Titration of acetic acid w/ NaOH problem

In summary, the concentration of the diluted acetic acid solution can be calculated using the initial volume and number of moles of the original solution. The concentration of the original acetic acid solution can also be determined using the equivalence point volume and pH. The pKa and Ka for acetic acid can be found using the formula pH = pKa + log([base/acid]). However, the pKa value of 4.4 may have a slight error due to the given information.
  • #1
Xezdeth
2
0
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?
 
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  • #2
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
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).
 

Related to Titration of acetic acid w/ NaOH problem

1. What is the purpose of titration in this problem?

Titration is used to determine the concentration of a solution, in this case the concentration of acetic acid, by reacting it with a known concentration of another solution, in this case NaOH.

2. How do you calculate the molarity of acetic acid from the titration data?

The molarity of acetic acid can be calculated by using the formula M1V1 = M2V2, where M1 is the molarity of NaOH, V1 is the volume of NaOH used, M2 is the molarity of acetic acid, and V2 is the volume of acetic acid. By plugging in the known values, the molarity of acetic acid can be determined.

3. What indicator is used in this titration?

Phenolphthalein is commonly used as the indicator in titrations involving acetic acid and NaOH. It changes from colorless to pink at the endpoint, which is when all the acetic acid has been neutralized by the NaOH.

4. How do you know when the endpoint of the titration has been reached?

The endpoint is reached when the indicator changes color, in this case from colorless to pink. It is important to add the NaOH solution slowly near the endpoint to ensure accuracy.

5. What are some potential sources of error in this titration?

Some potential sources of error in this titration include inaccurate measurement of volumes, improper mixing of solutions, and incorrect interpretation of endpoint due to human error or color perception. It is important to perform multiple trials and take the average for more accurate results.

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