Ammonia has a Kb of 1.8x10^-5 (NOT the Ka)
Kb = [NH4+][OH-] / [NH3]
Let [OH-] = [OH-] = x
Then,
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)
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The approximate % ionization is:
[(0.00199) / (0.22)](100) = _______?
PartA:
Ammonia,is a weak base with a Ka value of 1.8x10^-5 .What is the pH of a 0.22M ammonia solution?
PartB:
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!!!
2 answers
thanx a tons =)
4 answers