Ammonia,is a weak base with a Ka value of 1.8x10^-5 .What is the pH of a 0.22M ammonia solution?

What is the percent ionization of ammonia at this concentration?

I got the answer for PartA which is pH=11.3. but i cant slove PartB..HELP!!!

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  1. Ammonia has a Kb of 1.8x10^-5 (NOT the Ka)
    Kb = [NH4+][OH-] / [NH3]
    Let [OH-] = [OH-] = x
    1.8x10^-5 = x^2/(0.22-x)
    We can get an adequate approximate solution by assuming
    1.8x10^-5 = x^2/(0.22)
    x = [OH-] = 0.00199
    pOH = -log(0.00199) = 2.7
    pH = 11.3 (your answer)
    The approximate % ionization is:
    [(0.00199) / (0.22)](100) = _______?

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    posted by GK
  2. thanx a tons =)

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    posted by chem

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