By titration, 17.6 ml of aqueous h2so4 neutralized 31.4 ml of 0.0111 m lioh solution. what was the molarity of the aqueous acid solution?

The balanced equation for the reaction between LiOH and H₂SO₄ is 
2LiOH + H₂SO₄ → Li₂SO₄ + 2H₂O

Molarity (M) = moles (mol) / volume of the solution (L)

Molarity of LiOH                            = 0.0111 M
Hence, moles of LiOH in 31.4 mL = molarity x volume of the solution
                                                      = 0.0111 M x 31.4 x 10⁻³ L                                                                                               = 3.4854 x 10⁻⁴ mol

The stoichiometric ratio between LiOH and H₂SO₄ is 2 : 1

Hence, reacted moles of H₂SO₄  = moles of LiOH / 2
                                                     = 3.4854 x 10⁻⁴ mol / 2
                                                     = 1.7427 x 10⁻⁴ mol

Those moles were in 17.6 mL of H₂SO₄ solution.
Hence, the molarity of H₂SO₄ = 1.7427 x 10⁻⁴ mol / 17.6 x 10⁻³ L
                                                  = 9.901 x 10⁻³ M

The  molarity of   the aqueous  acid solution  is  9.9   x10^-3  M

   calculation
calculate the moles of LiOH  used

moles =  molarity  x volume in liters
molarity =0.0111 M
volume  in liters = 31.4/1000=  0.0314 liters

moles is therefore = 0.0111  x 0.0314 = 3.485  x10 ^-4  moles  of LiOH

write the equation for neutralization to help to find the moles of H2SO4

= H2SO4 + 2 LiOH →Li2SO4  + 2H2O 

by  use of mole ratio between H2SO4  to LiOH which is  1:2 the moles of  H2SO4 is therefore =  3.485x10^-4  x1/2 = 1.743 x10^-4 moles

molarity   of LiOH=moles/volume in liters

volume in liters = 17.6/1000 =0.0176 liters

molarity  is =  (1 .743 x10^-4) ÷0.0176  =9.9 x10 ^-3 M

Ithink the answer is a or d.