Answer
$K_b = 2.6 \times 10^{-6} $
Work Step by Step
1. Calculate the molar mass:
$ C_{17}H_{21}O_4N $ : ( 1.008 $\times$ 21 )+ ( 12.01 $\times$ 17 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 4 )= 303.35 g/mol
2. Use the information as conversion factors to find the molarity of this solution:
$$\frac{ 0.17 g \space C_{17}H_{21}O_4N }{100 \space mL \space solution} \times \frac{1000 \space mL}{1 \space L} \times \frac{1 \space mol \space C_{17}H_{21}O_4N }{ 303.35 \space g \space C_{17}H_{21}O_4N } = 5.6 \times 10^{-3} \space M$$
3. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ C_{17}H_{21}O_4N ]& [ C_{17}H_{22}O_4N^+ ]& [ OH^- ]\\
Initial& 5.6 \times 10^{-3} & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 5.6 \times 10^{-3} -x& 0 +x& 0 +x\\
\end{vmatrix}$$
5. Write the expression for $K_b$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_b = \frac{[Products]}{[Reactants]} = \frac{[ C_{17}H_{22}O_4N^+ ][ OH^- ]}{[ C_{17}H_{21}O_4N ]}$$
$$K_b = \frac{(x)(x)}{[ C_{17}H_{21}O_4N ]_{initial} - x}$$
6. Using the pH, find the $OH^-$ concentration:
$$[H_3O^+] = 10^{-pH} = 10^{-10.08} = 8.3 \times 10^{-11} \space M$$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 8.3 \times 10^{-11} } = 1.2 \times 10^{-4} \space M$$
$ x = [OH^-] = 1.2 \times 10^{-4} $
7. Substitute the value of x and calculate the $K_b$:
$$K_b = \frac{( 1.2 \times 10^{-4} )^2}{ 5.6 \times 10^{-3} - 1.2 \times 10^{-4} }$$
$K_b = 2.6 \times 10^{-6} $
