## Chemistry: The Central Science (13th Edition)

$[C_{10}H_{15}ONH^+] = [OH^-]\approx 2.138 \times 10^{-3}M$ $[C_{10}H_{15}ON](equilibrium) \approx 0.033762M$
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: $[C_{10}H_{15}ON](equilibrium) = [C_{10}H_{15}ON](initial) - [OH^-]$ $[C_{10}H_{15}ON](equilibrium) \approx 0.0359 - 2.138 \times 10^{-3}$ $[C_{10}H_{15}ON](equilibrium) \approx 0.033762$