Question

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].

Answer 1

            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