Question

You prepare a buffer solution from 10.0 mL of 0.100 M MOPS (3-morpholinopropane-1-sulfonic acid) and 10.0 mL of 0.074 M NaOH. Next, you add 1.00 mL of 6.65 × 10-4 M lidocaine to this mixture.
Denoting lidocaine as L, calculate the fraction of lidocaine present in the form LH+ .

MOPS Ka= 6.3x10^-8
Lidocaine Kb= 8.7x10^-7

Answer 1

10.0 mL of 0.100 M MOPS is 0.01L x 0.1 = 0.001 moles

10.0 mL of 0.074 M NaOH is 0.01 L x 0.074 = 0.00074 moles in 20 mL is 0.04M

After reaction with NaOH acid form is 0.001-0.00074 = 0.00026 in 20 mL is 0.01M

pKa = -log Ka = - log 6.3 x 10^-8 = 7.2

So

pH = pKa + log base/acid

pH = 7.2 + log 0.04/0.01

pH = 7.8

lidocaine as L Kb = 8.7 x 10^-7 Ka = 1 x 10^-14/ 8.7 x 10^-7 = 1.15 x 10^-8; pKa = - log Ka = 7.94

pH = pKa + log base/acid

7.8 = 7.94 + log base/acid

log base/acid = -0.14

base/acid = 10^-0.14 = 0.725

If base= 0.725 acid form = 1

So LH form is 1/1.725 = 0.58 fraction is LH form