Question

# Cocaine is a weak organic base whose molecular formula is $C_{17}H_{21}NO_4.$ An aqueous solution of cocaine was found to have a pH of 8.53 and an osmotic pressure of 52.7 torr at $15 ^ { \circ } \mathrm { C }$ Calculate $K_b$ for cocaine.

Solution

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The reaction is

$\mathrm{C_{17}H_{21}NO_4(aq) + H_2O(l) \rightleftharpoons C_{17}H_{21}NO_4H^+(aq) + OH^-(aq)}$

The pH of a solution is 8.53

The osmotic pressure is $P = 52.7 \mathrm{torr} \cdot \frac { 1 \mathrm{atm} } { 760 \mathrm{torr} } = 0.0693 \mathrm{atm}$

The temperature is $T = 15^{o}C = (273 + 15) K = 288 \mathrm{K}$

$R = 0.0821 \mathrm{L \cdot atm / mol \cdot K }$

Let us calculate the $K_b$ for cocaine.

First, let us find the initial concentration of cocaine

\begin{align*} PV &= nRT\\ \frac { n } { V } &= \frac { P } { RT }\\ [C_{17}H_{21}NO_4] &= \frac { 0.0693 \mathrm{atm} } { 0.0821 \mathrm{L \cdot atm / mol \cdot K } \cdot 288 \mathrm{K} }\\ &= 2.93 \cdot 10^{-3} \mathrm{M} \end{align*}

Now let us find $[OH^-]$

\begin{align*} pOH &= 14 - pH\\ &= 14 - 8.53\\ &= 5.47\\ [OH^-] &= 10^{-pOH}\\ &= 10^{-5.47}\\ &= 3.388 \cdot 10^{-6} \mathrm{M} \end{align*}

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