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chemistry

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A 0.942 M sample of carbonic acid, H2CO3, has a measured hydronium ion concentration of 6.36 ´ 10–4 M. Calculate the acid-ionization constant of carbonic acid. The equilibrium equation is:
H2CO3(aq) + H2O(l) <--><--> H3O+(aq) + HCO-3(aq)

  • chemistry - ,

    Ka = (H3O^+)(HCO3^-)(H2CO3)

    Set up an ICE chart, substitute into Ka expression, and solve for Ka.
    Post your work if you get stuck.

  • chemistry - ,

    what is an ICE chart?

  • chemistry - ,

    ICE.
    I = initial
    C = change
    E = equilibrium.

    H2CO3 + H2O ==> H3O^+ + HCO3^-

    Ka = (H3O^+)(HCO3^-)/(H2CO3).

    initial:
    H2CO3 = 0.942 M
    (H3O^+) = 0
    (HCO3^-) = 0

    equilibrium:
    (H3O^+) = 6.36 x 10^-4 M
    (HCO3^-) = 6.36 x 10^-4 M
    (H2CO3) = 0.942-6.36 x 10^-4 = 0.942

    This solution assumes Ka is the only ionization constant for H2CO3; in reality, H2CO3 has a k1 and a k2 but we are ignoring k2. The error is small.

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