/*! 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 104 Write out the stepwise \(K_{\mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 104

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.

Short Answer

Expert verified
The stepwise dissociation reactions and corresponding Ka expressions for citric acid are: 1. \( \mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} (aq) + \mathrm{H}_{2} \mathrm{O} (l) \longleftrightarrow \mathrm{H}_{3} \mathrm{O}_{\mathrm{a}} + \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-} (aq)\) $$K_{a1} = \dfrac{[\mathrm{H}_{3} \mathrm{O}_{\mathrm{aq}}][\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-}]}{[\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}]}$$ 2. \( \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-} (aq) + \mathrm{H}_{2} \mathrm{O} (l) \longleftrightarrow \mathrm{H}_{3} \mathrm{O}_{\mathrm{a}^{+}} + \mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-} (aq)\) $$K_{a2} = \dfrac{[\mathrm{H}_{3} \mathrm{O}_{\mathrm{aq}}^+][\mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}]}{[\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-}]}$$ 3. \( \mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-} (aq) + \mathrm{H}_{2} \mathrm{O} (l) \longleftrightarrow \mathrm{H}_{3} \mathrm{O}_{\mathrm{a}}^{+} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{3-} (aq)\) $$K_{a3} = \dfrac{[\mathrm{H}_{3} \mathrm{O}_{\mathrm{aq}}^+][\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{3-}]}{[\mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}]}$$

Step by step solution

01

Identify acidic protons

Citric acid has three acidic protons in its structure. These protons are attached to oxygen atoms in the carboxylic acid (COOH) functional groups, and can dissociate in water to form a hydronium ion (H鈧僌鈦�) and corresponding citrate ion.
02

First dissociation reaction

The first dissociation reaction involves one acidic proton being removed from the citric acid molecule, forming the hydrogen di-citrate ion. The balanced chemical equation for the first dissociation reaction is: $$ \mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} \left( \mathrm{aq} \right) + \mathrm{H}_{2} \mathrm{O} \left( \mathrm{l} \right) \longleftrightarrow \mathrm{H}_{3} \mathrm{O}_{\mathrm{a}} + \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-} \left( \mathrm{aq} \right) $$
03

First Ka expression

The acid dissociation constant (Ka) expression for the first dissociation step is: $$ K_{a1} = \dfrac{[\mathrm{H}_{3} \mathrm{O}_{\mathrm{aq}}][\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-}]}{[\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}]} $$
04

Second dissociation reaction

The second dissociation reaction involves another acidic proton being removed, this time from the hydrogen di-citrate ion, forming the dihydrogen citrate ion. The balanced chemical equation for the second dissociation reaction is: $$ \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-} \left( \mathrm{aq} \right) + \mathrm{H}_{2} \mathrm{O} \left( \mathrm{l} \right) \longleftrightarrow \mathrm{H}_{3} \mathrm{O}_{\mathrm{a}^{+}} + \mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-} \left( \mathrm{aq} \right) $$
05

Second Ka expression

The acid dissociation constant (Ka) expression for the second dissociation step is: $$ K_{a2} = \dfrac{[\mathrm{H}_{3} \mathrm{O}_{\mathrm{aq}}^+][\mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}]}{[\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{-}]} $$
06

Third dissociation reaction

The third and final dissociation reaction involves the last acidic proton being removed, from the dihydrogen citrate ion, forming the tricitrate ion. The balanced chemical equation for the third dissociation reaction is: $$ \mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-} \left( \mathrm{aq} \right) + \mathrm{H}_{2} \mathrm{O} \left( \mathrm{l} \right) \longleftrightarrow \mathrm{H}_{3} \mathrm{O}_{\mathrm{a}}^{+} + \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{3-} \left( \mathrm{aq} \right) $$
07

Third Ka expression

The acid dissociation constant (Ka) expression for the third dissociation step is: $$ K_{a3} = \dfrac{[\mathrm{H}_{3} \mathrm{O}_{\mathrm{aq}}^+][\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{3-}]}{[\mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{2-}]} $$ These are the final stepwise dissociation reactions and Ka expressions for citric acid.

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

Calculate the \(\mathrm{pH}\) of an aqueous solution containing \(1.0 \times 10^{-2} M\) \(\mathrm{HCl}, 1.0 \times 10^{-2} \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\), and \(1.0 \times 10^{-2} \mathrm{M} \mathrm{HCN}\).

What are the major species present in \(0.250 \mathrm{M}\) solutions of each of the following acids? Calculate the \(\mathrm{pH}\) of each of these solutions. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

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 \(14.3 .\) a. \(\mathrm{KCl}\) c. \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) e. \(\mathrm{NH}_{4} \mathrm{~F}\) b. \(\mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) d. \(\mathrm{KF}\) f. \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{CN}\)

A codeine-containing cough syrup lists codeine sulfate as a major ingredient instead of codeine. The Merck Index gives \(\mathrm{C}_{36} \mathrm{H}_{44} \mathrm{~N}_{2} \mathrm{O}_{10} \mathrm{~S}\) as the formula for codeine sulfate. Describe the composition of codeine sulfate. (See Exercise 143.) Why is codeine sulfate used instead of codeine?

For the following, mix equal volumes of one solution from Group I with one solution from Group II to achieve the indicated \(\mathrm{pH}\). Calculate the \(\mathrm{pH}\) of each solution. Group I: \(0.20 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}, 0.20 \mathrm{MHCl}, 0.20 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}, 0.20\) \(M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NHCl}\) Group II: \(0.20 \mathrm{M} \mathrm{KOI}, 0.20 \mathrm{M} \mathrm{NaCN}, 0.20 \mathrm{M} \mathrm{KOCl}, 0.20 \mathrm{M}\) \(\mathrm{NaNO}_{2}\) a. the solution with the lowest \(\mathrm{pH}\) b. the solution with the highest \(\mathrm{pH}\) c. the solution with the \(\mathrm{pH}\) closest to \(7.00\)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Inorganic 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.