The reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O applied to an acid-base tritration. The equivalence point occurs when 26.66 mL of 0.1923 M NaOH is added to a 42.45 mL portion of H2SO4. Calculate the original [H2SO4].
We make use of ratios to solve this question. moles NaOH = c · V = 0.1923 mmol/mL · 26.66 mL = 5.126718 mmol moles H2SO4 = 5.126718 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.563359 mmol Hence [H2SO4]= n/V = 2.563359 mmol / 42.45 mL = 0.06039 M The answer to this question is [H2SO4] = 0.06039 M
