## Chemistry and Chemical Reactivity (9th Edition)

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