Answer
(a) $[OH^-] = 2.9\times 10^{- 5}$
(b) $[NH{_4}^+] \approx 0.102M$
(c) $[Cl^-] = 0.102M$
(d) $[H_3O^+] = 3.4 \times 10^{- 10}$
Work Step by Step
(a)
1. Drawing the ICE table we get these concentrations at the equilibrium:
$N{H_3}(aq) + H_2O(l) \lt -- \gt NH{_4}^+(aq) + OH^-(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[N{H_3}] = 0.164 M - x$
$[NH{_4}^+] = 0.102M + x$
$[OH^-] = 0 + x$
2. Calculate 'x' using the $K_b$ expression.
$ 1.8\times 10^{- 5} = \frac{[NH{_4}^+][OH^-]}{[N{H_3}]}$
$ 1.8\times 10^{- 5} = \frac{( 0.102 + x )* x}{ 0.164 - x}$
Considering 'x' has a very small value.
$ 1.8\times 10^{- 5} = \frac{ 0.102 * x}{ 0.164}$
$ 1.8\times 10^{- 5} = 0.622x$
$\frac{ 1.8\times 10^{- 5}}{ 0.622} = x$
$x = 2.9\times 10^{- 5}$
Percent ionization: $\frac{ 2.9\times 10^{- 5}}{ 0.164} \times 100\% = 0.0176\%$
x = $[OH^-]$
(b)
$[NH{_4}^+] = 0.102M + x = 0.102M + 2.9\times 10^{- 5}M \approx 0.102M$
(c)
The chloride ion did not react with anything so:
$[Cl^-]_{eq} = [Cl^-]_{initial} = 0.102M$
(d)
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.9 \times 10^{- 5} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.9 \times 10^{- 5}}$
$[H_3O^+] = 3.4 \times 10^{- 10}M$
