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Non Life Sciences => Chemistry => Topic started by: dgt20 on 26/03/2018 11:17:20

Title: How to calculate pH from kB?
Post by: dgt20 on 26/03/2018 11:17:20
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
Title: Re: How to calculate pH from kB?
Post by: chiralSPO on 26/03/2018 14:42:29
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!
Title: Re: How to calculate pH from kB?
Post by: Constantinut on 02/04/2018 06:11:22
Thanks for the way it is calculated, it is very beneficial to me. It helped me a lot.
Title: Re: How to calculate pH from kB?
Post by: aricjoshua on 27/09/2021 09:13:21
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.