I donât have any idea on how to get the Ka from my 1/4 and 3/4 equivalence points.

We started with 25 mL of an unknown weak acid and titrated it with 0.10045 M NaOH. I have my 1/4 equivalence point at 5.5 mL of NaOH added with a pH of 4.3 and my 3/4 equivalence point at 16.5 mL of NaOH added with a pH of 5.25. Thanks! My graph and question in the lab manual are attached below.

Titration of An Unknown Weak Acid egivalonce p 23 7mL a pt of 8.6 6.5 mL pH of 525 10 15 25 30 35 Volume of NaOH added Iml)

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Jarrod RobelLv2

12 Jan 2019

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Could someone please help me with this chemistry problem? Thank you!

After performing a titration using 20 mL of unknown weak acid with 0.3830M NaOH, I obtained the following results:

0 mL added NaOH = pH 3.80

Final amount of added NaOH = 8.59 mL = pH 12.20

At Equivalence Point: Volume of NaOH = 4.835 mL; pH = approx 8.40

At the Halfway Point; Volume of NaOH = 2.418 mL: pH = 5.15

Ka = 7.08 x 10^-6

Initial [HA] = 0.0926M (determined by M1V1=M2V2 at the equivalence point)

Molar Mass = 1079.8 g/mol (determined by using original mass 1.9977 g/1.85 x 10^-3 moles

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Now I am asked the following questions in order to check my results for consistency;

Using your experimental Ka value, the concentration of your unknown acid solution and the concentration of the standardized NaOH solution, CALCULATE the pH of the following points on your titration curve:

1) The beginning of the titration before any any NaOH was added

2) 25% of the way to the equivalence point (25% of 4.835 mL is 1.21 mL)

3) 75% of the way to the equivalence point

4) The equivalence point (remember, this is found using Ka and concentrations of acid and base)

5) 20% of the way to the equivalence point (1.20 x 4.835= 5.80 mL)

I've found some info on the internet that says when in the buffer region, the moles of A- = moles of base added and that moles of HA = original moles of HA - moles of base added. Where I got confused was when the equilibrium expression was written as Ka = [H+] [mols Base Added/Volume] Ã· [mols Acid Left/Volume]. I don't know what volumes to use because I always think it makes more sense to use total volume of acid plus base mL's. But maybe you are supposed to use only the volume of each (acid/base) separately, in which case I think the volume of acid would be 20 mL. Maybe this is a totally wrong way to do it- I don't know! Please Help!!!!

A lab experiment was conducted utilizing software to general a titration curve. 25 mL of an unknown analyte was titrated with .1015 M of NaOH solution.