1.
• Predict and explain aqueous solubilities of conjugate acids and bases. For example. In the
experiment of Seperating Panacetin. We used asprin, for our unknowns phenacetin or acetnailide
and sucrose. I dont understand how to go about answering or understanding the objective
question.
Solution
Any organic compound (solute) that has an acidic functional group like -COOH may dissolve in
basic aqueous solvents like sodium bicarbonate solvent, as the compound form the
corresponding conjugate base or salt which is polar( thus soluble in polar solvent like water.)
Organic compound-COOH +NaHCO3 ----->Organic compound-COONa +H2CO3
In the experiment for separation of components of panacetin (drug)-containing components-
aspirin,sucrose and an unknown component (could be acetanilide or phenacetin)
Going by the structure of the components-
Aspirin has -COOH (acidic ) functional group ,so dissolves in sodium bicarbonate dissolved in
water.
sucrose is a polar molecule ,that is completely insoluble in non-polar solvent like CH2Cl2 or
dichloromethane
Unknowns ,acetanilide or phenacetin has amide functional group -NH-CO-CH3,so its slightly
polar only ,thus dissolves in CH2Cl2 or dichloromethane,polar aprotic solvent completely
Firstly, panacetin sample is dissolved in dichloromethane ,so that sucrose is precipitated,as it is
insoluble.It can be separated by filtration.Then , sodium bicarbonate solution is added to the
organic solution of dichloromethane.
The aspirin dissolves in sodium bicarbonate solution,forming conjugate base or salt.Two
different layers are formed thus, organic layer is separeated from the aqueous layer.Then
sufficient HCl is added to aqueous layer with aspirin so that aspirin is re-formed.The aqueous
layer was cooled and the precipitate of aspirin separated by filtration

2.
-COONa +HCl ---> -COOH +NaCl
The organic layer has the unknown in it, so the solvent is evaporated to get te unknown as solid.