Question

To a 25.00 mL volumetric flask, a lab technician adds a 0.300 g sample of a weak monoprotic acid, HA, and dilutes to the mark with distilled water. The technician then titrates this weak acid solution with 0.0867 M KOH. She reaches the endpoint after adding 41.23 mL of the KOH solution. Determine the number of moles of the weak acid in the solution. moles of weak acid: mol Determine the molar mass of the weak acid. molar mass = g/mol After the technician adds 18.27 mL of the KOH solution, the pH of the mixture is 4.39. Determine the pKa of the weak acid. pKa =

Answer 1

1 mole weak acid = 1 mole KOH

No of mole of KOH consumed to reach endpoint = M*V

= 0.0867*41.23

= 3.575 mmol

No of mole of weak acid reacted = 3.575 mmole = 3.575*10^-3 moles

Molarmass of weak acid = w/n = 0.3/(3.575*10^-3) = 83.92 g/mol

after addition of 18.27 ml KOH solution.

pH = pka + log(salt(or) conjugate base/acid)

No of mole of KOH consumed = salt formed = 18.27*0.0867 = 1.584 mmole

No of mole of weak acid left unreacted = 3.575 - 1.584 = 1.991 mmole

pH = 4.39

4.39 = pka + log(1.584/1.991)

pka of weak acid = 4.49