A 20.0 mL sample of 0.50 M H2C6H6O6 (ascorbic acid, a diprotic acid) was titrate

Question

                    A 20.0 mL sample of 0.50 M H2C6H6O6 (ascorbic acid, a diprotic acid) was titrated with .50 M NaOH. The following data were gathered during the titration. What                     is Ka2 for ascorbic acid?                 

                    mL NaOH added and Corresponding pH:
                

                    10.00 mL, pH of 4.17                 

                    20.00 mL, pH of 5.21                 

                    30.00 mL, pH of 11.55                 

                                     40.00 mL, pH of 12.8

Explanation / Answer

At 1st equivalence point:

H2C6H6O6 + NaOH => HC6H6O6- + Na+ + H2O


Moles of NaOH added = moles of H2C6H6O6 = 20.0/1000 x 0.50 = 0.01 mol

Volume of NaOH added = moles/concentration of NaOH

= 0.01/0.50 = 0.02 L = 20.00 mL


At 2nd equivalence point:

H2C6H6O6 + 2 NaOH => C6H6O62- + 2 Na+ + 2 H2O


Moles of NaOH added = 2 x moles of H2C6H6O6 = 2 x 20.0/1000 x 0.50 = 0.02 mol

Volume of NaOH added = moles/concentration of NaOH

= 0.02/0.50 = 0.04 L = 40.00 mL


At halfway to second equivalence point (corresponding to 30.00 mL):

[HC6H6O6-] = [C6H6O62-] => [C6H6O62-]/[HC6H6O6-] = 1

pH = pKa2 + log([C6H6O62-]/[HC6H6O6-])

= pKa2 + log(1)

= pKa2 = -log Ka2 = 11.55



Ka2 = 10^(-pKa2) = 10^(-11.55)

= 2.82 x 10^(-12) = 2.8 x 10^(-12)

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