH of the buffer solution is \( 5.3 \), which matches Option (3).