Answer
a.
$H_2O$: Bronsted-Lowry Acid.
$C_5H_5N$: Bronsted-Lowry Base.
$OH^-$: Conjugate base.
$C_5H_5NH^+$: Conjugate acid.
b.
$HNO_3$: Bronsted-Lowry Acid.
$H_2O$: Bronsted-Lowry Base.
$N{O_3}^-$: 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. $H_2O$ is donating a proton $H^+$ to $C_5H_5N$:
$H_2O$: Bronsted-Lowry Acid.
$C_5H_5N$: Bronsted-Lowry Base.
b. $HNO_3$ is donating a proton to $H_2O$:
$HNO_3$: Bronsted-Lowry Acid.
$H_2O$: Bronsted-Lowry Base.
2. On 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, $H_2O$ produces $OH^-$. And after receiving a proton, $C_5H_5N$ produces $C_5H_5NH^+$.
$OH^-$: Conjugate base.
$C_5H_5NH^+$: Conjugate acid.
b. 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.
