Post a New Question

chemistry

posted by .

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?

  • chemistry -

    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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question