Answer
$pH = 9.107$
Work Step by Step
1. Draw the ICE table (It is in the end of this answer)
2. Calculate "x" using the $K_b$ expression:
$ 1.76\times 10^{- 5} = \frac{[N{H_4}^+][OH^-]}{[N{H_3}]}$
$ 1.76\times 10^{- 5} = \frac{ (0.22 + x )* x}{ 0.16 - x}$
Considering 'x' has a very small value.
$ 1.76\times 10^{- 5} = \frac{ 0.22 * x}{ 0.16}$
$ 1.76\times 10^{- 5} = 1.375x$
$\frac{ 1.76\times 10^{- 5}}{ 1.375} = x$
$x = 1.28\times 10^{- 5}$
Percent ionization: $\frac{ 1.28\times 10^{- 5}}{ 0.16} \times 100\% = 8\times 10^{- 3}\%$
- Greater than $5\%$, correct approximation.
$[OH^-] = x$
3. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 1.28 \times 10^{- 5})$
$pOH = 4.893$
$pH + pOH = 14$
$pH + 4.893 = 14$
$pH = 9.107$
