Question

A weak acid with a Ka of 1.8 × 10–5 is titrated with a strong base. During the titration,12.5 mL of 0.10 M NaOH is added to 50.0 mL of 0.100 M acetic acid. What is the pH after the addition of the NaOH?

A.
7.22

B.
5.13

C.
4.26

D.
8.59

I know without any addition, the pH is 2.87. I don't know what it would be after addition of base. Thank you so much!

Answer 1

This is a buffer solution problem. Let's let HAc be acetic acid. Then
millimoles HAc initially = mL x M = 50 mL x 0.1 M = 5
millimoles NaOH added = 12.5 mL x 0.1 M = 1.25
.....................HAc + NaOH ==> NaAc + H2O
initial..............5...........0..............0..............0
added......................1.25..................................
change........-1.25...-1.25.............+1.25......1.25
equilibrium...3.75.......0..................1.25......1.25

Plug the equilibrium line into the Henderson-Hasselbalch buffer equation and solve for pH.
The H-H equation is pH = pKa + log (base concn/acid concn). The base is the salt NaAc and the acid is HAc remaining. For pKa find pKa = -log Ka.
Post your work if you get stuck.

Answer 2

ROBERT

Answer 3

Is the answer 4.26?

Answer 4

The answer is 4.26