From the statement and the information available in the question, the pH of the solution is 12.2.
What is Ka?
The term Ka refers to the acid dissociation constant of a substance, we can obtain this by setting up the ICE table as shown;
C6H5O^- + H2O ⇄ C6H5OH + OH^-
I 0.250 0 0
C -X +X +X
E 0.250 -x x x
Now;
Kb = 1 * 10^-14/ 1.0 x 10-10
Kb = 1 * 10^-4
Kb = [C6H5OH] [OH^-]/[C6H5O^-]
1 * 10^-4 = x^2/ 0.250 -x
1 * 10^-4 (0.250 -x) = x^2
2.5 * 10^-4 - 1 * 10^-4 x = x^2
x^2 + 1 * 10^-4 x - 2.5 * 10^-4 = 0
x=0.016 M
Given that; [C6H5OH] = [OH^-] =x
pOH = -log(0.016 M) = 1.8
pH = 14 - 1.8 = 12.2
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