Answer
$pH = 10.96 $
Work Step by Step
- $[BOH]_{initial} = 0.24M$
Since it is a pure solution:
- $[OH^-] = [B^+] = x$
- $[BOH] = [BOH]_{inital} - x$
1. Now, use the Kb value and equation to find the 'x' value.
$Kb = 3.5 \times 10^{- 6}= \frac{x * x}{ 2.4 \times 10^{- 1}- x}$
$Kb = 3.5 \times 10^{- 6}= \frac{x^2}{ 2.4 \times 10^{- 1}- x}$
$ 8.4 \times 10^{- 7} - 3.5 \times 10^{- 6}x = x^2$
$ 8.4 \times 10^{- 7} - 3.5 \times 10^{- 6}x - x^2 = 0$
$\Delta = (- 3.5 \times 10^{- 6})^2 - 4 * (-1) *( 8.4 \times 10^{- 7})$
$\Delta = 1.22 \times 10^{- 11} + 3.36 \times 10^{- 6} = 3.36 \times 10^{- 6}$
$x_1 = \frac{ - (- 3.5 \times 10^{- 6})+ \sqrt { 3.36 \times 10^{- 6}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 3.5 \times 10^{- 6})- \sqrt { 3.36 \times 10^{- 6}}}{2*(-1)}$
$x_1 = - 9.18 \times 10^{- 4} (Negative)$
$x_2 = 9.14 \times 10^{- 4}$
The concentration can't be negative, therefore: x = $9.14 \times 10^{- 4}M$
$[OH^-] = x = 9.14 \times 10^{- 4}M$
2. Calculate the pOH, and then the pH.
- $pOH = -log[OH^-] = -log(9.14 \times 10^{- 4}) = 3.04$
- $pH + pOH = 14$
- $pH + 3.04 = 14$
- $pH = 14 - 3.04 = 10.96$