Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 102

Calculate the percentage of pyridine $\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)\( that forms pyridinium ion, \)\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+},\( in a \)0.10-M$ aqueous solution of pyridine \(\left(K_{\mathrm{b}}=1.7 \times 10^{-9}\right)\)

Short Answer

Expert verified
In a 0.10-M aqueous solution of pyridine, approximately 0.00412% of the pyridine forms pyridinium ion.

Step by step solution

01

Write the equilibrium equation and expression

When pyridine dissolves in water, it undergoes a protonation reaction which forms pyridinium ion: C5H5N + H2O ⇌ C5H5NH+ + OH- The equilibrium constant expression for Kb can be written as: \(K_b = \frac{[C_{5}H_{5}NH^{+}][OH^-]}{[C_{5}H_{5}N]}\) where [C5H5NH+] is the concentration of pyridinium ion, [OH-] is the concentration of hydroxide ions, and [C5H5N] is the concentration of pyridine at equilibrium.
02

Set up an ICE table for the equilibrium problem

To find the equilibrium concentrations of pyridine and pyridinium ion, we can set up an Initial, Change, Equilibrium (ICE) table to keep track of the concentrations: | | C5H5N | H2O | C5H5NH+ | OH- | |--------|-------|-----|---------|-----| |Initial | 0.10 | - | 0 | 0 | |Change | -x | - | +x | +x | |Eqm | 0.10-x| - | x | x |
03

Substitute the equilibrium concentrations into the Kb expression and solve

Substitute the equilibrium concentrations from the ICE table into the Kb expression: \(K_b = \frac{[C_{5}H_{5}NH^{+}][OH^-]}{[C_{5}H_{5}N]} = \frac{x^2}{0.10-x}\) The given Kb value is 1.7 x 10⁻⁹. \(1.7 \times 10^{-9} = \frac{x^2}{0.10-x}\) Since Kb is very small, we can assume x is also very small and therefore, (0.10-x) ≈ 0.10: \(1.7 \times 10^{-9} ≈ \frac{x^2}{0.10}\) Now, solve for x: \(x^2 = 1.7 \times 10^{-9} \times 0.10\) \(x^2 = 1.7 \times 10^{-10}\) \(x = \sqrt{1.7 \times 10^{-10}} ≈ 4.12 \times 10^{-6} M\) So, at equilibrium, [C5H5NH+] = x = 4.12 x 10⁻⁶ M
04

Calculate the percentage of pyridine that forms pyridinium ion

Now that we have the equilibrium concentration of pyridinium ion, we can calculate the percentage of pyridine that forms pyridinium ion: Percentage = \(\frac{[C_{5}H_{5}NH^{+}]}{[C_{5}H_{5}N]_{initial}} \times 100\%\) Percentage = \(\frac{4.12 \times 10^{-6}}{0.10} \times 100\%\) Percentage ≈ \(4.12 \times 10^{-5} \times 100\% = 0.00412\%\) Thus, 0.00412% of the pyridine forms pyridinium ion in a 0.10-M aqueous solution of pyridine.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Making use of the assumptions we ordinarily make in calculating the \(\mathrm{pH}\) of an aqueous solution of a weak acid, calculate the pH of a \(1.0 \times 10^{-6}-\mathrm{M}\) solution of hypobromous acid \(\left(\mathrm{HBrO}, K_{\mathrm{a}}=2 \times 10^{-9}\right) .\) What is wrong with your answer? Why is it wrong? Without trying to solve the problem, explain what has to be included to solve the problem correctly.

Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced equations for the reactions causing the solution to be acidic or basic. The relevant \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values are found in Tables 14.2 and \(14.3 .\) $\begin{array}{ll}{\text { a. } \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}} & {\text { d. } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{ClO}_{2}} \\ {\text { b. } \mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}} & {\text { e. } \mathrm{NH}_{4} \mathrm{F}} \\ {\text { c. } \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{O} \mathrm{l}} & {\text { f. } \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{CN}}\end{array}$

Consider a \(0.67-M\) solution of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)$ a. Which of the following are major species in the solution? i. \(C_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) ii. \(\mathrm{H}^{+}\) ii.. \(\mathrm{OH}^{-}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}+\) b. Calculate the \(\mathrm{pH}\) of this solution.

Calculate the \(\mathrm{pH}\) of a $0.20-M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\( solution \)\left(K_{\mathrm{b}}=\right.\( \)5.6 \times 10^{-4} )$

Calculate the pH of the following solutions: a. 1.2\(M \mathrm{CaBr}_{2}\) b. 0.84$M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}\left(K_{\mathrm{b}} \text { for } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}=3.8 \times 10^{-10}\right)$ c. 0.57$M \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}} \text { for } \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)$
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free