Question:
Calculate the pOH of \(0.01\) M monobasic acid that is completely dissociated, at \(298\) K.
Calculate the pOH of \(0.01\) M monobasic acid that is completely dissociated, at \(298\) K.
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Complete dissociation gives [H+] = 0.01 M, so pH = 2 and pOH = 14 - pH.
Updated On: Oct 1, 2026
- \(02\)
- \(12\)
- \(03\)
- \(11\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
A monobasic acid gives one \(\text{H}^+\) per molecule. If it dissociates completely, the hydrogen ion concentration equals the acid concentration.
Step 2: Key Formulas:
\[ \text{pH} = -\log[\text{H}^+],\quad \text{pH} + \text{pOH} = 14 \text{ at } 298 \text{ K} \]
Step 3: Detailed Explanation:
\([\text{H}^+] = 0.01 = 10^{-2}\) M, so \(\text{pH} = 2\).
Then \(\text{pOH} = 14 - 2 = 12\).
Check: \([\text{OH}^-] = K_w/[\text{H}^+] = 10^{-14}/10^{-2} = 10^{-12}\) M, so \(\text{pOH} = 12\).
Step 4: Why the other options are wrong.
The value 02 (A) is the pH, not the pOH, a common slip. Value 03 (C) and 11 (D) would need acid concentrations of 0.001 M or 0.0001 M, which is not the case here.
Final Answer:
The pOH is 12, option (B). \[ \boxed{\text{pOH} = 12} \]
A monobasic acid gives one \(\text{H}^+\) per molecule. If it dissociates completely, the hydrogen ion concentration equals the acid concentration.
Step 2: Key Formulas:
\[ \text{pH} = -\log[\text{H}^+],\quad \text{pH} + \text{pOH} = 14 \text{ at } 298 \text{ K} \]
Step 3: Detailed Explanation:
\([\text{H}^+] = 0.01 = 10^{-2}\) M, so \(\text{pH} = 2\).
Then \(\text{pOH} = 14 - 2 = 12\).
Check: \([\text{OH}^-] = K_w/[\text{H}^+] = 10^{-14}/10^{-2} = 10^{-12}\) M, so \(\text{pOH} = 12\).
Step 4: Why the other options are wrong.
The value 02 (A) is the pH, not the pOH, a common slip. Value 03 (C) and 11 (D) would need acid concentrations of 0.001 M or 0.0001 M, which is not the case here.
Final Answer:
The pOH is 12, option (B). \[ \boxed{\text{pOH} = 12} \]
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