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}$
$[H_3O^+] = \frac{10^{-14}}{ 4.69 \times 10^{- 5}}$
$[H_3O^+] = 2.132 \times 10^{- 10}$
