Calculate pH

Question

[NH₄Br]_(aq) = 0.020M , pKa of NH₄⁺= 9.24

 

Given that the question is asking for pH, I know I need to find the [H₃O⁺], so I've set up the reaction to look like this:

NH₄Br (aq) + H₂O ⇌ H₃O⁺ + NH₃Br^-

 

This way I found pH= 5.5, using an ICE table, calculating Ka from pKa, then taking -log[H₃O⁺] to find the pH.

 

However, the fact that I was given the pKa for NH₄⁺ is making me think I set up the reaction wrong... Should one of my products be NH₄⁺ since I was given the pKa for that? 

If done that way, then would I find the [OH^-] instead of [H₃O⁺] then work from there to find pH?

 

Thanks for your help.

 

Answer 1

NH4Br is an acidic salt (from a strong acid and a weak base), so the pH should be <7.

You have set it up correctly using the hydrolysis of NH4⁺ and the Ka for NH4⁺ (an acid). 

NH4⁺ + H2O ==> NH3 + H3O⁺

(NH4OH + H+)

Ka = [NH3][H3O+]/[NH4+]

From pKa find Ka and solve for H3O+ 

Obtain pH as -log [H30+]