Given that the structures of cocaine (on the right side of the equation) and its free base ( on the left side) with an H+ :
a. Would you expect K for the cocaine to be greater or less than 1.0? Explain.
b. Would K for the free base be higher or lower than for cocaine? Explain.
c. If you were going to extract cocaine into dichloromethane (an immiscible organic solvent) from
an aqueous solution, would you make the aqueous solution acidic or basic? Explain why.
depending on the reaction taking place, the k values would be different: if the Cocaine is a product, then the pH is being lowered (due to the addition of the H+ ion). Thus the concentration of *C cocaine* would be higher than the *C freebase*. Meaning the K value would be lower than 1.0 (since the C cocaine is the denominator in the equation).
To answer these questions, we need to consider the concept of equilibrium constant (K) and the effect of pH on the protonation and deprotonation of cocaine.
a. The equilibrium constant (K) is a measure of the extent of a reaction at equilibrium. It is calculated as the ratio of product concentrations to reactant concentrations, each raised to their respective stoichiometric coefficients. In this case, we are comparing the acid form of cocaine (on the right side of the equation) with the free base form (on the left side).
If the acid form of cocaine predominates, meaning it is more stable, the equilibrium will favor the right side of the equation, resulting in a larger concentration of the acid form. In contrast, if the free base form is more stable, the equilibrium will favor the left side, resulting in a larger concentration of the free base.
Since an acid is being added to the free base to obtain the acidic form of cocaine, it implies that the acid form is more stable. Therefore, K for cocaine (acid form) is expected to be greater than 1.0.
b. The equilibrium constant is a function of temperature and the stoichiometry of the reaction. pH, on the other hand, is a measure of the concentration of H+ ions in a solution. When cocaine is present in an acidic solution, it is in the acidic (protonated) form. When cocaine is present in a basic solution, it is in the basic (deprotonated) form, also known as the free base.
The equilibrium constant (K) is dependent on the concentration of the reactants and products. In an acidic solution, the concentration of H+ ions is high, favoring the formation of the acidic form of cocaine, as indicated in the equation you provided. Therefore, the concentration of the acidic form will be higher, resulting in a larger value of K for cocaine compared to the free base.
c. To extract cocaine into dichloromethane from an aqueous solution, we can use the concept of acid-base extraction. In this method, the basic (deprotonated) form of cocaine, also known as the free base, is more soluble in non-polar organic solvents like dichloromethane. Therefore, we need to convert cocaine into its free base form before extraction.
To convert cocaine to its free base, we need to make the aqueous solution basic. This can be achieved by adding an alkali or a base such as sodium hydroxide (NaOH) or ammonia (NH3) to the solution. The base will deprotonate the acidic form of cocaine, resulting in the formation of the water-insoluble free base. This allows the free base to be extracted into the dichloromethane layer, leaving behind any water-soluble impurities in the aqueous layer.
In summary, to extract cocaine into dichloromethane from an aqueous solution, the aqueous solution should be made basic by adding an alkali or base. This conversion to the basic form facilitates better extraction of the free base into the organic solvent.