Answer
$[OH^-] = [H_2NN{H_3}^+] = 2.449 \times 10^{- 3}M $
$[H_2NNH_2] \approx 2M$
$pH = 11.389$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [H_2NN{H_3}^+] = x$
-$[H_2NNH_2] = [H_2NNH_2]_{initial} - x = 2 - x$
For approximation, we consider: $[H_2NNH_2] = 2M$
2. Now, use the Kb value and equation to find the 'x' value.
$Ka = \frac{[OH^-][H_2NN{H_3}^+]}{ [H_2NNH_2]}$
$Ka = 3 \times 10^{- 6}= \frac{x * x}{ 2}$
$Ka = 3 \times 10^{- 6}= \frac{x^2}{ 2}$
$ 6 \times 10^{- 6} = x^2$
$x = 2.449 \times 10^{- 3}$
Percent ionization: $\frac{ 2.449 \times 10^{- 3}}{ 2} \times 100\% = 0.1225\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [H_2NN{H_3}^+] = x = 2.449 \times 10^{- 3}M $
$[H_2NNH_2] \approx 2M$
$pOH = -log[OH^-]$
$pOH = -log( 2.449 \times 10^{- 3})$
$pOH = 2.611$
$pH + pOH = 14$
$pH + 2.611 = 14$
$pH = 11.389$
