An ecobotanist separates the components of a tropical bark extract by chromatogr

Question

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 arrythmic heartbeat. Quinidine has two basic nitrogens (K_b1 = 4.0 times 10^-6, K_b2 = 1.0 times 10^-10). To measure the concentration, she carries out a titration. Because of the low solubility of quinidine, she protonates both nitrogens with excess HCI and titrates the acidified solution with standardized base. A 33.85-mg sample of quinidine (mm = 324.41 g/mol) is acidified with 6.55 mL of 0.150 M HCL. a) How many milliliters of 0.0133 M NaOH are needed to titrate the excess HCI? 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?

Explanation / Answer

quinidine concentration = 33.85 mg/324.41 g/mol = 0.1043 mmol

moles of HCl needed = 2 x 0.1043 = 0.2086 mmol

Total HCl moles available = 0.150 M x 6.55 ml = 0.9825 mmol

excess moles of HCl = 0.9825 - 0.2086 = 0.7739 mmol

a) volume of NaOH needed = 0.7739 mmol//0.0133 M = 58.18 ml

b) moles of NaOH is greater than what is needed to reach Ist and IInd equivalence point

c) pH at Ist equivalence point

= 1/2(pKa1 + pKa2)

= 1/2(8.6 + 4)

= 6.3

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