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solubility question


gammagirl

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The pka of valproic acid is 5. The pka of cyclohexanol is 17. Valproic acid is fully protonated in acid and water-insoluble. In the base, valproic acid becomes COO- and water soluble. Why does cyclohexanol remain neutral in bases and acids and insoluble? Please explain.

 

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First, Hello again. Okay, is it that by using Henderson-Hasselbach, if you put cyclohexanol(pKa is 17) in a base, ph 10, for example, 10^ph-pka= 10^10-17 is 10^-7 is less than 1 and an acid solution,ph=3, 10^3-17=10^-14 is less than 1. Meaning, the acid is predominant or protonated species and does not ionize in acid or a base solution, therefore, no reaction? 

Edited by gammagirl
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Sorry, just saw your reply.

Why Henderson-Hasselbalch? This isn't a buffer question. I don't really follow what you are saying, anyway. 

The pKa of cyclohexanol is 17. This means that below a pH of 17, it remains mostly in the acid form (HA). Above pH 17, the majority of it will be deprotonated to the conjugate base (A-). Valproic acid has a pKa of 4.8. Similarly, this means that below pH 4.8 the compound remains mostly intact as HA, and above that pH it is mostly deprotonated to the conjugate base. From this we would say that valproic acid is vastly more acidic (i.e. More willing to give up H+) than cyclohexanol is. Why this is comes down to the conjugate bases and their relative stabilities. Do you think the conjugate base of valproic acid is more stable or less stable than the one you drew for cyclohexanol? Why? What is different about their structures that would lead to this?

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