How is the strength of an acid in water assessed?

1 Answer
Apr 26, 2017

We must specify a WATER solvent...............

Explanation:

And thus we address the extent of the following equilibrium.......

#HA(aq) + H_2O(l) rightleftharpoons H_3O^(+) + A^(-)#

For the typical mineral acids, #HX# #(X!=F, X=Cl, Br, I)#, #H_2SO_4#, #HClO_4#, #HNO_3#, this equilibrium LIES strongly to the RIGHT. The extent of the equilibrium is an experimental phenomenon. For other mineral acids, e.g. #H_3PO_4#, #H_2SO_3#, the equilibrium lies LESS strongly to the right.

For carboxylic acids, the given equilibrium lies to the left, and significant quantities of the acid would remain at equilibrium:

#RCO_2H(aq)+H_2O(l)rightleftharpoonsRCO_2^(-) + H_3O^+#

For #"acetic acid"#, #pK_a=4.76#

Ordinary phenol does act as a weak acid.........

#pK_a""_(C_6H_5OH)=9.95#

#C_6H_5OH(aq) + H_2O(l) rightleftharpoons C_6H_5O^(-) + H_3O^+#

Nitro substitution on the phenol ring, greatly enhances the acidity (this is arguably an entropy effect). Picric acid, #2,4,5-(O_2N)_3C_6H_2OH# has #pK_a=0.38#.

On the other hand, water is a fairly weak acid. And alcohols are unknown to exhibit acidic behaviour in water.

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