From the calculation and data available in the question, the pH of the solution is 2.8 .
The term Kb is used to refer to the base dissociation constant of a solution. Here we have the Kb as 1.7 × 10⁻⁹ but;
Ka = Kw/Kb = 1 * 10^-14/1.7 × 10⁻⁹ = 5.9 * 10^-6
We have to set up the ICE table as follows;
C5H5NH^+(aq) + H2O(l) <-----> C5H5NH(aq) + H3O^+(aq)
I 0.420 0 0
C -x +x +x
E 0.420 - x x x
Ka = [ C5H5NH] [H3O^+]/[C5H5NH^+]
5.9 * 10^-6 = x^2/ 0.420 - x
5.9 * 10^-6 (0.420 - x ) = x^2
2.478 * 10^-6 - 5.9 * 10^-6x = x^2
x^2 + 5.9 * 10^-6x - 2.478 * 10^-6 = 0
x=0.00157 M
Since;
[ C5H5NH] = [H3O^+] = x = 0.00157 M
pH = -log(0.00157 M)
pH = 2.8
Learn more about pKa:https://brainly.com/question/13178964?
