Answer
From the strongest to the weakest:
$HNO_3$, $C_5H_5NH^+$, $N{H_4}^+$, $H_2O$
Work Step by Step
- $HNO_3$ is the only strong acid (Ka > 1.0) in the list, so it is the stronger:
- $H_2O$ has a $K_a = 10^{-14}$
Analyzing table 13-3, we get that:
- $Kb (NH_3) = 1.8 \times 10^{-5}$, therefore:
- Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its Ka by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 1.8\times 10^{- 5} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_a = 5.556\times 10^{- 10}$
and
- $Kb(C_5H_5N) = 1.7 \times 10^{-9}$
- Since $C_5H_5NH^+$ is the conjugate acid of $C_5H_5N$ , we can calculate its kb by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 1.7\times 10^{- 9} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.7\times 10^{- 9}}$
$K_a = 5.882\times 10^{- 6}$
Therefore, ordering from the higher ka value, to the lower:
$HNO_3$, $C_5H_5NH^+$, $N{H_4}^+$, $H_2O$
