Log out Go to App
Go to App

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.

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 pH of human blood is steady at a value of approximately 7.4 owing to the following equilibrium reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) $$ Acids formed during normal cellular respiration react with the \(\mathrm{HCO}_{3}^{-}\) to form carbonic acid, which is in equilibrium with \(\mathrm{CO}_{2}(a q)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) During vigorous exercise, a person's \(\mathrm{H}_{2} \mathrm{CO}_{3}\) blood levels were $26.3 \mathrm{mM},\( whereas his \)\mathrm{CO}_{2}\( levels were 1.63 \)\mathrm{mM}$ . On resting, the \(\mathrm{H}_{2} \mathrm{CO}_{3}\) levels declined to 24.9 \(\mathrm{m} M\) . What was the \(\mathrm{CO}_{2}\) blood level at rest?

Calculate the pH of a \(0.010-M\) solution of iodic acid (HIO $_{3}, K_{\mathrm{a}}\( \)=0.17 )$

Is an aqueous solution of NaHSO_ acidic, basic, or neutral? What reaction occurs with water? Calculate the pH of a \(0.10-M\) solution of NaHSO.

Arrange the following 0.10\(M\) solutions in order from most acidic to most basic. See Appendix 5 for \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values. $$ \mathrm{CaBr}_{2}, \mathrm{KNO}_{2}, \mathrm{HClO}_{4}, \quad \mathrm{HNO}_{2}, \quad \mathrm{HONH}_{3} \mathrm{ClO}_{4} $$

Classify each of the following as a strong acid, weak acid, strong base, or weak base in aqueous solution. a. \(\mathrm{HNO}_{2}\) b. HNO \(_{3}\) c. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) d. \(\mathrm{NaOH}\) e. \(\mathrm{NH}_{3}\) f. \(\mathrm{HF}\) g. \(\mathrm{HC}-\mathrm{OH}\) h. \(\mathrm{Ca}(\mathrm{OH})_{2}\) i. \(\mathrm{H}_{2} \mathrm{SO}_{4}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

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