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!!!
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) = _______?
thanx a tons =)
To solve Part B, we need to calculate the percent ionization of ammonia at a concentration of 0.22M. The percent ionization is defined as the ratio of the concentration of the ionized form to the initial concentration, multiplied by 100.
First, we need to determine the concentration of the ionized form of ammonia. In water, ammonia (NH3) partially ionizes to form ammonium ions (NH4+) and hydroxide ions (OH-).
The balanced equation for the ionization of ammonia in water is:
NH3 + H2O ⇌ NH4+ + OH-
Since ammonia is a weak base, we can assume that the concentration of OH- ions formed is equal to the concentration of NH4+ ions formed. Therefore, we only need to determine the concentration of the NH4+ ions.
For each mole of ammonia that ionizes, one mole of NH4+ is formed. So, the concentration of NH4+ ions is equal to the concentration of the ionized ammonia.
Given that the concentration of ammonia is 0.22M, we can assume that the concentration of NH4+ ions is also 0.22M.
Percent ionization = (concentration of NH4+ ions / initial concentration of NH3) x 100
= (0.22M / 0.22M) x 100
= 100%
Therefore, the percent ionization of ammonia at a concentration of 0.22M is 100%.