Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 14 - Acids and Bases - Exercises - Page 704: 63

Answer

a. pH = 2.00; The major species are: $HNO_2$ and $H_2O$. a. pH = 2.68; The major species are: $CH_3CO_2H$ and $H_2O$.

Work Step by Step

a. 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $HNO_2(aq) + H_2O(l) \lt -- \gt NO_2^-(aq) + H_3O^+(aq) $ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $NO_2^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M -$[H_3O^+] = [NO_2^-] = 0 + x = x$ -$[HNO_2] = [HNO_2]_{initial} - x$ For approximation, we are going to consider $[HNO_2]_{initial} = [HNO_2]$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][NO_2^-]}{ [HNO_2]}$ $Ka = 4 \times 10^{- 4}= \frac{x * x}{ 0.250}$ $Ka = 4 \times 10^{- 4}= \frac{x^2}{ 0.250}$ $x^2 = 0.250 \times 4 \times 10^{-4} $ $x = \sqrt { 0.250 \times 4 \times 10^{-4}} = 0.01 $ Percent dissociation: $\frac{ 1 \times 10^{- 2}}{ 0.25} \times 100\% = 4\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [NO_2^-] = x = 1 \times 10^{- 2}M $ And, since 'x' has a very small value (compared to the initial concentration): $[HNO_2] \approx 0.250M$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 0.010)$ $pH = 2.00$ The species with a significant concentration in this solution are: $HNO_2$ and $H_2O$. ----------------------- 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $CH_3CO_2H(aq) + H_2O(l) \lt -- \gt CH_3CO_2^-(aq) + H_3O^+(aq) $ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $CH_3CO_2^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M -$[H_3O^+] = [CH_3CO_2^-] = 0 + x = x$ -$[CH_3CO_2H] = [CH_3CO_2H]_{initial} - x$ For approximation, we are going to consider $[CH_3CO_2H]_{initial} = [CH_3CO_2H]$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][CH_3CO_2^-]}{ [CH_3CO_2H]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.250}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.250}$ $x^2 = 0.250 \times 1.8 \times 10^{-5} $ $x = \sqrt { 0.250 \times 1.8 \times 10^{-5}} = 2.1 \times 10^{-3} $ Percent dissociation: $\frac{ 2.1 \times 10^{- 3}}{ 0.250} \times 100\% = 0.85\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [CH_3CO_2^-] = x = 2.1 \times 10^{- 3}M $ And, since 'x' has a very small value (compared to the initial concentration): $[CH_3CO_2H] \approx 0.250M$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 2.1 \times 10^{- 3})$ $pH = 2.68$ The species with a significant concentration in this solution are: $CH_3CO_2H$ and $H_2O$.
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