## Chemistry: Atoms First (2nd Edition)

- $NH_3$, $OH^-$, $N{H_4}^+$ and $H_2O$ - $[OH^-] = 1.643 \times 10^{- 3}M$ and $pH = 11.216$
- Since $NH_3$ is a weak base, the ionization reaction is an equilibrium reaction: $NH_3(aq) + H_2O(l) \lt -- \gt N{H_4}^+(aq) + OH^-(aq)$ Therefore, both reactants and products are present in the solution. 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [{NH_4}^+] = x$ -$[NH_3] = [NH_3]_{initial} - x = 0.15 - x$ For approximation, we consider: $[NH_3] = 0.15M$ 2. Now, use the Kb value and equation to find the 'x' value. $Ka = \frac{[OH^-][{NH_4}^+]}{ [NH_3]}$ $Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.15}$ $Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.15}$ $2.7 \times 10^{- 6} = x^2$ $x = 1.643 \times 10^{- 3}$ Percent ionization: $\frac{ 1.643 \times 10^{- 3}}{ 0.15} \times 100\% = 1.095\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [{NH_4}^+] = x = 1.643 \times 10^{- 3}M$ $[NH_3] \approx 0.15M$ 3. Calculate the pH: $pOH = -log[OH^-]$ $pOH = -log( 1.643 \times 10^{- 3})$ $pOH = 2.784$ $pH + pOH = 14$ $pH + 2.784 = 14$ $pH = 11.216$ 