chemistry

What is the pH of an aqueous solution of a weakly acidic drug that is 40% dissociated and has a pKa =4.7?

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  1. Call the weak acid HA.
    ............HA ==> H^+ + A^-
    initial.....X......0.......0
    change...-0.4x...0.4x....0.4x
    equil...0.6x....0.4x.....0.4x

    Ka (H^+)(A^-)/(HA)

    Substitute the equil line into the Ka exparession and solve for x, then (H^+) + 0.4x. Convert to pH = -log(H^+)

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