Boric acid is written as H3BO3 or B(OH)3...but if it is written as B(OH)3...why is it still an acid and not a base?

Consider nitric acid, i.e. #HNO_3#...the way I would write the Lewis structure is #HO-stackrel+N(=O)O^(-)#...and this is more or less a strong acid...that dissociates according to the equation...

#HO-stackrel+N(=O)O^(-)(aq)+H_2O(l)rarrH_3O^+ + ""^(-)O_2stackrel+N(=O)#

Any resonance isomer of the nitrate anion has 3 atoms with formal charge. But the parent is manifestly an hydroxide....

And we could got to a stronger acid...#"sulfuric acid"#:

#(HO)_2S(=O)_2(aq) + 2H_2Orarr2H_3O^+ +""^(-)(O-)_2S(=O)_2#

As written this is another hydroxide..

Of course the sulfate anion is resonance stabilized but I think you get the idea. Are you happy with this....? And of course there is also #"perchloric acid"#...#HOCl(=O)_3#...another strong acid....these ARE ALL HYDROXIDES....but the #H-O# bond is attenuated and weak...and so the proton pops off to torment the solvent.

And so another generalization....the hydroxides of non-metals feature fairly weak bonds between hydrogen and the oxygen...on the other hand, for metal hydroxides, the #H-O# bond is strong.

It can accept electron density into the empty #2p# orbital of boron, OR it can dissociate an #"H"^(+)#. Analogize with #"BH"_3#:

Those #"OH"^-# cannot dissociate on their own. Instead, we have that it accepts an #"OH"^(-)# from water's autoionization as a Lewis acid:

#2"H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + cancel("OH"^(-)(aq))#
#ul("B"("OH")_3(aq) + cancel("OH"^(-)(aq)) rightleftharpoons "B"("OH")_4^(-)(aq))#
#"B"("OH")_3(aq) + 2"H"_2"O"(l) \rightleftharpoons "B"("OH")_4^(-)(aq) + "H"_3"O"^(+)(aq)#

#K_(a,L) = 7.3 xx 10^(-10)#

So the equilibrium constant for the addition of #"OH"^(-)# onto #"B"("OH")_3# (step 2) is quite thermodynamically favorable, at #7.3 xx 10^4# at #25^@ "C"#, making it quite the Lewis acid.

Or, since it has #"OH"^(-)# on it, it can also dissociate an #"H"^(+)#, apparently... which is how we normally know it---as a Bronsted-Lowry acid.

#"B"("OH")_3(aq) + "H"_2"O"(l) rightleftharpoons "BO"("OH")_2^(-)(aq) + "H"_3"O"^(+)(aq)#

or

#"H"_3"BO"_3(aq) + "H"_2"O"(l) rightleftharpoons "H"_2"BO"_3^(-)(aq) + "H"_3"O"^(+)(aq)#

#K_(a,BL) = 5.8 xx 10^(-10)#

However, it's not much of a Bronsted-Lowry acid (#K_(a,BL)# is small).