Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 54

Approximately 4 metric tons of quinoline, \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{N},\) is produced annually. The principal source of quinoline is coal tar. Quinoline is a weak base in water. A handbook gives \(K_{\mathrm{a}}=6.3 \times 10^{-10}\) for protonated quinoline, \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{NH}^{+} .\) Write the ionization reaction for \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{NH}^{+}\) and calculate \(\mathrm{p} \bar{K}_{\mathrm{b}}\) for quinoline.

Short Answer

Expert verified
The ionization reaction for protonated quinoline is: \(C_{9}H_{7}NH^{+} (aq) \rightarrow C_{9}H_{7}N (aq) + H^{+}(aq)\). The pKb for quinoline is 4.80.

Step by step solution

01

Write the ionization reaction for protonated quinoline

Considering that quinoline acts as a weak base in water, when it gets protonated, it will act as an acid in solution.The ionization of protonated quinoline (a Bronsted acid) can be written as follows: \[C_{9}H_{7}NH^{+} (aq) \rightarrow C_{9}H_{7}N (aq) + H^{+}(aq)\]This indicates that protonated quinoline donates a hydrogen ion (or proton) to become quinoline.
02

Relationship between Ka, Kb, and Kw

The relationship between the acid dissociation constant (Ka) for an acid, the base dissociation constant (Kb) for its conjugate base and the ion product of water (Kw) is given by the expression: \[Kw = Ka \cdot Kb\]At 25 degrees Celsius, the value of Kw is 1.0 x 10^-14.
03

Calculate Kb for quinoline

Using the equation from step 2, we can calculate Kb for quinoline as follows: \[Kb = \frac{Kw}{Ka} = \frac{1.0 x 10^{-14}}{6.3 x 10^{-10}} = 1.59 x 10^{-5}\]
04

Calculate pKb for quinoline

pKb is the negative logarithm to base 10 of the Kb value. It can be calculated as follows: \[pKb = -\log(1.59 x 10^{-5}) = 4.80\]

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 solubility of 1 -naphthylamine, \(\mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{2}, \mathrm{a}\) substance used in the manufacture of dyes, is given in a handbook as 1 g per \(590 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\). What is the approximate \(\mathrm{pH}\) of a saturated aqueous solution of 1-naphthylamine? $$\begin{array}{r} \mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{7} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-} \\ \mathrm{p} K_{\mathrm{b}}=3.92 \end{array}$$

Predict whether a solution of each of the following salts is acidic, basic, or pH neutral: (a) KCl; (b) KF; (c) \(\mathrm{NaNO}_{3} ;\) (d) \(\mathrm{Ca}(\mathrm{OCl})_{2} ;\) (e) \(\mathrm{NH}_{4} \mathrm{NO}_{2}\)

\(50.00 \mathrm{mL}\) of \(0.0155 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) is mixed with \(75.00 \mathrm{mL}\) of 0.0106 M KOH(aq). What is the pH of the final solution?

The antimalarial drug quinine, \(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{O}_{2} \mathrm{N}_{2},\) is a diprotic base with a water solubility of \(1.00 \mathrm{g} / 1900 \mathrm{mL}\) of solution. (a) Write equations for the ionization equilibria corresponding to \(\mathrm{p} K_{\mathrm{b}_{1}}=6.0\) and \(\mathrm{p} K_{\mathrm{b}_{2}}=9.8\) (b) What is the \(\mathrm{pH}\) of saturated aqueous quinine?

Predict which is the stronger acid: (a) \(\mathrm{HClO}_{2}\) or \(\mathrm{HClO}_{3} ;(\mathrm{b}) \mathrm{H}_{2} \mathrm{CO}_{3}\) or \(\mathrm{HNO}_{2} ;(\mathrm{c}) \mathrm{H}_{2} \mathrm{SiO}_{3}\) or \(\mathrm{H}_{3} \mathrm{PO}_{4}\) Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

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