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Ch.16 - Acid-Base Equilibria

Chapter 16, Problem 73a

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: C10H15ON1aq2 + H2O1l2 Δ C10H15ONH+1aq2 + OH-1aq2 A 0.035 M solution of ephedrine has a pH of 11.33. (a) What are the equilibrium concentrations of C10H15ON, C10H15ONH+, and OH-?

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Hello. In this problem we are told that ketamine is a weak organic base. That is used as a dissociative anesthetic for a 0.48 moller ketamine solution with a ph of 12.55. But are the concentrations of are reacting to products at equilibrium. Let's begin then by finding the concentration of hydroxide ions at equilibrium. Call them that the P. O. H. Equal to 14 -2 p. So our p. o. h. works out to 1.45 and our hydroxide ion concentration is equal to the intend to the negative P. O. H. So this works out to then 0.03548 Moeller And step two. Then let's create an ice table. So the first step in our ice table we need to write a reaction so that simplify our notation for our base ketamine and its conjugate acid. So called ketamine be for our base and its conjugate acid, B. H. Plus. So then we have our base ketamine undergoes hydraulic sis from the conjugate acid and I dropped it irons. So we have that initial change in equilibrium. Initially we have 0.48. All of our base ketamine ignore water since it's a pure liquid and we initially have none of the conjugate acid or hydroxide ions are change then is minus X. And then plus X. And plus X equilibrium. Then is the initial plus. To change from the ice table, then we know that X. Is equal to our hydroxide ion concentration Which is equal to 0.03548 moller. Finding the concentration of our reactant, so our base which is ketamine, It is equal to 0. -0.03548. So this works out to 0.013 smaller. The concentration of hydroxide is equal to the concentration of our conjugate acid, which are both equal to X. So at equilibrium concentration of our base is 0.13 moller, and the concentration of the conjugate acid and hydroxide ions is 0.35 moller. This corresponds then, to answer C. Thanks for watching. Hope this helps.