Answer
$pH = 8.89$
Work Step by Step
1. Drawing the ICE table we get these concentrations at the equilibrium:
$NH_3(aq) + H_2O(l) \lt -- \gt N{H_4}^+(aq) + OH^-(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[NH_3] = 0.15 M - x$
$[N{H_4}^+] = 0.35M + x$
$[OH^-] = 0 + x$
2. Calculate 'x' using the $K_b$ expression.
$ 1.8\times 10^{- 5} = \frac{[N{H_4}^+][OH^-]}{[NH_3]}$
$ 1.8\times 10^{- 5} = \frac{( 0.35 + x )* x}{ 0.15 - x}$
Considering 'x' has a very small value.
$ 1.8\times 10^{- 5} = \frac{ 0.35 * x}{ 0.15}$
$ 1.8\times 10^{- 5} = 2.3x$
$\frac{ 1.8\times 10^{- 5}}{ 2.3} = x$
$x = 7.7\times 10^{- 6}$
Percent ionization: $\frac{ 7.7\times 10^{- 6}}{ 0.15} \times 100\% = 5.1\times 10^{- 3}\%$
x = $[OH^-]$
3. Calculate the pOH:
$pOH = -log[OH^-]$
$pOH = -log( 7.7 \times 10^{- 6})$
$pOH = 5.11$
4. Find the pH:
$pH + pOH = 14$
$pH + 5.11 = 14$
$pH = 8.89$
