Question

Cocaine is a weak organic base whose molecular formula is C17H21NO4.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 15C15 ^ { \circ } \mathrm { C } Calculate KbK_b for cocaine.

Solution

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The reaction is

C17H21NO4(aq)+H2O(l)C17H21NO4H+(aq)+OH(aq)\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=52.7torr1atm760torr=0.0693atmP = 52.7 \mathrm{torr} \cdot \frac { 1 \mathrm{atm} } { 760 \mathrm{torr} } = 0.0693 \mathrm{atm}

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

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

Let us calculate the KbK_b for cocaine.

First, let us find the initial concentration of cocaine

PV=nRTnV=PRT[C17H21NO4]=0.0693atm0.0821Latm/molK288K=2.93103M\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][OH^-]

pOH=14pH=148.53=5.47[OH]=10pOH=105.47=3.388106M\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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