# Titration of acetic acid w/ NaOH problem

• Xezdeth

#### Xezdeth

1) Given: 0.100 M NaOH, 5 mL concentrated acetic acid. 30 mL of distilled water was added to dilute the concentrated acetic acid.

Calculated: Equivalence point (Vol = 32.5 mL, pH = 9.3)
1/2 equivalence = 32.5 mL/2 = 16.25, pH = 4.4

1. Calculate the concentration (M) of the diluted acetic acid solution

2. Calculate the concentration of the original acetic acid solution

3. Determine pKa and Ka for the acetic acid.

3) For 1., I tried calculating the [H+] by finding the pH when the volume of NaOH was 0, then finding the number of moles by using volume. Then using the number of moles over the initial volume (5mL). Then I realized that acetic acid doesn't dissociate fully in water, and the [H+] doesn't equal to [CH3COOH].

For 3, I used the formula pH = pKa + log([base/acid]).
log(1/1) becomes 0.
-4.4 = long(Ka)
Ka = 10^-4.4
Ka = 3.98*10^-5

I checked the Ka for acetic acid online and got a value of 1.8*10^-5, which has a difference of 45%. Did I do something wrong here?