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10 July, 22:09

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

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  1. 11 July, 01:05
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    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
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