Answer
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.
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. $ 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.
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