Answer
a. $HI$ is a Bronsted-Lowry acid and $H_2O$ is a Bronsted-Lowry base.
b. $H_2O$ is a Bronsted-Lowry acid and $F^-$ is a Bronsted-Lowry base.
c. $H_2S$ is a Bronsted-Lowry acid and $CH_3CH_2NH_2$ is a Bronsted-Lowry base.
Work Step by Step
a. HI transforms into $I^-$ when it loses a proton $(H^+)$, therefore, it is considered an acid in this case. $H_2O$ gains a proton to produce $H_3O^+$, making it a base.
b. $H_2O$ transforms into $OH^-$ when it loses a proton $(H^+)$, therefore, it is considered an acid in this case. $F^-$ gains a proton to produce $HF$, making it a base.
c. $H_2S$ transforms into $HS^-$ when it loses a proton $(H^+)$, therefore, it is considered an acid in this case. $CH_3CH_2NH_2$ gains a proton to produce $CH_3CH_2NH_3^+$, making it a base.
