Answer
The pH of that $C_2H_5NH_3$ solution is equal to 12.00
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$C_2H_5NH_2(aq) + H_2O(l) \lt -- \gt C_2H_5NH_3^+(aq) + OH^-(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $C_2H_5NH_3^+$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M
-$[OH^-] = [C_2H_5NH_3^+] = 0 + x = x$
-$[C_2H_5NH_2] = [C_2H_5NH_2]_{initial} - x $
For approximation, we are going to consider $[C_2H_5NH_2]_{initial} = [C_2H_5NH_2]$
2. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][C_2H_5NH_3^+]}{ [C_2H_5NH_2]}$
$Kb = 5.6 \times 10^{- 4}= \frac{x * x}{ 0.20}$
$Kb = 5.6 \times 10^{- 4}= \frac{x^2}{ 0.20}$
$x = \sqrt {5.6\times 10^{-4} \times 0.20} = 1.1 \times 10^{-2}$
Percent ionization: $\frac{ 1.1 \times 10^{- 2}}{ 0.20} \times 100\% = 5.5\%$
%ionization > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 5.6 \times 10^{- 4}= \frac{x^2}{ 0.2- x}$
$ 1.1 \times 10^{- 4} - 5.6 \times 10^{- 4}x = x^2$
$ 1.1 \times 10^{- 4} - 5.6 \times 10^{- 4}x - x^2 = 0$
Bhaskara:
$\Delta = (- 5.6 \times 10^{- 4})^2 - 4 * (-1) *( 1.1 \times 10^{- 4})$
$\Delta = 3.1 \times 10^{- 7} + 4.5 \times 10^{- 4} = 4.5 \times 10^{- 4}$
$x_1 = \frac{ - (- 5.6 \times 10^{- 4})+ \sqrt { 4.5 \times 10^{- 4}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 5.6 \times 10^{- 4})- \sqrt { 4.5 \times 10^{- 4}}}{2*(-1)}$
$x_1 = - 1.1 \times 10^{- 2} (Negative)$
$x_2 = 1.0 \times 10^{- 2}$
- The concentration can't be negative, so $[OH^-]$ = $x_2$
3. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 0.010)$
$pOH = 2.00$
4. Find the pH:
$pH + pOH = 14$
$pH + 2.00 = 14$
$pH = 12.00$
