Chemistry: The Molecular Science (5th Edition)

$pH = 7.449$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[OH^-] = [C_{10}H_{15}N{H_3}^+] = x$ -$[C_{10}H_{15}NH_2] = [C_{10}H_{15}NH_2]_{initial} - x = 1 \times 10^{- 3} - x$ For approximation, we consider: $[C_{10}H_{15}NH_2] = 1 \times 10^{- 3}M$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][C_{10}H_{15}N{H_3}^+]}{ [C_{10}H_{15}NH_2]}$ $Kb = 7.9 \times 10^{- 11}= \frac{x * x}{ 1\times 10^{- 3}}$ $Kb = 7.9 \times 10^{- 11}= \frac{x^2}{ 1\times 10^{- 3}}$ $7.9 \times 10^{- 14} = x^2$ $x = 2.811 \times 10^{- 7}$ Percent ionization: $\frac{ 2.811 \times 10^{- 7}}{ 1\times 10^{- 3}} \times 100\% = 0.02811\%$ %ionization < 5% : Right approximation. Therefore: $[OH^-] = [C_{10}H_{15}N{H_3}^+] = x = 2.811 \times 10^{- 7}M$ $[C_{10}H_{15}NH_2] \approx 0.001M$ 3. Calculate the pH. $pOH = -log[OH^-]$ $pOH = -log( 2.811 \times 10^{- 7})$ $pOH = 6.551$ $pH + pOH = 14$ $pH + 6.551 = 14$ $pH = 7.449$