Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 100

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 \(0.10 \mathrm{M}\) 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 \(0.243\%\) of the pyridine forms the pyridinium ion.

Step by step solution

01

Establish base ionization equation

The ionization equation for pyridine in water looks like this: C5H5N + H2O ⇌ C5H5NH+ + OH- Here, pyridine reacts with water to form the pyridinium ion and hydroxide ions.
02

Write the Ka expression and substitute Kb

We can write the Ka expression for this reaction: Ka = [C5H5NH+][OH-] / [C5H5N] However, we are given Kb, the base dissociation constant. We can use the relationship Ka * Kb = Kw, where Kw is the ion-product constant for water, to find Ka: Kw = 1.0 x 10^{-14} Ka = Kw / Kb = 1.0 x 10^{-14} / 1.7 x 10^{-9} = 5.88 x 10^{-6} Now we can rewrite the Ka expression with this value: 5.88 x 10^{-6} = [C5H5NH+][OH-] / [C5H5N]
03

Plug in the initial concentrations and solve for equilibrium concentrations

The initial concentration of pyridine is 0.10 M. Let x be the concentration of C5H5NH+ and OH- at equilibrium. The equilibrium concentration of pyridine becomes (0.10 - x). Now we can plug these values into the Ka expression: 5.88 x 10^{-6} = (x)(x) / (0.10 - x)
04

Solve the quadratic equation for x

There are two possible approaches for solving this equation for x. One is to use the quadratic formula and the other is to make the approximation that x << 0.10, and therefore (0.10 - x) ≈ 0.10. Since the Kb value is very small, we can make this approximation to simplify the problem: 5.88 x 10^{-6} ≈ x^2 / 0.10 x^2 ≈ 5.88 x 10^{-7} x ≈ √(5.88 x 10^{-7}) ≈ 2.43 x 10^{-4} So the equilibrium concentration of C5H5NH+ is 2.43 x 10^{-4} M.
05

Calculate the percentage of pyridine that forms pyridinium ions

Now that we have the equilibrium concentration of C5H5NH+, we can find the percentage of pyridine that forms pyridinium ions: Percentage of pyridine ionized = (Concentration of C5H5NH+ / Initial concentration of C5H5N) * 100% Percentage of pyridine ionized ≈ (2.43 x 10^{-4} M / 0.10 M) * 100% ≈ 0.243% In a 0.10 M aqueous solution of pyridine, approximately 0.243% of the pyridine forms the pyridinium ion.

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

The \(\mathrm{pH}\) of a \(1.00 \times 10^{-2} \mathrm{M}\) solution of cyanic acid (HOCN) is \(2.77\) at \(25^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{a}}\) for \(\mathrm{HOCN}\) from this result.

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

How would you prepare \(1600 \mathrm{~mL}\) of a \(\mathrm{pH}=1.50\) solution using concentrated \((12 M) \mathrm{HCl} ?\)

A solution is prepared by dissolving \(0.56 \mathrm{~g}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make \(1.0 \mathrm{~L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and the \(\mathrm{pH}\) of this solution.

Write out the stepwise \(K_{\mathrm{a}}\) reactions for citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\), a triprotic acid.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Nuclear 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