Answer
a. $[OH^-] = 8.9 \times 10^{-3}M$
$[H_3O^+] = 1.1 \times 10^{-12}M$
$pH = 11.96$
b. $[OH^-] = 4.7 \times 10^{-5}M$
$[H_3O^+] = 2.1 \times 10^{-10}M$
$pH = 9.68$
Work Step by Step
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
$pH = -log[H_3O^+]$
$pH = -log( 1.1 \times 10^{- 12})$
$pH = 11.96$
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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
$pH = -log[H_3O^+]$
$pH = -log( 2.1 \times 10^{- 10})$
$pH = 9.68$
