## a)
$pKb = -log(Kb)=-log(1.6 \times 10^{-6})=5.79588$
$pKa = 14 - pKb=8.2041$

## b)
$k_{b}=\frac{[\text { Product }]}{[\text { reactant }]}=\frac{\left[\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3} \mathrm{H}^{+}\right][\mathrm{OH}^-]}{\left[\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right]}$

$1.6 \times 10^{-6}=\frac{[x][x]}{[0.05-x]}$

The value of $k_b$ is very low, so $0.05-x \approx 0.05$

Solve the equation above, we have:

+ $x = 2.2828427125 \times 10^{-4}$

The equilibrium concentrations:

+ $[\mathrm{OH}^-]=\left[\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3} \mathrm{H}^{+}\right]= 2.2828427125 \times 10^{-4}$

+ $\left[\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3} \right]=0.05-2.2828427125 \times 10^{-4}=0.049717 M$

## c)

$\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-log(2.2828427125 \times 10^{-4})=3.548$

$\mathrm{pH}=14-\mathrm{pOH}=10.452$