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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.