How much of a 0.250 M lithium hydroxide is required to neutralize 20.0 mL of 0.345M chlorous acid?

27.6mL of LiOH 0.250M

Explanation:

The reaction of lithium hydroxide (LiOH) with chlorous acid (HClO₂) is:

LiOH + HClO₂ → LiClO₂ + H₂O

That means, 1 mole of hydroxide reacts per mole of acid

Moles of 20.0 mL = 0.0200L of 0.345M chlorous acid are:

0.0200L ₓ (0.345mol / L) = 6.90x10⁻³ moles of HClO₂

To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:

6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =

27.6mL of LiOH 0.250M