Question

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?

Answer 1

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

Answer 2

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