Go Back   Chemistry Help Forum > Chemistry Forums > Undergraduate Chemistry Forum

Undergraduate Chemistry Forum Ask your undergraduate chemistry questions here.

Thread Tools Display Modes
Old Nov 3rd 2016, 08:44 PM   #1
Junior Member
Join Date: Nov 2016
Location: Los Angeles
Posts: 1
ilovechem is on a distinguished road
pKa of an unknown acid

I titrated 10 mL an unknown weak acid with a strong base (.0755 M) and found the equivalence point to be 12.745 mL of base.

To find the molarity of the unknown weak acid solution, I used the equivalence 12.745 to find the moles of NaOH which will equal the moles of acid (which ended up being 9.6225e-4 moles). Then 9.6225e-4 moles acid over the 10 mL = .096 M. Although the 10 mL of acid was diluted in 40 mL of water so I'm not sure that this is correct or if the water even matters.

I figured to get the pKa, I would use the HH equation at half equivalence point (since pKa = pH at half equivalence).

So if the equivalence needs 12.745 mL of base then the half equivalence needs 6.3725 mL.

original concentration of weak acid .096 M * .01 L = .00096 moles
.0755 M NaOH * .0063725 = .00048 moles
So at this point, conj base = weak acid moles and pH = pKa.

According to the graph of my data, the pH at 5.98 mL of OH added is 5.05 and at 6.52 mL it's 5.11. For 6.3725 I just estimated that the pH was about 5.09 and used

5.09 = pKa

and then the Ka would equal 10 ^(-5.09) = 8.128e-6.

In the given Ka values of the possible weak acids that this could be, Potassium hydrogen phthalate comes closest at 3.9e-6 ... is this close enough for did I make a mistake?

Last edited by ilovechem; Nov 3rd 2016 at 10:50 PM.
ilovechem is offline   Reply With Quote

  Chemistry Help Forum > Chemistry Forums > Undergraduate Chemistry Forum

Thread Tools
Display Modes

Facebook Twitter Google+ RSS Feed