## Chemistry 9th Edition

a. $[OH^-] = 8.9 \times 10^{-3}M$ $[H_3O^+] = 1.1 \times 10^{-12}M$ $pH = 11.96$ b. $[OH^-] = 4.7 \times 10^{-5}M$ $[H_3O^+] = 2.1 \times 10^{-10}M$ $pH = 9.68$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $(C_2H_5)_3N(aq) + H_2O(l) \lt -- \gt (C_2H_5)_3NH^+(aq) + OH^-(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $(C_2H_5)_3NH^+$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M -$[OH^-] = [(C_2H_5)_3NH^+] = 0 + x = x$ -$[(C_2H_5)_3N] = [(C_2H_5)_3N]_{initial} - x$ For approximation, we are going to consider $[(C_2H_5)_3N]_{initial} = [(C_2H_5)_3N]$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][(C_2H_5)_3NH^+]}{ [(C_2H_5)_3N]}$ $Kb = 4.0 \times 10^{- 4}= \frac{x * x}{ 0.20}$ $Kb = 4.0 \times 10^{- 4}= \frac{x^2}{ 0.20}$ $x^2 = 4.0 \times 10^{-4} \times 0.20$ $x = \sqrt { 4.0 \times 10^{-4} \times 0.20} = 8.9 \times 10^{-3}$ Percent ionization: $\frac{ 8.9 \times 10^{- 3}}{ 0.20} \times 100\% = 4.4\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [(C_2H_5)_3NH^+] = x = 8.9 \times 10^{- 3}M$ $[(C_2H_5)_3N] \approx 0.20M$ 3. Calculate the hydronium ion concentration: $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $8.9 \times 10^{- 3} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 8.9 \times 10^{- 3}}$ $[H_3O^+] = 1.1 \times 10^{- 12}$ 4. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.1 \times 10^{- 12})$ $pH = 11.96$ ------------------ b. 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $HONH_2(aq) + H_2O(l) \lt -- \gt HONH_3^+(aq) + OH^-(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $HONH_3^+$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M -$[OH^-] = [HONH_3^+] = 0 + x = x$ -$[HONH_2] = [HONH_2]_{initial} - x$ For approximation, we are going to consider $[HONH_2]_{initial} = [HONH_2]$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][HONH_3^+]}{ [HONH_2]}$ $Kb = 1.1 \times 10^{- 8}= \frac{x * x}{ 0.20}$ $Kb = 1.1 \times 10^{- 8}= \frac{x^2}{ 0.20}$ $x^2 = 1.1 \times 10^{-8} \times 0.20$ $x = \sqrt { 1.1 \times 10^{-8} \times 0.20} = 4.7 \times 10^{-5}$ Percent ionization: $\frac{ 4.7 \times 10^{- 5}}{ 0.20} \times 100\% = 0.024\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [HONH_3^+] = x = 4.7 \times 10^{- 5}M$ $[HONH_2] \approx 0.2M$ 3. Calculate the hydronium ion concentration: $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $4.7 \times 10^{- 5} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 4.7 \times 10^{- 5}}$ $[H_3O^+] = 2.1 \times 10^{- 10}$ 4. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 2.1 \times 10^{- 10})$ $pH = 9.68$