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In a 1.760 M aqueous solution of a monoprotic acid, 3.21% of the acid is ionized. What is the value of it's Ka?

x=1.760M * 3.21/100 - .0565M

Ka expression is Ka= {[H+][A-]} / [HA]

  • Chemistry -

    Did you make a typo; that = 0.0565M and not -.
    All of that looks good to me.

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