Question

A 0.28 −mol sample of a weak acid with an unknown pKa was combined with 12.0 mL of 3.00 M KOH and the resulting solution was diluted to 1.500 L. The measured pH of the solution was 3.85.

What is the pKa of the weak acid?

Answer 1

Moles KOH used = 3.00 molar times .012 Liters = .036 moles

Moles of weak acid neutralized will also be .036 K0H + HA = K+ + A- + H20

Moles remaining weak acid .28 moles minus .036 = .244 moles.

I will call the weak acid HA. Moles of A- created will be .036 when .036 moles of HA was reacted with .036 moles of KOH

so in this l.5 liters solution we have .036 moles of A- and .244 moles of HA

and knowing the pH we can find the (H+) concentration

(H+) = antilog (-pH) or antilog (-3.85) which = 1.41 x l0^-4 moles per liter

now to find the Ka which = (H+)(A-) over (HA)

We also need the molarity of (A-) and (HA)

Molarity (A-) = .036 moles A- over l.5 liters = .024

Molarity (HA) = .244 moles over l.5 liters = 0.162

Ka = (1.4 x l0^-4)(.024) over (.162) = 2.07 x l0^-5

pKa = -log (Ka) = -log (2.07 x l0^-5) which = 4.68