You carry out the titration of a weak diprotic solid acid. You take 178.0 mg of

Question

You carry out the titration of a weak diprotic solid acid.  You take 178.0 mg of the solid acid and dissolve it in about 50 mL of distilled water.  You titrate with 0.095 M NaOH and arrive at an endpoint at 25.20 ml.  You know that at the final pH both protons have been titrated.  What is the formula weight of the acid?  At what volumes would you determine the pKa You carry out the titration of a weak diprotic solid acid.  You take 178.0 mg of the solid acid and dissolve it in about 50 mL of distilled water.  You titrate with 0.095 M NaOH and arrive at an endpoint at 25.20 ml.  You know that at the final pH both protons have been titrated.  What is the formula weight of the acid?  At what volumes would you determine the pKa

Explanation / Answer

Let the acid be H2A

At the endpoint (2nd equivalence point): H2A + 2 NaOH => Na2A + 2 H2O


Moles of NaOH = volume x concentration of NaOH

= 25.20/1000 x 0.095 = 0.002394 mol


Moles of H2A = 1/2 x moles of NaOH

= 1/2 x 0.002394 = 0.001197 mol


Formula weight = mass/moles of H2A

= 0.1780/0.001197

= 148.7 g/mol (or approximately 149 g/mol)



At first equivalence point, total volume = 25.20/2 = 12.60 mL

pH = pKa1 + log([HA-]/[H2A] = pKa1 when [HA-] = [H2A] (half of first equivalence point)

Thus volume for pH = pKa1 = 12.60/2 = 6.30 mL


At second equivalence point, total volume = 25.20 mL = (12.60 + 12.60) mL

pH = pKa2 + log([A2-]/[HA-] = pKa2 when [A2-] = [HA-] (half of second equivalence point)

Thus volume for pH = pKa2 = 12.60 + 12.60/2 = 18.90 mL

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