Q17 / 150 · MHT CET 2026 (20 April Shift 1) · Chemistry
Calculate the pOH of $0.01$ M monobasic acid that is completely dissociated, at $298$ K.
- A$02$
- Correct answerB$12$
- C$03$
- D$11$
Solution
For a completely dissociated monobasic acid, the concentration of hydrogen ions is equal to the concentration of the acid.
$[H^+] = 0.01$ M $= 10^{-2}$ M
The pH of the solution is calculated as:
$\text{pH} = -\log[H^+] = -\log(10^{-2}) = 2$
At $298$ K, the relationship between pH and pOH is:
$\text{pH} + \text{pOH} = 14$
Substituting the value of pH:
$\text{pOH} = 14 - 2 = 12$
Answer: $12$
$[H^+] = 0.01$ M $= 10^{-2}$ M
The pH of the solution is calculated as:
$\text{pH} = -\log[H^+] = -\log(10^{-2}) = 2$
At $298$ K, the relationship between pH and pOH is:
$\text{pH} + \text{pOH} = 14$
Substituting the value of pH:
$\text{pOH} = 14 - 2 = 12$
Answer: $12$
Hence, the correct answer is option: 'B'.