Buffer Solution

Question

Starting with a 1.00 L of a buffer thatis 1.00 M in acetic acid and 1.00 M in sodium acetate, calculate the pH after the addition of 0.100 mole of NaOH. (assume that the addition does not change the volume of the solution)

Answer 1

The Henderson Hasselbalch equation can be used here.  That is

pH = pKa + log [salt]/[acid]

In this problem, OH- is added, so it will react with the weak acid, acetic acid, to form sodium acetate.  Thus, the initial [acetic acid] will decrease, and the initial [sodium acetate] will increase.  It is easiest to set up an ICE table.  Let HAc be acetic acid and NaAc is sodium acetate or just Ac- is acetate since Na is a spectator.

 

......HAc + OH- ===> H2O + Ac-

I....1 M.....0..............0.........1 M.....

C.. -0.1....+0.1..................+0.1... 

E...0.9.......0.......................1.1.....

 

pH = 4.74 + log [1.1]/[0.9] = 4.74 + log 1.22

pH = 4.74 + 0.087

pH = 4.83

 

Looking at it without the ICE table

0.1 moles NaOH + 1 mole HAc ==>0.1 mole NaAc and 0.9 moles HAc

Adding the 0.1 mole NaAc to the already present 1.0 mole NAc = 1.1 moles NaAc

Since 0.1 moles HAc reacted with the NaOH, that leaves 0.9 moles HAc

Taking 4.74 as the pKa for acetic acid, plug these values into the Henderson Hasselbalch equation.

Solve for pH