JEE Main 2020 — Ionic Equilibrium Question with Solution
From: JEE Main 2020 (Online) 6th September Morning Slot
Question
Arrange the following solutions in the
decreasing order of pOH :
(A) 0.01 M HCl
(B) 0.01 M NaOH
(C) 0.01 M CH3COONa
(D) 0.01 M NaCl
(A) 0.01 M HCl
(B) 0.01 M NaOH
(C) 0.01 M CH3COONa
(D) 0.01 M NaCl
Choose an option
Show full solutionCorrect option: B
Correct answer
B(A) > (D) > (C) > (B)
Step-by-step explanation
(A) 10–2 M HCl [H+] = 10–2 M pH = 2 pOH = 14 - 2 = 12
(B) 10–2 M NaOH [OH–] = 10–2 M pOH = 2
(C) 10–2 M CH3COO–Na+ [OH+] > 10–7 pOH < 7
(D) 10–2 M NaCl Neutral pOH = 7
Order of pOH value A > D > C > B
(B) 10–2 M NaOH [OH–] = 10–2 M pOH = 2
(C) 10–2 M CH3COO–Na+ [OH+] > 10–7 pOH < 7
(D) 10–2 M NaCl Neutral pOH = 7
Order of pOH value A > D > C > B
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