## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - Study Questions - Page 677: 3

#### Answer

$pH = 9.25$

#### Work Step by Step

1. Since $N{H_4}^+$ is the conjugate acid of $N{H_3}$ , we can calculate its $K_a$ by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.6\times 10^{- 10}$ 2. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 5.6 \times 10^{- 10})$ $pKa = 9.25$ 3. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.2}{0.2}$ - 1: It is. 4. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.2}{5.6 \times 10^{-10}} = 3.6\times 10^{8}$ - $\frac{0.2}{5.6 \times 10^{-10}} = 3.6\times 10^{8}$ 5. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 9.25 + log(\frac{0.2}{0.2})$ $pH = 9.25 + log(1)$ $pH = 9.25 + 0$ $pH = 9.25$

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