Question:
Determine the \(\mathrm{pH}\) of \(0.024\mathrm{\ M}\) hydroxylamine hydrochloride solution. (\(K_b\) of hydroxyl amine is \(10^{-8}\))
Determine the \(\mathrm{pH}\) of \(0.024\mathrm{\ M}\) hydroxylamine hydrochloride solution. (\(K_b\) of hydroxyl amine is \(10^{-8}\))
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Salt of weak base and strong acid: \([\mathrm{H}^+] = \sqrt{(K_w/K_b) \cdot C}\).
Updated On: Apr 23, 2026
- \(3.4\)
- \(6.2\)
- \(3.8\)
- \(5.6\)
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Formula / Definition}
\[ K_h = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-8}} = 10^{-6} \]
Step 2: Calculation / Simplification}
\([\mathrm{H}^+] = \sqrt{K_h \cdot C} = \sqrt{10^{-6} \times 0.024} = \sqrt{2.4 \times 10^{-8}} = 1.55 \times 10^{-4}\)
\(\mathrm{pH} = -\log(1.55 \times 10^{-4}) = 4 - 0.19 = 3.81 \approx 3.8\)
Step 3: Final Answer
\[ 3.8 \]
\[ K_h = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-8}} = 10^{-6} \]
Step 2: Calculation / Simplification}
\([\mathrm{H}^+] = \sqrt{K_h \cdot C} = \sqrt{10^{-6} \times 0.024} = \sqrt{2.4 \times 10^{-8}} = 1.55 \times 10^{-4}\)
\(\mathrm{pH} = -\log(1.55 \times 10^{-4}) = 4 - 0.19 = 3.81 \approx 3.8\)
Step 3: Final Answer
\[ 3.8 \]
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