Answer
$pH = 7.449$
Work Step by Step
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.
$pOH = -log[OH^-]$
$pOH = -log( 2.811 \times 10^{- 7})$
$pOH = 6.551$
$pH + pOH = 14$
$pH + 6.551 = 14$
$pH = 7.449$
