CHEM 4511 PS 2 F10 Solutions

CHEM 4511 PS 2 F10 Solutions - CHEM 4511 PS 2 Fall 10 1....

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Unformatted text preview: CHEM 4511 PS 2 Fall 10 1. Ethane is a nonpolar molecule and is unable to form any favorable interactions with H2O nonpolar; there are no potential dipole‐dipole or hydrogen bonding capabilities. Ethanol, on the other hand, is a polar molecule, can participate in hydrogen bonds with water 2. 3. (A) (B) (C) (D) 4. In stomach: ~99% of aspirin will be protonated form (pKa = 3.5 and pH = 1.5). Neutral molecule at this stage: Nonpolar and will more readily pass through the plasma membrane. Small Intestine: Greater than 99% will be deprotonated (pKa = 3.5 and pH = 6.0). Aspirin will be negatively charged and not pass as readily through the small intestine and the plasma membrane of the cells therein. 5. More soluble in 0.1 M NaOH ‐ Carboxylic acid (pKa of 4‐5) will deprotonate in basic conditions More soluble in 0.1 M NaOH ‐ pKa of phenol ~ 9 will deprotonate in a strongly basic solution More soluble in 0.1 M HCl, the pyridine nitrogen (pKa of 5.2) ‐ will be protonated in acidic solution Therefore, most absorption: occurs in stomach (at least according to this particular paradigm). Acid RCOOH RNH3+ H3PO4 H2CO3 Conjugate Base RCOO– RNH2 H2PO4– HCO3– * All three of these molecules have hydrophobic character. When ionized in an acidic or basic solution, they will then carry a charge and their water solubility will be increased. 6. (A) 8.6 – 10.6 (B) pH = pKa + log 9 – 9.6 = log ‐ 0.6 = log RNH RNH RNH RNH R–NH3+ RNH2 + H+ pKa = 9.6 “AH” Acid “A–“ Conjugate Base RNH RNH RNH 0.25 = RNH or = 5 total 4 out of 5 glycine molecules are in RNH3+ form (C) Assume negligible volume change at pH 9.0 ⇒ 0.08 M R‐NH3+ and 0.02 M R‐NH2 • Solve for this at pH 10 using above equation at pH 10 0.029 M RNH3+ and 0.071 M RNH2 • Need addition of 0.051 M –OH: 0.051 mol NaOH × µL N OH = 10.2 μl NaOH needed (D) pKa = 9.6 pH will be at 7.6 to give 99% RNH3+ (2 pH units lower than pKa) 7. pKa of H2PO4 : 6.86 pH = 6.86 + log pH = 7.0 . . M M ( H2PO4= HPO4– + H+ ) 8. pH – pKa = log A O– A OH 5.0 – 4.76 = log 0.24 = log . A O– A OH A O– A OH 1 out of 2.73 AcOH molecules or 37 % will be in AcOH form. = A O– A OH Since 0.2 M total of buffer…. Need: 0.073 M AcOH + 0.126 M NaOAc 0.2 M 9. pH = ‐log [H+] [H+] = – pH at start: [H+] = – 7.65 [H+] = 2.2 × 10‐8 M H+ present [H+] = – 6.87 [H+] = 1.35 × 10‐7 M H+ present *Assume complete ionization Convert to mols added: 15 mL × . L at end: The change in [H+] is the amount of [H+] added (1.35 × 10‐7 M) – (2.2 × 10‐8 M) = 1.13 × 10‐7 M H+ added = 1.7 × 10‐9 mol H+ added or 1.7 × 10‐9 mol Acetylcholine ...
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