# Chemical Forums

## Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: NewtoAtoms on February 14, 2009, 09:16:16 AM

Title: pH of NH4Br
Post by: NewtoAtoms on February 14, 2009, 09:16:16 AM
Calculate the pH of 0.015 M ammonium Bromide solution

This is my leg work and what I know:

a.  NH4Br <----H20------> NH4+ + Br-
The Br- is a neutral ion salt
The NH4+ is an acidic ion, therefore this solution would be slightly acidic

b.  the major species at equillibrium would be NH4Br, NH4+, Br-

c.  Ka = [NH4+][Br-] / [NH4Br]

d.
NH4Br            NH4+      Br-
Initial            0.015             0            0
Change          -x                 +x          +x
--------------------------------------------
Equilibrium    0.015 - x          x             x

therefore

Ka =
• / [0.015-x]

HOWEVER for each question I have solved thus far, I have a value of Ka. However I can't find the Ka of this reaction ANYWHERE.  Can anyone give me assistance on something that I am missing, or not doing correctly.

Thank you for your help and time!
Title: Re: pH of NH4Br
Post by: Astrokel on February 14, 2009, 09:46:02 AM
As if you would have notice, your ICE table doesn't consist of anything related to [H+]/[OH-] which is required to calculate pH of the solution. Also you have pointed out NH4+ is acidic while Br- is neutral ion, so which ion will undergo hydrolysis? As your ICE table suggests you considered both ion to undergo hydrolysis. So probably you can guess which Ka value you will used.
Title: Re: pH of NH4Br
Post by: NewtoAtoms on February 15, 2009, 12:24:20 PM
Okay, that doesn't make a lot of sense to me, but I am going to continue working through the questions to learn, learn, learn..

Therefore I know that the Br- won't affect pH, however NH4+ will.... therefore I must continue the hydrolysis??

NH4+ + H20 ----------> NH3 + H30+

Ka NH4+ = 5.4 x 10-10.

therefore:

NH4+     NH3     H+ (H30+)
Initial      0.15 M     O        O
Change     -x          +x       +x
----------------------------------------
Equili.    0.15 - x      x         x

Ka = [NH3][H+] / [NH4+]
5.4 x 10-10 = (x)(x) / 0.15 -x

0.15 - x ~ 0.15

5.4 x 10-10 (0.15) = x2
8.1 x 10-11 = x2
9.00 x 10-6 = x

THEREFORE

H+ = 9.00 x 10-6

pH = - log (9.00 x 10-6)
pH = 5.06

IS THIS LOGIC CORRECT?!?!?
I have just taken your words, tried to understand them and apply them to my question.