Naked Science Forum
Non Life Sciences => Chemistry => Topic started by: dgt20 on 26/03/2018 11:17:20
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What is the pH of a 0.15 M solution of weak acid ammonium bromide? The Kb value for ammonia is 1.8 × 10^–5
I keep getting:
x=√(1.8*10^-5)(0.15)
x= 5.1*10^-3
-log(5.1*10^-3)
=pOH = 2.79
14-2.79
pH=11.71
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Ammonium bromide is an acid, so solutions of it should have a pH less than 7.
The Kb value provided refers to the equilibrium:
NH3 + H2O ↔ NH4+ + OH–
Kb = [NH4+]*[OH–]/[NH3]
However, because you are adding NH4+, you need to figure out how much of it will dissociate back to neutral NH3 (protonating OH–), which means you also need to take into account Kw to find the Ka of NH4+
Kb*Ka = Kw
so:
Ka = Kw/Kb
= [H+]*[OH–]*[NH3]/([NH4+]*[OH–])
the [OH–] terms cancel, leaving:
Ka = [H+]*[NH3]/[NH4+]
and because we know the numerical values of Kw and Kb, we can calculate the numerical value of Ka, and then solve the expression below for [H+]:
Ka = [H+]*[NH3]/[NH4+]
=(x)*(x)/(0.15–x)
and then pH is –log10(x)
good luck!
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Thanks for the way it is calculated, it is very beneficial to me. It helped me a lot.
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Thanks for the clear explanation of this chemistry exercise. However, because you're adding NH4+, you'll need to calculate how much of it will dissociate back to neutral NH3 (protonating OH–), which means you'll need to include Kw when calculating NH4+ Ka.