If the kb of a weak base is 3.3 ã 10-6, what is the ph of a 0.17 m solution of this base?

The relation between base equilibrium constant and degree of dissociation is

C x^2 = Kb

x is degree of dissociation

C x^2 = 3.3 * 10^-6

(0.17) x^2 = 3.3 * 10^-6

x = 4.24 * 10^-3

Concentration of hydroxyl ion is

[OH-] = c x = 0.17 * 4.24 * 10^-3 = 0.72 * 10^-3 M

PoH = - log [OH-] = 2.85

PH = 14 - pOH = 11.15

Hence, pH of solution is 11.15

b.final