We are asked to find Kb for C5H5N given that the Ka of its conjugate acid (C5H5NH+) is 5.9x10-6
The dissociation of C5H5NH+ with water will appear as:
C5H5NH+ (aq) + H2O (l) ⇌ C5H5NH (aq) + OH- (aq) Ka = 5.9x10-6
Recall that Ka and Kb are connected by the autoionization constant of water (Kw) by the following equation:
What is Kb for pyridine (C5H5N) if Ka for its conjugate acid is 5.9 x 10-6 at 25°C?
A) 1.7 x 10-9
B) 5.9 x 108
C) 1 x 1014
D) 1.7 x 1019
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