# What is the pH of the solution if 0.015 mol of HCl is added to a buffer containing 0.014 mol of RCOOH and 0.033 mol of RCOONa (source of RCOO-)? Ka = 2.8E-5

I have tried the following:
adding 0.014 + 0.015 = 0.029

ph = pka + log (Base/Acid)
pH = -log(2.8E-5) + log(0.033/0.029)
pH = 4.61

pH = pka + log (Base/Acid)
pH = -log(2.8E-5) + log(0.033/0.014)
pH = 4.93
(i was desparate, i knew that this wasn't the answer)

i have tried to do and ICE diagram, where x ends up = 2.459E-5.
add that to RCOO- to equal 0.03302459, RCOOH = 0.029 (acid)

pH = -log(2.8E-5) + log(0.03302459/0.029)
pH = 4.61

Lynn

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1. ICE is the way to go.
Technically, one should use concentrations but in situations like this the volume (M = mols/L) cancels and we can use mols directly. The equation is and we use mols instead of molarity. Some profs will count off if you don't use M. If your prof does that, then everywhere you plug in mols into the Henderson-Hasselbalch equation, just plug in mol/v, the v cancels, and you go from there.
........RCOO^- + H^+ ==> RCOOH
I......0.033.....0.......0.014