Answer
$[OH^{-}]$=$[C_{10}H_{15}ONH^{+}]$=0.00214 M
$[C_{10}H_{15}ON]$= 0.0329M
$K_{b}$=$1.39*10^{-4}$
Work Step by Step
pOH=2.67
pOH=-log($[OH^{-}]$)
$[OH^{-}]$= $10^{-pOH}$
$[OH^{-}]$=$10^{-2.67}$=0.00214M
$[C_{10}H_{15}ON]$= 0.035-0.00214= 0.0329M
$K_{b}$=$\frac{[OH^{-}]*[C_{10}H_{15}ONH^{+}]}{[C_{10}H_{15}ON]}$=$\frac{0.00214*0.00214}{0.0329}$= 1.39*$10^{-4}$
