Neutralizing acidified seawater

Hi,
Sorry but your question is not properly defined and therefor I am almost sure you'll not get an answer.  In order to answer properly your question one would need the following data:
1.  Composition of your sea water (not all "sea water" are
     the same).
2.  A bit strange that the PH is only slightly above neutral. 
     Your "sea water" were probably taken next to a river.....  
3.  What is the concentration of your HCl.
4.  What is the concentration of the NaOh solution.
5.  Sea water contain mainly NaOH and several salts
     (e.g: MgCl2) The main reaction is: 
     HCl + NaOH -----  NaCl + H2O
Best regrds,
                                           

Note that sea water is no one specific solution.  Please be more specific.

Seawater contains as much KCl as NaCl, CaCl, and H/2\CO/3.  Water sampled from different areas will, of course, have radically different composition.  However:

Your sample has a Ph of 7.6 (quite alkaline for Pacific water) but load an erlenmeyer flask with 10Ml of your sea water.  Using a burette and a Ph meter loaded w/ 12.1 M HCl, you will add about 1.2-1.3 ml HCl solution to reach 7.0 Ph. 

Then the fun starts. 

You will find an assortment of carbonates, alkali metals, inert metals, ammonals, nitrates/nitrites, and organic acids that will throw off the best calculation. 

Reduce your end solution to the solutes and then, identify your first unknown.