Find the theoretical pH of .01M and .05M benzoic acid buffer when 2ml of .1MNaOH was added to each buffer.

Answer 1

This problem relies on the Henderson-Hasselbalch Equation which relates the pKa of a buffer to changes in the solution such as the addition of strong acids or strong bases.

So the first thing to do in this problem is to find the pKa of benzoic acid. That can be done using a table that already exists.

The pKa of benzoic acid is 4.202.

So the next step is to figure out how the 2 mL of 0.1 M NaOH affected the acid and it's conjugate base. The equation is:

pH = pKa + log([A-]/[HA])

Where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid. For the sake of this problem we will say the NaOH is added to a liter of each of the buffer solutions.

When the base is added it will react with the buffer to increase the amount of the conjugate base and decrease the amount of the acid.

Moles of NaOH introduced to the buffer solution:

.002 L * 0.1 M = 0.0004 moles of OH-

So this is how 0.0004 moles of OH- affects the solution:

benzoic acid before NaOH: 1 L * 0.01 M = 0.01 moles H+

benzoic acid after NaOH: 0.01 moles H+ - 0.0004 moles = 0.0096 moles H+

benzoate (conjugate base) before NaOH: 1 L * 0.01 M = 0.01 moles

benzoate after NaOH: 0.01 moles + 0.0004 = 0.0104 moles

Change in pH:

pH = 4.202 - log(.0096/.0104)
pH = 4.237

You can do the one with the 0.05 M benzoic acid.

Answer 2

I made a couple of mistakes

1: Equation is:

pH = pKa + log([A-]/[HA]

Notice addition instead of subtraction

2: Change in pH:

pH = 4.202 + log(.0096/.0104)
pH = 4.167