Question

Problem 16.73

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base:
C10H15ON(aq)+H2O(l)←−→C10H15ONH+(aq)+OH−(aq)
A 0.035 M solution of ephedrine has a pH of 11.33.

Part A

What is the equilibrium concentration of C10H15ON?

Express your answer using two significant figures.

[C10H15ON] =		M

SubmitMy AnswersGive Up

Part B

What is the equilibrium concentration of C10H15ONH+?

Express your answer using two significant figures.

[C10H15ONH+]=		M

SubmitMy AnswersGive Up

Part C

What is the equilibrium concentration of OH−?

Express your answer using two significant figures.

[OH−] =		M

SubmitMy AnswersGive Up

Part D

Calculate Kb for ephedrine.

Express your answer using two significant figures.

Kb =

Answer 1

Answer – Given, [C₁₀H₁₅ON] = 0.035 M , pH = 11.33

We know,

pH + pOH = 14

so, pOH = 14- pH

              = 14- 11.33

              = 2.67

So, [OH^-] = 10^-pOH

                 = 10^-2.67

                          = 0.00214 M

   We know

   C₁₀H₁₅ON + H₂O ------> C₁₀H₁₅ONH⁺ + OH^-

I   0.035                              0                 0

C -x                                 +x              +x

E 0.035-x                           +x               +x

We know, at equilibrium, [OH^-] = x = 0.00214 M

Part A) [C₁₀H₁₅ON] = 0.035 -x

                              = 0.035-0.00214

                               = 0.033 M

Part B)

We know, [C₁₀H₁₅ONH⁺] = x = 0.0021 M

Part C) , [OH^-] = x = 0.0021 M    

Part D)

Kb = [C₁₀H₁₅ONH⁺][OH^-] / [C₁₀H₁₅ON]

      = 0.00214 *0.00214 / 0.033

       = 1.4* 10^-4