#### Answer

$pH \approx 9.602$

#### 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.
-$[OH^-] = [Conj. Acid] = x$
-$[Base] = [Base]_{initial} - x = 1 \times 10^{- 3} - x$
For approximation, we consider: $[Base] = 1 \times 10^{- 3}M$
2. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][Conj. Acid]}{ [Base]}$
$Kb = 1.6 \times 10^{- 6}= \frac{x * x}{ 1\times 10^{- 3}}$
$Kb = 1.6 \times 10^{- 6}= \frac{x^2}{ 1\times 10^{- 3}}$
$ 1.6 \times 10^{- 9} = x^2$
$x = 4 \times 10^{- 5}$
Percent ionization: $\frac{ 4 \times 10^{- 5}}{ 1\times 10^{- 3}} \times 100\% = 4\%$
Since the percent ionization is less than 5 percent, this is a good approximation.
Therefore: $[OH^-] = [Conj. Acid] = x = 4 \times 10^{- 5}M $
$[Base] \approx 0.001M$
3. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 4 \times 10^{- 5})$
$pOH = 4.398$
$pH + pOH = 14$
$pH + 4.398 = 14$
$pH = 9.602$