General Chemistry: Principles and Modern Applications (10th Edition)

(a) $[OH^-] = 2.9\times 10^{- 5}$ (b) $[NH{_4}^+] \approx 0.102M$ (c) $[Cl^-] = 0.102M$ (d) $[H_3O^+] = 3.4 \times 10^{- 10}$
(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) $[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $2.9 \times 10^{- 5} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 2.9 \times 10^{- 5}}$ $[H_3O^+] = 3.4 \times 10^{- 10}M$