Answer
$[C_{10}H_{15}ONH^+] = [OH^-]\approx 2.138 \times 10^{-3}M$
$[C_{10}H_{15}ON](equilibrium) \approx 0.033762M$
Work Step by Step
1. Find the pOH:
$pOH = 14 - pH$
$pOH = 14 - 11.33$
$pOH = 2.67$
2. Calculate $[OH^-]$
$[OH^-] = 10^{-pH}$
$[OH^-] = 10^{-2.67}$
$[OH^-] \approx 2.138 \times 10^{-3}$
3. $[C_{10}H_{15}ONH^+] = [OH^-]\approx 2.138 \times 10^{-3} $
4. The concentration of $[C_{10}H_{15}ON] $ in equilibrium can be calculated:
$ [C_{10}H_{15}ON](equilibrium) = [C_{10}H_{15}ON](initial) - [OH^-]$
$[C_{10}H_{15}ON](equilibrium) \approx 0.0359 - 2.138 \times 10^{-3}$
$[C_{10}H_{15}ON](equilibrium) \approx 0.033762$
