Calculate the change in pH when 4.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq).

Question

Delta pH=?

Calculate the change in pH when 4.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.
Delta pH=?

Answer 1

When you post these problems it would help if you gave the pKa or pKb since th number I look up in the book may not be the same as that in your text. I will use 1.75E-6 for Kb or NH3 which makes pKa = 9.24
millimoles NH3 = 100 x 0.1 = 10.
mmols NH4Cl = 100 x 0.1 = 10.
Initial pH = 9.24 + log(10/10) = 9.24.

When 4.00 mL of 0.1M HCl is added that adds 4.00 x 0.100 = 0.4 millimol acid.
........NH3 + H^+ ==> NH4^+
I.......10.0...0......10.0..
add acid.......0.400.......
C.....-0.40..-0.40...0.40...
E.......9.60...0.......10.4.....

Substitute the equil line into the HH equation and solve for pH.
Then subtract from the original pH to find the difference.
The addition of NaOH is done the same way. Don't forget that the difference will be + for one of them and - for the other one.