## Chemistry (4th Edition)

(a) $pH = 11.28$ (b) $pH = 9.08$
(a) 1. We have these concentrations at equilibrium: -$[OH^-] = [N{H_4}^+] = x$ -$[NH_3] = [NH_3]_{initial} - x = 0.2 - x$ For approximation, we consider: $[NH_3] = 0.2M$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][N{H_4}^+]}{ [NH_3]}$ $Kb = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.2}$ $Kb = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.2}$ $3.6 \times 10^{- 6} = x^2$ $x = 1.9 \times 10^{- 3}$ Percent ionization: $\frac{ 1.9 \times 10^{- 3}}{ 0.2} \times 100\% = 0.95\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [N{H_4}^+] = x = 1.9 \times 10^{- 3}M$ $[NH_3] \approx 0.2M$ 3. Calculate the pOH: $pOH = -log[OH^-]$ $pOH = -log( 1.9 \times 10^{- 3})$ $pOH = 2.72$ 4. Find the pH: $pH + pOH = 14$ $pH + 2.72 = 14$ $pH = 11.28$ ------ (b) 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 $[NH_3] = 0.2 M - x$ $[N{H_4}^+] = 0.3M + x$ $[OH^-] = 0 + x$ 2. Calculate 'x' using the $K_b$ expression. $1.8\times 10^{- 5} = \frac{[N{H_4}^+][OH^-]}{[NH_3]}$ $1.8\times 10^{- 5} = \frac{( 0.3 + x )* x}{ 0.2 - x}$ Considering 'x' has a very small value. $1.8\times 10^{- 5} = \frac{ 0.3 * x}{ 0.2}$ $1.8\times 10^{- 5} = 1.5x$ $\frac{ 1.8\times 10^{- 5}}{ 1.5} = x$ $x = 1.2\times 10^{- 5}$ Percent ionization: $\frac{ 1.2\times 10^{- 5}}{ 0.2} \times 100\% = 6\times 10^{- 3}\%$ x = $[OH^-]$ 3. Calculate the pOH: $pOH = -log[OH^-]$ $pOH = -log( 1.2 \times 10^{- 5})$ $pOH = 4.92$ 4. Find the pH: $pH + pOH = 14$ $pH + 4.92 = 14$ $pH = 9.08$