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21 August, 07:22

Calculate the percent ionization of nitrous acid in a solution that is 0.139 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 ⋅ 10-4.

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  1. 21 August, 08:06
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    Ok first, we have to create a balanced equation for the dissolution of nitrous acid.

    HNO2 H (+) + NO2 (-)

    Next, create an ICE table

    HNO2 H + NO2-

    []i 0.139M 0M 0M

    Δ[] - x + x + x

    []f 0.139-x x x

    Then, using the concentration equation, you get

    4.5x10^-4 = [H+][NO2-]/[HNO2]

    4.5x10^-4 = x*x /.139 - x

    However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,

    assume 0.139-x ≈ 0.139

    4.5x10^-4 = x^2/0.139

    Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.

    We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.

    Then to find percent dissociation, you do final concentration/initial concentration.

    0.007M/0.139M =.0503 or

    ≈5.03% dissociation.
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