Answer : pH = 11.5
Explanation : Given [tex] K_{b}[/tex] = 1.8 X [tex] 10^{-5} [/tex] and C = 0.563;
So by using the formula of pOH;
pOH = - log [tex][ \sqrt{ K_{b} X C} ][/tex] = -log (3.18 X [tex] 10^{-3} [/tex] X 0.563) = 2.49.
Here, we got pOH = 2.49,
Now, we know pH = 14 - pOH
Therefore, pH = 14 - 2.49 = 11.5
