Question

Calculate the change in pH when 6.00 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). Calculate the change in pH when 6.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.

Answer 1

First what do we have?
100 mL x 0.1M NH3 = 10 millimoles.
100 mL x 0.1M NH4Cl = 10 mmoles.
6.00 mL x 0.1M HCl = 0.6 mmol.
6.00 mL x 0.1M NaOH = 0.6 mmol.
I'll do one to show you how to do it.

.........NH3 + H^+ ==> NH4^+
initial..10.....0.......10
add............0.6
change...-0.6..-0.6.....+0.6
equil....9.4.....0......10.6

pH = pKa + log (base/acid)
pH = pKa + log (9.4/10.6)
Solve for pH and take the difference between that and the pH of the original solution to find the change in pH. pH of original solution is done by HH equation, too.

I would workup a new ICE chart for the addition of NaOH as follows:
.........NH4^+ + OH^- ==> NH3 + H2O
.........10......0.........10
add..............0.6..............
change....-0.6...-0.6.......+0.6
equil.....9.4.....0........0.6
etc.

Answer 2

DrBob222, I understand the work, however why do we use mols? aren't we trying to find the concentrations?