Consider the reaction h2so4 + 2 naoh -> na2so4 + 2h2o. in an acid-base tritration, the stoichiometric point happens when 54.32 ml of 0.3561 m naoh is added to a 67.21 ml portion of h2so4. what is the [h2so4]?

This question involves a standard acid-base titration.
            moles NaOH = c · V = 0.3561 mmol/mL · 54.32 mL = 19.343352 mmol
           moles H2SO4 = 19.343352 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 9.671676 mmol
Hence
               [H2SO4]= n/V = 9.671676 mmol / 67.21 mL = 0.1439 M
The answer to this question is  [H2SO4] = 0.1439 M

Well to my expert knowledge and my exquisite calculations i think the the negative ion # 412 is the key to sucess

Idk the answer sorry