Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 16 - Exercises - Page 771: 35

Answer

a. $ H_2CO_3 $: Bronsted-Lowry Acid. $ H_2O $: Bronsted-Lowry Base. $ HC{O_3}^- $: Conjugate base. $ H_3O^+ $: Conjugate acid. b. $ H_2O $: Bronsted-Lowry Acid. $ NH_3 $: Bronsted-Lowry Base. $ OH^- $: Conjugate base. $ NH{_4}^{+} $: Conjugate acid. c. $ HNO_3 $: Bronsted-Lowry Acid. $ H_2O $: Bronsted-Lowry Base. $ N{O_3}^- $: Conjugate base. $ H_3O^+ $: Conjugate acid. d. $ H_2O $: Bronsted-Lowry Acid. $ C_5H_5N $: Bronsted-Lowry Base. $ OH^- $: Conjugate base. $ C_5H_5NH^+ $: Conjugate acid.

Work Step by Step

1. Identify which compound donates a proton and which receives one; the first will be the acid, and the second is the base. a. $ H_2CO_3 $ is donating a proton $(H^+)$ to $ H_2O $: $ H_2CO_3 $: Bronsted-Lowry Acid. $ H_2O $: Bronsted-Lowry Base. b. $ H_2O $ is donating a proton $(H^+)$ to $ NH_3 $: $ H_2O $: Bronsted-Lowry Acid. $ NH_3 $: Bronsted-Lowry Base. c. $ HNO_3 $ is donating a proton $(H^+)$ to $ H_2O $: $ HNO_3 $: Bronsted-Lowry Acid. $ H_2O $: Bronsted-Lowry Base. d. $ H_2O $ is donating a proton $(H^+)$ to $ C_5H_5N $: $ H_2O $: Bronsted-Lowry Acid. $ C_5H_5N $: Bronsted-Lowry Base. 2. On the products side, the result of the acid after donating a proton is the conjugate base, and the result of the base after receiving one proton is the conjugate acid. a. After donating a proton, $ H_2CO_3 $ becomes $ HC{O_3}^- $. And after receiving a proton, $ H_2O $ becomes $ H_3O^+ $: $ HC{O_3}^- $: Conjugate base. $ H_3O^+ $: Conjugate acid. b. After donating a proton, $ H_2O $ becomes $ OH^- $. And after receiving a proton, $ NH_3 $ becomes $ NH{_4}^{+} $: $ OH^- $: Conjugate base. $ NH{_4}^{+} $: Conjugate acid. c. After donating a proton, $ HNO_3 $ becomes $ N{O_3}^- $. And after receiving a proton, $ H_2O $ becomes $ H_3O^+ $: $ N{O_3}^- $: Conjugate base. $ H_3O^+ $: Conjugate acid. d. After donating a proton, $ H_2O $ becomes $ OH^- $. And after receiving a proton, $ C_5H_5N $ becomes $ C_5H_5NH^+ $: $ OH^- $: Conjugate base. $ C_5H_5NH^+ $: Conjugate acid. ---
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