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 to control arrhythmic heartbeat. Quinidine has two basic nitrogens $$pKb1=5.4$$ and $$pKb2=10$$.To measure the concentration, she carries out a titration. Because of the low solubility of quinidine in water, she first protonates both nitrogens with excess HCl and titrates the acidified solution with standardized base. A $$32.4$$ mg sample of quinidine is acidified with $$7$$ mL of $$0.15$$ M HCl.The volume of $$0.1$$ M NaOH required to titrate the excess HCl isThe pH at the first equivalence point is

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 to control arrhythmic heartbeat. Quinidine has two basic nitrogens $$pKb1=5.4$$  and  $$pKb2=10$$.To measure the concentration, she carries out a titration. Because of the low solubility of quinidine in water, she first protonates both nitrogens with excess HCl and titrates the acidified solution with standardized base. A $$32.4$$ mg sample of quinidine  is acidified with $$7$$ mL of $$0.15$$ M HCl.The volume of $$0.1$$ M NaOH required to titrate the excess HCl isThe pH at the first equivalence point is

Infinity Learn · Accepted Answer

$$nquinidine=32.4$$×$$10$$−$$3324=10$$−$$4nHCl$$ $$added=7$$×$$10$$−$$3$$×$$0.15=10.5$$×$$10$$−$$4nHCl$$ reqd. to neutralize $$quinidine=2$$×$$10$$−$$4$$ $$(as$$ it is diacidic $$base)nHCl$$ $$unreacted=10.5$$×$$10$$−$$4$$−$$2$$×$$10$$−$$4$$ $$=$$ $$8.5$$×$$10$$−$$4nNaOH$$ for unreacted $$HCl=8.5$$×$$10$$−$$4=0.1$$×V×$$10$$−$$3$$⇒$$V=8.5$$ mLLet $$QH22+$$ is the fully protonated form of quinidine.$$QH22+$$⇌$$pKa1=14$$−$$10=4QH$$ $$+$$⇌$$pKa2=14$$−$$5.4=8.6QAt$$ first equivalence point there will only be $$QH+$$.⇒$$pH=pKa1+pKa22=6.3$$

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