calculate the change in ph for 1l of buffer solution containing NH4OH and NH4Cl upon addition of 1) 0.02 moles of HCl

Question

2) 0.02 moles of NaOH

I got the initial pH as 9.2504
But now Im stuck cus I dunno what happens when we add HCl and NaOH?
Please help! :)

Answer 1

How did you get the initial pH when you don't know how much base/acid you started with.

Answer 2

Actually I got the pH of initial solution containing NH4OH and NH4Cl. Question is to find the pH change when we add acid and a base respectively. So I know the initial one but dunno what to do after that...

Answer 3

And moles of NH4OH = 0.1
moles of NH4Cl = 0.1
My bad :(
Thanks.

Answer 4

NH4OH initially = 0.1 mol
NH4Cl initially = 0.1 mol

When adding HCl the equation is
......NH3 + H^+ ==> NH4^+

When adding NaOH the equation is and I will do this one.
......NH4^+ OH^- ==> NH3 + H2O
I.....0.1....0.......0.1.......
add........0.02.............
C....-.02..-0.02.....0.02
E....0.08....0.......0.12

Now plug the E line into the HH equation and solve for the new pH. Then find the difference between this new pH and what you found at the beginning.

Adding the acid is the same way. I have used NH3 and not NH4OH. NH4OH actually is NH3 + H2O ==> NH4OH but technically NH4OH doesn't exist.

Answer 5

Oh that worked!
Thank you so much sir!
You literally saved me.
Thanks again :)