## Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning

# Chapter 13 - Exercises - Page 575d: 86

#### 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: $HNO_3$, $C_5H_5NH^+$, $N{H_4}^+$, $H_2O$

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