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Chapter 16: Problem 73

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: $$ \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ A \(0.035 \mathrm{M}\) solution of ephedrine has a \(\mathrm{pH}\) of 11.33 . (a) What are the equilibrium concentrations of $\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}, \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+},\( and \)\mathrm{OH}^{-} ?$ (b) Calculate \(K_{b}\) for ephedrine.

Short Answer

Expert verified
The equilibrium concentrations for the given reaction are approximately: [C₁₀H₁₅ON] ≈ 0.03287 M [C₁₀H₁₅ONH⁺] ≈ 2.130 × 10^(-3) M [OH⁻] ≈ 0.00426 M And the Kb value for ephedrine is Kb ≈ 1.294 × 10^(-5).

Step by step solution

01

Find the pOH and [OH-]

From the information given, we know that the pH of the solution is 11.33. To find the concentration of hydroxide ions, we should first find the pOH of the solution: pOH = 14.00 - pH pOH = 14.00 - 11.33 pOH = 2.67 Now, we can use the pOH to find the concentration of hydroxide ions: [OH⁻] = 10^(-pOH) [OH⁻] = 10^(-2.67) = 2.130 × 10^(-3) M
02

Set up the ICE table

Next, set up an ICE (Initial, Change, Equilibrium) table for the given reaction: \t{C_{10}H_{15}ON (aq) + H2O (l)}{\rightleftharpoons C_{10}H_{15}ONH^{+}(aq) + OH⁻(aq)} I: 0.035 M (initial concentration) -- 0 0.00213 M C: -x -- +x +x E: 0.035 - x -- x 0.00213 + x Note that initially, there are 0 concentration of C₁₀H₁₅ONH⁺ because it has not formed yet.
03

Write expressions for x and Kb

Since we have the equilibrium concentration of OH⁻, we can use it to find x: x = [OH⁻] at equilibrium - initial [OH⁻] x = (0.00213 + x) - 0.00213 x = 2.130 × 10^(-3) M Now, write the Kb expression for ephedrine: Kb = \(\dfrac{[C₁₀H₁₅ONH^{+}][OH^{-}]}{[C₁₀H₁₅ON]}\) Substitute the equilibrium concentrations from the ICE table: Kb = \(\dfrac{(x)(0.00213 + x)}{(0.035 - x)}\)
04

Calculate the Kb value

Since x is small compared to 0.035, we can assume (0.035 - x) ≈ 0.035: Kb = \(\dfrac{(2.130 \times 10^{-3})(2.130 \times 10^{-3} + x)}{0.035}\) Plug in the value of x: Kb = \(\dfrac{(2.130 \times 10^{-3})(2.130 \times 10^{-3} + 2.130 \times 10^{-3})}{0.035}\) Kb ≈ 1.294 × 10^(-5)
05

Find the equilibrium concentrations

Now that we have the value of x, we can find the equilibrium concentrations of C₁₀H₁₅ON, C₁₀H₁₅ONH⁺, and OH⁻. [C₁₀H₁₅ON] = 0.035 - x ≈ 0.035 - 2.130 × 10^(-3) ≈ 0.03287 M [C₁₀H₁₅ONH⁺] = x ≈ 2.130 × 10^(-3) M [OH⁻] ≈ 0.00213 + x ≈ 0.00213 + 2.130 × 10^(-3) ≈ 0.00426 M So, the equilibrium concentrations are: [C₁₀H₁₅ON] ≈ 0.03287 M [C₁₀H₁₅ONH⁺] ≈ 2.130 × 10^(-3) M [OH⁻] ≈ 0.00426 M And the Kb value for ephedrine is Kb ≈ 1.294 × 10^(-5).

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Most popular questions from this chapter

At the boiling point of water $\left(100^{\circ} \mathrm{C}\right), K_{w}=5.6 \times 10^{-13} .\( Calculate \)\left[\mathrm{H}^{+}\right]$ and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

Codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a weak organic base. A \(5.0 \times 10^{-3} \mathrm{M}\) solution of codeine has a pH of 9.95. Calculate the value of \(K_{b}\) for this substance. What is the \(\mathrm{p} K_{b}\) for this base?

Ammonia, \(\mathrm{NH}_{3}\), acts as an Arrhenius base, a Brønsted-Lowry base, and a Lewis base, in aqueous solution. Write out the reaction \(\mathrm{NH}_{3}\) undergoes with water and explain what properties of ammonia correspond to each of the three definitions of "base."

Indicate whether each of the following statements is correct or incorrect. (a) Every Brønsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Bronsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) \(\mathrm{K}^{+}\) ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.

Identify the Lewis acid and Lewis base in each of the following reactions: (a) $\mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ (b) $\mathrm{FeBr}_{3}(s)+\mathrm{Br}^{-}(a q) \rightleftharpoons \mathrm{FeBr}_{4}^{-}(a q)$ (c) $\mathrm{Zn}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)$ (d) $\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)$
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