Answer
$[H^+] = 1.73 \times 10^{- 12}M$
$pH = 11.76$
Work Step by Step
1. Using the ICE table, we got:
$[OH^-] and [C_2H_5NH_3^+] = x$
and
$[C_2H_5NH_2] = 0.085 - x$
2. Now, use the Kb value and equation to find the 'x' value.
$Kb = 4.3 \times 10^{- 4}= \frac{x * x}{ 8.5 \times 10^{- 2}- x}$
$Kb = 4.3 \times 10^{- 4}= \frac{x^2}{ 8.5 \times 10^{- 2}- x}$
$ 3.65 \times 10^{- 5} - 4.3 \times 10^{- 4}x = x^2$
$ 3.65 \times 10^{- 5} - 4.3 \times 10^{- 4}x - x^2 = 0$
$\Delta = (- 4.3 \times 10^{- 4})^2 - 4 * (-1) *( 3.65 \times 10^{- 5})$
$\Delta = 1.84 \times 10^{- 7} + 1.46 \times 10^{- 4} = 1.46 \times 10^{- 4}$
$x_1 = \frac{ - (- 4.29 \times 10^{- 4})+ \sqrt { 1.46 \times 10^{- 4}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 4.29 \times 10^{- 4})- \sqrt { 1.46 \times 10^{- 4}}}{2*(-1)}$
$x_1 = - 6.26 \times 10^{- 3} (Negative)$
$x_2 = 5.83 \times 10^{- 3}$
Therefore: $[OH^-] = 5.83 \times 10^{- 3}$
3.Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 5.83 \times 10^{- 3})$
$pOH = 2.234$
$pH + pOH = 14$
$pH + 2.234 = 14$
$pH = 11.76$
4. Now, find the $[H^+]$
$[H^+] = 10^{-pH}$
$[H^+] = 10^{- 11.76}$
$[H^+] = 1.73 \times 10^{- 12}$
