Question

A weak base with a concentration of 1.3 mol/L has a percent ionization of 0.72%. What is the Kb of this weak base?

could someone help me with this question or how to get started?

Answer 1

Call the weak base BOH.
BOH <==> B^+ + OH^-

Kb = (B^+)(OH^-)/(BOH)

If 0.72% ionized, that means(H^+) must be 1.3 M x 0.0072 = ??
Of course (OH^-) must be the same.
If BOH is 0.72% ionized, it must be 100% - 0.72% unionized.
Substitute into Kb expression and solve for Kb. Post your work if you get stuck.