In a titration of 35.00 ml of 0.737 m h2so4, ml of a 0.827 m koh solution is required for neutralization.

The reaction between H₂SO₄ and KOH is as follows;
2KOH  + H₂SO₄ --> K₂SO₄  + 2H₂O
stoichiometry of KOH to H₂SO₄ is 2:1
the number of H₂SO₄ moles that have reacted = 0.737 M /1000mL x 35 mL 
                                                                          = 0.0258 mol
According to stoichiometry, number of KOH moles that have reacted at point of neutralization = number of H₂SO₄ moles x 2
Therefore number of KOH moles = 0.0516 mol
The molarity of KOH = 0.827 mol/L 
The volume of 0.827 mol - 1 L 
Therefore volume for 0.0516  mol  - 1/0.827 x 0.0516 = 0.0624 L
volume of KOH needed = 62.4 mL 

Electrons are emitted