 Chemistry Help Forum pKa of an unknown acid
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 Thread Tools Display Modes Nov 3rd 2016, 07:44 PM #1 ilovechem Junior Member   Join Date: Nov 2016 Location: Los Angeles Posts: 1 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 09:50 PM.  Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode 