How does the behaviour of a strong acid in water compare to that of a weak acid when both are dissolved in water?

1 Answer
Jan 6, 2017

Answer:

Consider the solution behaviour of an acid, in water.........

Explanation:

In water we know that the following equilbrium operates, the autoprotolysis of water:

#2H_2O(l)rightleftharpoonsH_3O^+ + ""^(-)OH#

Where, #K_(eq)=K_w=[H_3O^+][HO^-]=10^(-14)# at #298*K#

An acid is conceived to protonate the solvent, here the water, to form the hydronium ion, #H_3O^+#, the so-called characteristic cation of the solvent. This is an ionization reaction by definition:

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

This is an equilibrium reaction, which might be small (i.e. lie on the reactants side), or might be large (lie towards the products), and as with any equilibrium, we may measure its extent by #K_(eq),# #"the so-called equilibrium constant:"#

#K_(eq) = ([H_3O^+][A^-])/([HA])#

And again, large values of #K_(eq)# means that the product side is favoured, and small values mean that the reactant side is favoured.

And likewise, a #"strong acid"# is an acid which largely ionizes in solution, one for whom #K_"eq"# is large, and for which say a #1*mol*L^-1# in #HA# is stoichiometric in #H_3O^+# and #A^-#.

And the moral, a strong acid is an acid that almost completely undergoes this ionization reaction in water, and yet this behaviour is still governed by #K_w#. Examples of strong acids include #HX (X=Cl, Br, I, !=F), H_2SO_4, HClO_4, HNO_3#.