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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