The pH of a solution obtained by mixing 5 mL of 0.1 M NH₄OH solution with 250 mL of 0.1 M NH₄Cl solution is _____ × 10⁻². (Nearest integer) Given: $pK_b$ (NH₄OH) = 4.74 log 2 = 0.30 log 3 = 0.48 log 5 = 0.70
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Answer: 756
Key idea: Henderson-Hasselbalch equation for basic buffer (NH₄OH/NH₄Cl system)
- Calculate moles: $n_{NH_4OH} = 0.1 \times 5 = 0.5$ mmol and $n_{NH_4Cl} = 0.1 \times 250 = 25$ mmol
- This is a basic buffer with base (NH₄OH) and its conjugate acid (NH₄⁺ from NH₄Cl)
- Apply Henderson-Hasselbalch: $pOH = pK_b + \log\frac{[Salt]}{[Base]} = 4.74 + \log\frac{25}{0.5}$
- Calculate: $\log(50) = \log(5 \times 10) = 0.70 + 1 = 1.70$
- Find pOH: $pOH = 4.74 + 1.70 = 6.44$
- Calculate pH: $pH = 14 - 6.44 = 7.56 = 756 \times 10^{-2}$