JEE MAIN - Chemistry (2020 - 6th September Morning Slot - No. 16)
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
(B) > (C) > (D) > (A)
(A) > (D) > (C) > (B)
(A) > (C) > (D) > (B)
(B) > (D) > (C) > (A)
Explanation
(A) 10–2 M HCl $$ \Rightarrow $$ [H+] = 10–2 M $$ \Rightarrow $$ pH = 2 $$ \Rightarrow $$ pOH = 14 - 2 = 12
(B) 10–2 M NaOH $$ \Rightarrow $$ [OH–] = 10–2 M $$ \Rightarrow $$ pOH = 2
(C) 10–2 M CH3COO–Na+ $$ \Rightarrow $$ [OH+] > 10–7 $$ \Rightarrow $$ pOH < 7
(D) 10–2 M NaCl $$ \Rightarrow $$ Neutral pOH = 7
Order of pOH value A > D > C > B
(B) 10–2 M NaOH $$ \Rightarrow $$ [OH–] = 10–2 M $$ \Rightarrow $$ pOH = 2
(C) 10–2 M CH3COO–Na+ $$ \Rightarrow $$ [OH+] > 10–7 $$ \Rightarrow $$ pOH < 7
(D) 10–2 M NaCl $$ \Rightarrow $$ Neutral pOH = 7
Order of pOH value A > D > C > B
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