The pH of a solution obtained by mixing 5 mL of 0.1 M $NH_4OH$ solution with 250 mL of 0.1 M $NH_4Cl$ solution is ______ × 10⁻². (Nearest integer)
Given: $pK_b (NH_4OH)$ = 4.74
Given: $pK_b (NH_4OH)$ = 4.74
Correct Answer: 756
Solution and Explanation
Step 1: Understanding the Concept:
A mixture of a weak base ($NH_4OH$) and its salt ($NH_4Cl$) forms a basic buffer. The $pOH$ is calculated using the Henderson-Hasselbalch equation, and then converted to $pH$.
Step 2: Key Formula or Approach:
1. $pOH = pK_b + \log\left(\frac{[\text{Salt}]}{[\text{Base}]}\right)$ 2. $pH = 14 - pOH$
Step 3: Detailed Explanation:
1. Moles of Base ($NH_4OH$) = $5 \times 0.1 = 0.5 \text{ mmol}$ 2. Moles of Salt ($NH_4Cl$) = $250 \times 0.1 = 25 \text{ mmol}$ 3. Calculate $pOH$: \[ pOH = 4.74 + \log\left(\frac{25}{0.5}\right) = 4.74 + \log(50) \] \[ pOH = 4.74 + 1.699 = 6.439 \] 4. Calculate $pH$: \[ pH = 14 - 6.439 = 7.561 \] 5. Final value: $7.561 \times 100 = 756 \times 10^{-2}$.
Step 4: Final Answer:
The value is 756.
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