Titration of acetic acid w/ NaOH problem

  • Thread starter Xezdeth
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  • #1
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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 realised 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?
 

Answers and Replies

  • #2
Borek
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Finding concentration is just about following stoichiometry, not about finding concentration of some particular ion. You needed 32.5 mL of 0.100 M NaOH to fully neutralize the acid - that's all you need to know to calculate amount of the acid.

pKa of 4.4 is a bit off, but that's the best estimate you can do with a given information.
 
  • #3
epenguin
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As I understand you are given the information
"1/2 equivalence (this bit doesn't matter now) pH = 4.4"

The half equivalence point equals the pK always - you have worked out this general principle now. No error (but pK's do very with ionic strength and temperature).
 

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