Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 13 - Exercises - Page 575e: 97

Answer

a) $[H_3O^+] = 1.118 \times 10^{- 12}M$ $[OH^-] = 8.944 \times 10^{- 3}M $ $pH = 11.952$ b) $[H_3O^+] = 2.132 \times 10^{- 10}M$ $[OH^-] = 4.69 \times 10^{- 5}M $ $pH = 9.671M$

Work Step by Step

a) 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [(C_2H_5)_3NH^+] = x$ -$[(C_2H_5)_3N] = [(C_2H_5)_3N]_{initial} - x = 0.2 - x$ For approximation, we consider: $[(C_2H_5)_3N] = 0.2M$ 2. Now, use the Kb value and equation to find the 'x' value. $Ka = \frac{[OH^-][(C_2H_5)_3NH^+]}{ [(C_2H_5)_3N]}$ $Kb = 4 \times 10^{- 4}= \frac{x * x}{ 0.2}$ $Kb = 4 \times 10^{- 4}= \frac{x^2}{ 0.2}$ $ 8 \times 10^{- 5} = x^2$ $x = 8.944 \times 10^{- 3}$ Percent ionization: $\frac{ 8.944 \times 10^{- 3}}{ 0.2} \times 100\% = 4.472\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [(C_2H_5)_3NH^+] = x = 8.944 \times 10^{- 3}M $ $[(C_2H_5)_3N] \approx 0.2M$ 3. Calculate the pH: $pOH = -log[OH^-]$ $pOH = -log( 8.944 \times 10^{- 3})$ $pOH = 2.048$ $pH + pOH = 14$ $pH + 2.048 = 14$ $pH = 11.952$ 4. Find the hydronium concentration: $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $ 8.944 \times 10^{- 3} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 8.944 \times 10^{- 3}}$ $[H_3O^+] = 1.118 \times 10^{- 12}$ b) 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [HON{H_3}^+] = x$ -$[HONH_2] = [HONH_2]_{initial} - x = 0.2 - x$ For approximation, we consider: $[HONH_2] = 0.2M$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][HON{H_3}^+]}{ [HONH_2]}$ $Kb = 1.1 \times 10^{- 8}= \frac{x * x}{ 0.2}$ $Kb = 1.1 \times 10^{- 8}= \frac{x^2}{ 0.2}$ $ 2.2 \times 10^{- 9} = x^2$ $x = 4.69 \times 10^{- 5}$ Percent ionization: $\frac{ 4.69 \times 10^{- 5}}{ 0.2} \times 100\% = 0.02345\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [HON{H_3}^+] = x = 4.69 \times 10^{- 5}M $ $[HONH_2] \approx 0.2M$ 3. Calculate the pH: $pOH = -log[OH^-]$ $pOH = -log( 4.69 \times 10^{- 5})$ $pOH = 4.329$ $pH + pOH = 14$ $pH + 4.329 = 14$ $pH = 9.671$ 4. Find the hydronium concentration: $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $ 4.69 \times 10^{- 5} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 4.69 \times 10^{- 5}}$ $[H_3O^+] = 2.132 \times 10^{- 10}$
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