The reaction h2so4 + 2 naoh -> na2so4 + 2h2o applied to an acid-base tritration. the equivalence point occurs when 27.45 ml of 0.1857 m naoh is added to a 35.63 ml portion of h2so4. calculate the original [h2so4].

            moles NaOH = c · V = 0.1857 mmol/mL · 27.45 mL = 5.097465 mmol
            moles H2SO4 = 5.097465 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.5487325 mmol
Hence
            [H2SO4]= n/V = 2.5487325 mmol / 35.63 mL = 0.07153 M
The answer to this question is  [H2SO4] = 0.07153 M

Cause its being chemically applied?