Calculate the change in pH when 5.00 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). (I have several problems like this can some one work this through for me as a base line)?

Answer 1

millimoles NH3 = mL x M = 100 x 0.1 = 10
mmols NH4^+ = 100 x 0.1 = 10
mmols HCl added = 5 x 0.1 = 0.5

.......NH3 + H^+ ==> NH4^+
I.....10.....0........10
added.......0.5...............
C....-0.5..-0.5......+0.5
E.....9.5.....0......+10.5

pH = pKa + log(base)/(acid)
pH = pKa + log(9.5/10.5)
pH = ?