The pH of the solution as obtained from the calculations as 2.8.
What is Kb?
The term Kb is the base dissociation constant. We obtain Kb after setting up the ICE table as shown;
C5H5NH^+ (aq) + H2O(l) ⇄ C5H5N(aq) + H3O^+(aq)
I 0.42 0 0
C -x +x +x
E 0.42 - x x x
Since Kb = 1. 7 × 10⁻⁹
Ka = 1 * 10^-14/1. 7 × 10⁻⁹ = 5.88 * 10^-6
Ka = [C5H5N] [H3O^+]/[C5H5NH^+]
5.88 * 10^-6 = x^2/0.42 - x
5.88 * 10^-6 (0.42 - x) = x^2
2.47 * 10^-6 - 5.88 * 10^-6x = x^2
x^2 + 5.88 * 10^-6x - 2.47 * 10^-6 = 0
x=0.0016 M
But x = [C5H5N] = [H3O^+]
pH = - log [H3O^+]
pH = - log(0.0016 M)
pH = 2.8
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