## Chemistry: Molecular Approach (4th Edition)

a. $HI$: Bronsted-Lowry Acid. $H_2O$: Bronsted-Lowry Base. $I^-$: Conjugate base. $H_3O^+$: Conjugate acid. b. $H_2O$: Bronsted-Lowry Acid. $CH_3NH_2$: Bronsted-Lowry Base. $OH^-$: Conjugate base. $CH_3NH{_3}^{+}$: Conjugate acid. c. $H_2O$: Bronsted-Lowry Acid. $C{O_3}^{2-}$: Bronsted-Lowry Base. $OH^-$: Conjugate base. $HC{O_3}^-$: Conjugate acid. d. $HBr$: Bronsted-Lowry Acid. $H_2O$: Bronsted-Lowry Base. $Br^-$: Conjugate base. $H_3O^+$: Conjugate acid.
1. Identify which compound donates a proton and which receives one; the first will be the acid, and the second is the base. a. $HI$ is donating a proton $(H^+)$ to $H_2O$: $HI$: Bronsted-Lowry Acid. $H_2O$: Bronsted-Lowry Base. b. $H_2O$ is donating a proton $(H^+)$ to $CH_3NH_2$: $H_2O$: Bronsted-Lowry Acid. $CH_3NH_2$: Bronsted-Lowry Base. c. $H_2O$ is donating a proton $(H^+)$ to $C{O_3}^{2-}$: $H_2O$: Bronsted-Lowry Acid. $C{O_3}^{2-}$: Bronsted-Lowry Base. d. $HBr$ is donating a proton $(H^+)$ to $H_2O$: $HBr$: Bronsted-Lowry Acid. $H_2O$: Bronsted-Lowry Base. 2. In the products side, the result of the acid after donating a proton is the conjugate base, and the base after receiving one proton is the conjugate acid. a. After donating a proton, $HI$ becomes $I^-$. And after receiving a proton, $H_2O$ becomes $H_3O^+$: $I^-$: Conjugate base. $H_3O^+$: Conjugate acid. b. After donating a proton, $H_2O$ becomes $OH^-$. And after receiving a proton, $CH_3NH_2$ becomes $CH_3NH{_3}^{+}$: $OH^-$: Conjugate base. $CH_3NH{_3}^{+}$: Conjugate acid. c. After donating a proton, $H_2O$ becomes $OH^-$. And after receiving a proton, $C{O_3}^{2-}$ becomes $HC{O_3}^-$: $OH^-$: Conjugate base. $HC{O_3}^-$: Conjugate acid. d. After donating a proton, $HBr$ becomes $Br^-$. And after receiving a proton, $H_2O$ becomes $H_3O^+$: $Br^-$: Conjugate base. $H_3O^+$: Conjugate acid. ---