Calculate the pH of a buffer solution #2?

Question

A buffer solution is made with:

40.0 mLs of 0.100 M acetic acid

30.0 mLs of 0.100 M NaOH

10.0 mLs of Deionized water

Use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution.

**I know this is similar to my first question but I'm struggling with this so the more help the better :)

Answer 1

It is usually just stoichiometry when you do a titration. Notice how NaOH is a strong base and it reacts completely according to the equation:

NaOH(aq) + CH3COOH(aq) --> CH3COONa(aq) + H2O(l)

First we have to find out how many moles of acid and base there are.

.04L HAc *(.1moles/L) = .004moles of HAc

.03L NaOH *(.1moles/L) = .003moles of NaOH

Next we find out how many moles of Acetic acid are neutralized by NaOH. For every mole of acetic acid neutralized one mole of NaOH is used.

.003moles NaOH (1mol HAc/ 1mol NaOH) =

.003moles HAc react

Next subtract the total amount of HAc from the amount the reacts.

(.004moles of HAc total) - (.003moles of HAc react) = .001moles left

To find out how much conjugate base is produced after the neutralization. For every mole of NaOH that reacts one mole of acetate is produced.

.003moles NaOH *(1mol Ac-/1mol NaOH)= .003moles of Acetate

Now we need to find the concentrations of Acetic acid and acetate (conjugate base).

[HAc] = .001moles of HAc/ .08L = .0125M

[Ac-] = .003moles of Ac-/ .08L = .0375M

Ka of acetic acid = 1.8*10^-5

pH = pKa + log(base/acid)

pH = -log(1.8*10^-5) + log(.0375/.0125)

pH = 5.22