Consider the reaction h2so4 + 2 naoh -> na2so4 + 2h2o. in an acid-base tritration, the stoichiometric point happens when 45.56 ml of 0.3223 m naoh is added to a 59.34 ml portion of h2so4. what is the [h2so4]?

The balanced equation for the reaction between H₂SO₄ and NaOH is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
stoichiometric point is also called the equivalence point. this is the point at which equal quantities of acid and base are added and all the H⁺ ions and OH⁻ ions react with each other.
the number of NaOH moles reacted - molar concentration x volume
number of NaOH moles = 0.3223 mol/L x 0.04556 L = 0.01468 mol 
according to stoichiometry 
2 mol of NaOH reacts with 1 mol of H₂SO₄
then 0.01468 mol of NaOH reacts with - 1/2 x 0.01468 mol = 0.007340 mol 
the number of H₂SO₄ moles in 59.34 mL - 0.007340 mol 
Therefore number of H₂SO₄ moles in 1000 mL - 0.007340 mol / 59.34 x 1000 = 0.1237 mol
molar concentration of H₂SO₄ is 0.1237 M

Moles NaOH = c · V = 0.3223 mmol/mL · 45.56 mL = 14.683988 mmol
            moles H2SO4 = 14.683988 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 7.341994 mmol
Hence
            [H2SO4]= n/V = 7.341994 mmol / 59.34 mL = 0.1237 M
The answer to this question is  [H2SO4] = 0.1237 M