An ecobotanist separates the components of a tropical bark extract by chromatography. She discovers a large proportion of quinidine, a dextrorotatory isomer of quinine used for control of arrhythmic heartbeat. Quinidine has two basic nitrogens (Kb1 = 4.0 multiplied by 10-6 and Kb2 = 1.0 multiplied by 10-10). To measure the concentration, she carries out a titration. Because of the low solubility of quinidine, she first protonates both nitrogens with excess HCl and titrates the acidified solution with standardized base. A 32.80 mg sample of quinidine (M = 324.41 g/mol) is acidified with 6.65 mL of 0.150 M HCl.
(a) How many mL of 0.0127 M NaOH are needed to titrate the excess HCl?
(b) How many additional milliliters of titrant are needed to reach the first equivalence point of quinidine dihydrochloride?
(c) What is the pH at the first equivalence point?