/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 Write the reaction and the corre... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 85

Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. \(\mathrm{NH}_{3}\) b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)

Short Answer

Expert verified
a. The reaction for \(\mathrm{NH}_{3}\) acting as a base in water is: \[\mathrm{NH_3}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{NH_4^+}(aq) + \mathrm{OH^-}(aq)\] and the corresponding \(K_b\) expression is: \[K_b = \frac{[\mathrm{NH_4^+}][\mathrm{OH^-}]}{[\mathrm{NH_3}]}\] b. The reaction for \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\) acting as a base in water is: \[\mathrm{C}_5\mathrm{H}_5\mathrm{N}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{C}_5\mathrm{H}_5\mathrm{NH^+}(aq) + \mathrm{OH^-}(aq)\] and the corresponding \(K_b\) expression is: \[K_b = \frac{[\mathrm{C}_5\mathrm{H}_5\mathrm{NH^+}][\mathrm{OH^-}]}{[\mathrm{C}_5\mathrm{H}_5\mathrm{N}]}\]

Step by step solution

01

a. Reaction and \(K_b\) for \(\mathrm{NH}_{3}\)

Step 1: Write the chemical equation When \(\mathrm{NH}_3\) accepts a proton from water, it forms its conjugate acid, the ammonium ion (\(\mathrm{NH_4^+}\)) and hydroxide ions (\(\mathrm{OH^-}\)). \[\mathrm{NH_3}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{NH_4^+}(aq) + \mathrm{OH^-}(aq)\] Step 2: Write the \(K_b\) expression The base dissociation constant, \(K_b\), can be written as the ratio of the concentrations of the products to the reactants, excluding water since it is a pure liquid. \[K_b = \frac{[\mathrm{NH_4^+}][\mathrm{OH^-}]}{[\mathrm{NH_3}]}\]
02

b. Reaction and \(K_b\) for \(\mathrm{C}_5\mathrm{H}_5\mathrm{N}\)

Step 1: Write the chemical equation When \(\mathrm{C}_5\mathrm{H}_5\mathrm{N}\) accepts a proton from water, it forms its conjugate acid (\(\mathrm{C}_5\mathrm{H}_5\mathrm{NH^+}\)) and hydroxide ions (\(\mathrm{OH^-}\)). \[\mathrm{C}_5\mathrm{H}_5\mathrm{N}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{C}_5\mathrm{H}_5\mathrm{NH^+}(aq) + \mathrm{OH^-}(aq)\] Step 2: Write the \(K_b\) expression As with the previous example, the base dissociation constant, \(K_b\), can be written as the ratio of the concentrations of the products to the reactants, excluding water. \[K_b = \frac{[\mathrm{C}_5\mathrm{H}_5\mathrm{NH^+}][\mathrm{OH^-}]}{[\mathrm{C}_5\mathrm{H}_5\mathrm{N}]}\]

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=378.85 \mathrm{g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=\) \(8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C} ) .\) Calculate the \(\mathrm{pH}\) of a \(30.0-\mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at \(35.0^{\circ} \mathrm{C}\) is \(2.1 \times 10^{-14} .\)

Calculate the \(\mathrm{pH}\) of a \(0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(1.3 \times 10^{-3} )\)

Determine the pH of a \(0.50-M\) solution of \(\mathrm{NH}_{4} \mathrm{OCl.}\) . See Exercise \(181 .\) )

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 \(14.3 .\) \(\begin{array}{ll}{\text { a. } \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}} & {\text { d. } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{ClO}_{2}} \\ {\text { b. } \mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}} & {\text { e. } \mathrm{NH}_{4} \mathrm{F}} \\ {\text { c. } \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{O} \mathrm{l}} & {\text { f. } \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{CN}}\end{array}\)

Calculate the \(\mathrm{pH}\) and \(\left[\mathrm{S}^{2-}\right]\) in a \(0.10-M \mathrm{H}_{2} \mathrm{S}\) solution. Assume \(K_{\mathrm{a}_{1}}=1.0 \times 10^{-7} ; K_{\mathrm{a}_{2}}=1.0 \times 10^{-19}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

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.