Question

The following compounds are drawn in their acidic forms, and their pKa values are given. Draw...

The following compounds are drawn in their acidic forms, and their pKa values are given. Draw teh form in which each compound would predominantly exist at pH=8

CH3COOH CH3CH2OH CH3OH2+ CH3CH2NH3+

pKa=4.8 pKa=15.9 pKa=-2.5 pKa=11.2

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Answer #1

If the pH is lower than the pKa, then the compound will be protonated. If the pH is higher than the pKa, then the compound will be deprotonated. A further consideration is the charge on the compound. Acids are neutral when protonated and negatively charged (ionized) when deprotonated.

  1. In the case of CH3COOH, the PH 8 is higher than pKa 4.8, so it will get deprotonate and the structure is CH3COO-
  2. For CH3CH2OH, the pH 8 is lower than PKa 15.9, so it will get protonate and the structure is
  3. In the Case of CH3OH2+ , the pH is higher than pKa -2.5. So it will get deprotonate and the structure of the compound is CH3OH+
  4. For CH3CH2NH3+ , the PH 8 is lower than Pka 11.2, so it will get protonate and the structure is CH3CH2NH4+
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