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In a 1.070 M aqueous solution of a monoprotic acid, 4.81% of the acid is ionized. What is the value of its Ka?

  • chemistry -

    ........HA ==> H^+ + A^-
    Ka = (H^+)(A^-)/(HA)
    This tells you that the 1.070 HA ionized 4.81% which means
    (H^) = 1.070 x 0.0481 = ?
    (A^-) = same as (H^+).
    (HA) = [1.070 - (H^+)]

    Substitute these values into the Ka expression and solve for Ka.

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