## Chemistry 12th Edition

$pH = 8.89$
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 $[NH_3] = 0.15 M - x$ $[N{H_4}^+] = 0.35M + x$ $[OH^-] = 0 + x$ 2. Calculate 'x' using the $K_b$ expression. $1.8\times 10^{- 5} = \frac{[N{H_4}^+][OH^-]}{[NH_3]}$ $1.8\times 10^{- 5} = \frac{( 0.35 + x )* x}{ 0.15 - x}$ Considering 'x' has a very small value. $1.8\times 10^{- 5} = \frac{ 0.35 * x}{ 0.15}$ $1.8\times 10^{- 5} = 2.3x$ $\frac{ 1.8\times 10^{- 5}}{ 2.3} = x$ $x = 7.7\times 10^{- 6}$ Percent ionization: $\frac{ 7.7\times 10^{- 6}}{ 0.15} \times 100\% = 5.1\times 10^{- 3}\%$ x = $[OH^-]$ 3. Calculate the pOH: $pOH = -log[OH^-]$ $pOH = -log( 7.7 \times 10^{- 6})$ $pOH = 5.11$ 4. Find the pH: $pH + pOH = 14$ $pH + 5.11 = 14$ $pH = 8.89$