Indirect Titration of Sodium Bisulfite with Potassium Iodate

Question

I'm trying to develop a procedure to determine the concentration (in g/ml) of a solution of sodium bisulfite (NaHSO3). I have already determined the solution concentration using a direct titration with iodine. However, this method is not ideal because I do not have an iodine solution readily available (I made a small amount to test the solution strength as a control) so I would like to use a procedure for which I already have the test chemicals. The procedure I'm using is an indirect titration with potassium iodate (KIO3, 0.1 N) in an acidic solution (acidified with 0.1 N HCl) of potassium iodide (10% KI by mass, in excess) with a starch indicator. The reaction is as follows: KIO3 + 5KI (excess) + 6HCl -> 6KCl + 3I2 + 3H2O. This reaction releases iodine, which reacts with the sodium bisulfite according to the following reaction: 2NaHSO3 + 2H2O + 2I2 -> Na2SO4 + 4HI + H2SO4. I am able to get a consistent endpoint, but I'm having trouble converting the result to a sodium bisulfite concentration. As I understand it, I need to know the n-factor for the iodine in KIO3 to convert it to a molarity so that I can do stoichiometry to determine the amount of sodium bisulfite that reacted. I'm not familiar with n-factor or normality and what I could find online isn't helping much because the reaction is complicated. I thought the n-factor would be 5 but when I use that for the calculation my final result is consistently lower than the solution concentration. I know that I could determine this experimentally but I would really like to understand the calculation correctly. Would anyone be able to help me with this calculation? I'm happy to provide more details if needed.