Calculate pOH, [CH 3 NH 3 + ], [C 4 H 13 N 2 + ] & [C 4 H 14 N 2 2+ ] in an aqueous solution consisting of 0.2 M Methyl amine (K b = 4.1 × 10 –4 ) & 0.1 M Butane-1, 4-diamine (K b1 = 6.2 × 10 –4 M ; K b2 = 2.25 × 10 –5 ).
Text Solution
Verified by Experts1
1.92 ; 
; 
; 
Sol. Assuming α 2 << α 1 & α 2 << α due to common ion effect of OH – from first dissociation of same base (with α 1 ) & dissociation of CH 3 NH 2 (with α ) on second dissociation of base (with α 2 ), we have :
[OH – ] = 
= 
= 
∴ pOH = – log 0.012 ≈ 1.92
Now , K b = 
⇒ 4.1 × 10 –4 = 
[CH 3 
] ⇒ [CH 3 
] = 
× 10 –2 M
Again 
= 
⇒ 6.2 × 10 –4 = 
[C 4 H 13 
] ⇒ [C 4 H 13 
] = 
× 10 –2 M
And 
= 
⇒ 2.25 × 10 –5 = 
⇒ [C 4 H 14 
] = 
× 10 –6 M
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