What happens to an acid when it completely dries out?
When an aqueous solution of an acid is dried, water evaporates and the $\ce{H+}$ ions crystallize with the anionic part of the acid and form a crystal. 

If the acid used to make the solution was not of crystalline form then after evaporation of water the anion and cation would have combined to form a compound of some other type (may be a solid formed by co-ordinate bonds or a liquid or a gas like $\ce{CO2}$)

Initially, because of presence of $\ce{H2O}$ molecules, the $\ce{H+}$ ions were bonded with them to form $\ce{H3O+}$ molecules and the anionic part of the acid were solvated by these $\ce{H3O+}$ ions. This solvation happened because it liberated more energy than consumed in dissociation of the acid molecule (i.e. was thermodynamicaly favorable)

But, now since the water molecules are absent, the $\ce{H+}$ and $\ce{A-}$ (anions) will combine to form $\ce{AH (s)}$ as its formation will be favorable because there is no more water to compensate the energy required in there dissociation.

If there had been a solution of acid and base then the $\ce{H+}$ ions from the acid and $\ce{OH-}$ ions from the base would have already reacted to form $\ce{H2O}$ and hence the scenario will be the same as above instead we will also notice crystallization of the cation $\ce{(B+)}$ from the base $\ce{(BOH)}$ and the anion ($\ce{A-}$) from the acid ($\ce{AH}$) to form $\ce{BA (s)}$.

If it would had been an aqueous solution of only a base then also the water would have evaporated and crystals of the corresponding base would have formed.

To talk a bit more (though not necessary in the answer): Acidity is a measure of concentration of $\ce{H+}$ ions in water and now since there is no solution present but only the crystals, the crystals won't behave like an acidic solution until and unless solvent is added to it.