Question

Consider a weak base \(B\) of \(pK_b = 5.699\). \(x\) mL of \(0.02\) M HCl and \(y\) mL of \(0.02\) M weak base \(B\) are mixed to make \(100\) mL of a buffer of pH \(=9\) at \(25^\circ\text{C}\). The values of \(x\) and \(y\) respectively are

• A. \(x=42.7,\ y=57.3\)
• B. \(x=11.1,\ y=88.9\)
• C. \(x=85.7,\ y=14.3\)
• D. \(x=14.3,\ y=85.7\)

Hint:

For basic buffers, always convert \(pK_b\) to \(pK_a\) before using the Henderson–Hasselbalch equation.

Answer 1

The Correct Option is D

Concept: A buffer made of a weak base and its conjugate acid follows the Henderson–Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \log\!\left(\frac{[\text{Base}]}{[\text{Conjugate Acid}]}\right) \] For a base: \[ pK_a + pK_b = 14 \]
Step 1: Calculate \(pK_a\) of the conjugate acid \[ pK_a = 14 - 5.699 = 8.301 \]
Step 2: Apply Henderson–Hasselbalch equation \[ 9 = 8.301 + \log\!\left(\frac{[\text{B}]}{[\text{BH}^+]}\right) \] \[ \log\!\left(\frac{[\text{B}]}{[\text{BH}^+]}\right) = 0.699 \Rightarrow \frac{[\text{B}]}{[\text{BH}^+]} = 5 \]
Step 3: Express mole balance Moles of HCl added: \[ n_{\text{HCl}} = 0.02 \times \frac{x}{1000} \] Moles of base added: \[ n_B = 0.02 \times \frac{y}{1000} \] After neutralisation: \[ \text{BH}^+ = n_{\text{HCl}}, \quad \text{B remaining} = n_B - n_{\text{HCl}} \]
Step 4: Use ratio condition \[ \frac{n_B - n_{\text{HCl}}}{n_{\text{HCl}}} = 5 \Rightarrow \frac{y-x}{x} = 5 \Rightarrow y = 6x \]
Step 5: Use total volume condition \[ x + y = 100 \Rightarrow x + 6x = 100 \Rightarrow x = 14.3 \Rightarrow y = 85.7 \] Final Answer: \[ \boxed{x = 14.3\ \text{mL},\ y = 85.7\ \text{mL}} \]