What is the concentration of an naoh solution that requires 15.0 ml of a 0.750 m h2so4 solution to neutralize 17.5 ml of naoh?

So the first thing we must do is write a balanced equation for the reaction and we know the equation is balnced when all the species on the RHS is equal to the species on the LHS
                      2NaOH  +  H₂SO₄   →  Na₂SO₄  +  2H₂O

So now it's time to identify what reactant you know the most for from the question (volume & conc. of H₂SO₄) and use that info to find the unknown (conc. of NaOH)

If 1000 ml  of H₂SO₄ contain 0.750 mol   [0.750 M is the amount of moles in 
                                                                       1 L (1000 ml)]
then let 15 ml of H₂SO₄ contain x mol     [15 ml is the amount of the acid that                                                                                  took part in the reaction]
    
⇒  x  =  
       
         = 0.01125 mol

Mole ratio of NaOH  to  H₂SO₄  can be obtained from the balanced equation
                 0 2NaOH  +  1H₂SO₄   →  Na₂SO₄  +  2H₂O

    mole ratio of   NaOH  to  H₂SO₄  is 2 : 1

∴ if mole of of H₂SO₄   =  0.01125 mol
   then moles of NaOH = (0.01125 mol) × 2
                                      = 0.0225 mol

If 17.5 ml of NaOH contain 0.0225 mol      [this was given in the question]
then let 1000 ml of NaOH contain  x

⇒ x  =  
       
       = 1.286 mol

∴ concentration of NaOH is 1.286 mol/L 

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