Answer to Question #121974 in Organic Chemistry for ivynf

Question

you have been asked to identify a compound which is known to have an ionisable groups with a pka of 10.33, and one other ionisable group with a pka between 5 and 7. to 100mL of a 2.0M solution of this compound at a pH of 9.5 was added 40mL of a solution of 0.2MHCl acid. the pH changed to 6.85

Answer 1

pH = pKa + log ([A] / [HA])

pH goes to 6.85

9.5 = 10.33 + log ([A] / [HA])

10.33 - 9.5 = log ([A] / [HA])

0.83 = log ([A] / [HA])

so using the new pH,

6.85 = pKa + 0.83

6.85 - 0.83 = pKa

pKa = 6.02

Answer: pKa = 6.02.