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

Step 1: Understanding the Concept:
When a weak base and a strong acid react, a basic buffer is formed if the weak base is in excess. The pH of a basic buffer is calculated using the Henderson-Hasselbalch equation.
Step 2: Key Formula or Approach:
\[ pOH = pK_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right) \]
Step 3: Detailed Explanation:
Given \(pH = 9\), so \(pOH = 14 - 9 = 5\).
Reaction: \(B + HCl \rightarrow B^+Cl^-\)
Initial mmoles: \(Base = 0.02y\), \(Acid = 0.02x\)
After reaction: \(mmoles \text{ Salt formed} = 0.02x\), \(mmoles \text{ Base remaining} = 0.02y - 0.02x\)
Using Henderson equation:
\[ 5 = 5.699 + \log\left(\frac{0.02x}{0.02y - 0.02x}\right) \]
\[ -0.699 = \log\left(\frac{x}{y - x}\right) \]
Since \(\log 5 = 0.699\), then \(-0.699 = \log(1/5)\).
\[ \frac{x}{y - x} = \frac{1}{5} \implies 5x = y - x \implies y = 6x \]
We are given total volume \(x + y = 100 \text{ mL}\).
\[ x + 6x = 100 \implies 7x = 100 \implies x \approx 14.3 \text{ mL} \]
\[ y = 100 - 14.3 = 85.7 \text{ mL} \]
Step 4: Final Answer:
The values are \(x = 14.3\) and \(y = 85.7\).